Student solutions manual for ''Elementary Linear Algebra, 5th edition'' Stephen Andrilli & David Hec

STUDENT SOLUTION MANUAL Student Manual for

Elementary Linear Algebra Fifth Edition

Stephen Andrilli David Hecker

Dedication To all the students who have used the various editions of our book over the years

Table of Contents Preface …………………………………………………..iii Chapter 1 Solutions ……………………………………....1 Chapter 2 Solutions ……………………………………..27 Chapter 3 Solutions ……………………………………..77 Chapter 4 Solutions ……………………………………118 Chapter 5 Solutions ……………………………………161 Chapter 6 Solutions ……………………………………194 Chapter 7 Solutions ……………………………………220 Chapter 8 Solutions ……………………………………243 Chapter 9 Solutions ……………………………………285 Appendix B Solutions …………………………………313 Appendix C Solutions …………………………………315 Appendix D Solutions …………………………………316

ii

Preface This Student Solutions Manual is designed to accompany the fifth edition of Elementary Linear Algebra,

by Andrilli and Hecker. It contains detailed solutions for all of the exercises in the textbook marked with a star () or a triangle (). In the triangle exercises, the student is typically asked to prove a theorem that appears in the textbook.

The solutions presented are generally self-contained, except that a comment may appear at the beginning of an exercise that applies to all the solutions for each part of the exercise. For example, the solution to Exercise 1 in Section 1.1 begins with a description of the general strategy used to solve problems in each of parts (a) and (c). Then, the solution is given to each part separately, following the previously stated strategy. Therefore, you should always check for a comment at the heading of a problem before jumping to the part in which you are interested.

We hope you find these solutions helpful. You can find other useful information at the web site at which you obtained this manual.

Stephen Andrilli David Hecker August 2015 iii

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.1

Solutions to Selected Exercises Chapter 1 Section 1.1 (1) In each part, to find the vector, subtract the corresponding coordinates of the initial vector from the terminal vector. (a) In this case, [5 −1] − [−4 3] = [9 −4]. And so, the desired vector is [9 −4]. The distance between p √ the two points = k[9 −4]k = 92 + (−4)2 = 97. (c) In this case, [0 −3 2 −1 −1] − [1 −2 0 2 3] = [−1 −1 2 −3 −4]. And so, the desired vector is [−1 −1 2 −3 −4]. The distance between the two points = k[−1 −1 2 −3 −4]k = p √ (−1)2 + (−1)2 + 22 + (−3)2 + (−4)2 = 31

(2) In each part, the terminal point is found by adding the coordinates of the given vector to the corresponding coordinates of the initial point (1 1 1). (a) [1 1 1] + [2 3 1] = [3 4 2] (see Figure 1), so the terminal point = (3 4 2). (c) [1 1 1] + [0 −3 −1] = [1 −2 0] (see Figure 2), so the terminal point = (1 −2 0).

Figure 1

Figure 2

(3) In each part, the initial point is found by subtracting the coordinates of the given vector from the corresponding coordinates of the given terminal point. (a) [6 −9] − [−1 4] = [7 −13], so the initial point is (7 −13).

(c) [2 −1 −1 5 4] − [3 −4 0 1 −2] = [−1 3 −1 4 6], so the initial point = (−1 3 −1 4 6).

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

1

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.1

(4) (a) First, we find the vector v having the given initial and terminal point by subtracting: [10 −10 11] − [−4 7 2] = [14 −17 9] = v. Next, the desired point is found by adding 23 v (which is 23 the length of v) to the vector for the initial point (−4 7 2): £ ¤ £ 16 13 ¤ 34 [−4 7 2] + 23 v = [−4 7 2] + 23 [14 −17 9] = [−4 7 2] + 28 3 − 3 6 = 3 − 3 8 , ¡ 16 13 ¢ so the desired point is 3  − 3  8 .

1 (5) Let v represent the given vector. Then the desired unit vector u equals kvk v.

h i 1 1 v = k[3−56]k [3 −5 6] = √ 2 1 2 2 [3 −5 6] = √370  − √570  √670 ; (a) u = kvk 3 +(−5) +6 √ u is shorter than v since kuk = 1 ≤ 70 = kvk.

1 1 1 v = k[06−08]k [06 −08] = √ [06 −08] = [06 −08]; (c) u = kvk 2 2 (06) +(−08)

neither vector is longer because u = v.

(6) Two nonzero vectors are parallel if and only if one is a scalar multiple of the other. (a) [12 −16] and [9 −12] are parallel because 34 [12 −16] = [9 −12].

(c) [−2 3 1] and [6 −4 −3] are not parallel. To show why not, suppose they are. Then there would be a  ∈ R such that [−2 3 1] = [6 −4 −3]. Comparing first coordinates shows that −2 = 6, or  = −3. However comparing second coordinates shows that 3 = −4, or  = − 43 instead. But  can not have both values.

(7) (a) 3[−2 4 5] = [3(−2) 3(4) 3(5)] = [−6 12 15] (c) [−2 4 5] + [−5 −3 6] = [(−2 + (−5)) (4 + (−3)) (5 + 6)] = [−7 1 11]

(e) 4[−5 −3 6]−5[−2 4 5] = [4(−5) 4(−3) 4(6)]−[5(−2) 5(4) 5(5)] = [−20 −12 24]−[−10 20 25] = [(−20 − (−10)) (−12 − 20) (24 − 25)] = [−10 −32 −1] (8) (a) x+y = [−1 5]+[2 −4] = [(−1+2) (5−4)] = [1 1], x−y = [−1 5]−[2 −4] = [(−1−2) (5−(−4))] = [−3 9], y − x = [2 −4] − [−1 5] = [(2 − (−1)) ((−4) − 5)] = [3 −9] (see Figure 3, next page)

(c) x + y = [2 5 −3] + [−1 3 −2] = [(2 + (−1)) (5 + 3) ((−3) + (−2))] = [1 8 −5], x − y = [2 5 −3] − [−1 3 −2] = [(2 − (−1)) (5 − 3) ((−3) − (−2))] = [3 2 −1], y − x = [−1 3 −2] − [2 5 −3] = [((−1) − 2) (3 − 5) ((−2) − (−3))] = [−3 −2 1] (see Figure 4, next page)

(10) In each part, consider the center of the clock to be the origin. (a) At 12 PM, the tip of the minute hand is at (0 10). At 12 : 15 PM, the tip of the minute hand is at (10 0). To find the displacement vector, we subtract the vector for the initial point from the vector for the terminal point, yielding [10 0] − [0 10] = [10 −10].

(b) At 12 PM, the tip of the minute hand is at (0 10). At 12 : 40 PM, the minute hand makes a 210◦ angle with the positive -axis. So, as shown in Figure 1.10 in the textbook, hand makes h the ³ minute √ ´ ¡ ¢i 3 ◦ ◦ the vector v = [kvk cos  kvk sin ] = [10 cos(210 ) 10 sin(210 )] = 10 − 2  10 − 12 = √ [−5 3 −5]. To find the displacement vector, √ for the initial point from the √ we subtract the vector vector for the terminal point, yielding [−5 3 −5] − [0 10] = [−5 3 −15].

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

2

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Figure 3

Section 1.1

Figure 4

(13) As shown in Figure 1.10 in the textbook, for any vector v ∈ R2  v = [kvk cos  kvk sin ], where  is the angle v makes with the positive -axis. Now “southwest” corresponds to 225◦ , “east” corresponds to 0◦ , and northwest corresponds to 135◦ . Hence, if v1 , v2 , and v3 are the vectors corresponding to the three given movements, thenh i √

√

v1 = [1 cos(225◦ ) 1 sin(225◦ )] = − 22  − 22 , v2 = [05 cos(0◦ ) 05 sin(0◦ )] = [05 0], and h ³ √ ´ √ i v3 = [02 cos(135◦ ) 02 sin(135◦ )] = 02 − 22  02 22 . Hence, the result of all three putts is h √ h ³ √ ´ √ i √ i √ √ v1 +v2 +v3 = − 22  − 22 +[05 0]+ 02 − 22  02 22 = [05−06 2 −04 2] ≈ [−03485 −05657].

(14) As shown in Figure 1.10 in the textbook, for any vector v ∈ R2  v = [kvk cos  kvk sin ], where  is the angle v makes with the positive -axis. Now “northwestward” corresponds to 135◦ , and “southward” rower, and v2 corresponds to the current, then corresponds to 270◦ . Hence, if vh1 corresponds ³ √ ´ √ to i the £ √ √ ¤ 2 2 ◦ ◦ v1 = [4 cos(135 ) 4 sin(135 )] = 4 − 2  4 2 = −2 2 2 2 and v2 = [3 cos(270◦ ) 3 sin(270◦ )] = √ √ £ √ √ ¤ [0 −3]. Therefore, the rower’s net velocity = v1 +v2 = −2 2 2 2 +[0 −3] = [−2 2 −3+2 2]. The ° √ ° q √ √ ¢2 p √ √ ¡ resultant speed = °[−2 2 −3 + 2 2]° = (−2 2)2 + −3 + 2 2 = 25 − 12 2 ≈ 283 km/hr.

(16) If v1 corresponds to the rower, and v2 corresponds to the current and v3 corresponds to the resultant velocity, then v3 = v1 + v2 , or v1 = v3 − v2 . Now, as shown in Figure 1.10 in the textbook, for any vector v ∈ R2  v = [kvk cos  kvk sin ], where  is the angle v makes with the positive -axis. Also, corresponds to 45◦ . Thus, “westward” corresponds to 180◦h, and “northeastward” √ √ i √ ¤ £√ 2 2 and v3 = [8 cos(180◦ ) 8 sin(180◦ )] = [−8 0]. v2 = [2 cos(45◦ ) 2 sin(45◦ )] = 2 22  2 22 = √ √ £√ √ ¤ Hence, v1 = v3 − v2 = [−8 0] − 2 2 = [−8 − 2 − 2].

(17) Let f and a represent the resultant force and the resultant acceleration, respectively. Then f = ¤ £ [3−124] 4 48 16 = √ 2 4 2 2 [3 −12 4] = 13 [3 −12 4] = 12 f1 + f2 + f3 . Now, f1 = 4 k[3−124]k 13  − 13  13 , 3 +(−12) +4

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

3

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.1

¤ £ 0 − 85  − 65 , and ¤ £ ¤ £ [001] 48 16 8 6 f3 = 6 k[001k = 61 [0 0 1] = [0 0 6]. Therefore, f = f1 +f2 +f3 = 12 13  − 13  13 + 0 − 5  − 5 +[0 0 6] = £ 12 344 392 ¤ 1 1 12 344 392 13  − 65  65 . Finally, f = a . Since  = 20, acceleration = a =  f = 20 [ 13  − 65  65 ] ≈ [0−4−3] f2 = 2 k[0−4−3]k =√2

2 [0 −4 −3] = 25 [0 −4 −3] = 0 +(−4)2 +(−3)2

[00462 −02646 03015].

(20) As shown in Figure 1.10 in the textbook, for any vector v ∈ R2  v = [kvk cos  kvk sin ], where  is the angle v makes with the positive -axis. Thus, if we let  = kak and  = kbk, then h √ h √ i √ i a = [ cos(135◦ )  sin(135◦ )] = − 22   22 and b = [ cos(60◦ )  sin(60◦ )] =  12   23 . Now, since h √ h √ i √ i the weight is not moving, a + b + g = 0. That is, − 22   22 +  12   23 + [0 −] = [0 0]. Using √ √ the first coordinates gives us − 22 +  12 = 0. Hence,  = 2. Using the second coordinates gives us √ √ √ √ √ √  22 +  23 −  = 0. Substituting in  = 2 yields  22 + 2 23 −  = 0. Solving for  produces √ 2 √ . Plugging these values in to the formulas for a and b gives √ , and  = 2 = 1+  = √2 2 3 1+ 3 ( ) √ −    a = [ 1+√3  1+√3 ] and b = [ 1+√3  1+√33 ].

(22) In each part, let x = [1  2       ], y = [1  2       ], and  ∈ R. (b) x + (−x) = [1       ] + [−1      − ] = [(1 + (−1 ))     ( + (− ))] = [0     0] = 0.

Also, (−x) + x = x + (−x) (by part (1) of Theorem 1.3) = 0, by the above. (c) (x+y) = ([(1 +1 )     ( + )]) = [(1 +1 )     ( + )] = [(1 +1 )     ( + )] = [1       ] + [1       ] = x + y (23) If  = 0, we are done. Otherwise, x = 0 =⇒ ( 1 )(x) = 1 (0) =⇒ ( 1 · )x = 0 (by part (7) of Theorem 1.3) =⇒ x = 0. Thus either  = 0 or x = 0. p (25) (a) False. The length of [1  2  3 ] is 21 + 22 + 23 . The formula given in the problem is missing the square root. So, for example, the length of [0 3 4] is actually 5, not 02 + 32 + 42 = 25. (b) True. This is easily proved using parts (1) and (2) of Theorem 1.3. (c) True. [2 0 −3] = 2[1 0 0] + (−3)[0 0 1].

(d) False. Two nonzero vectors are parallel if and only if one is a scalar multiple of the other. To show why [3 −5 2] and [6 −10 5] are not parallel, suppose they are. Then there would be a  ∈ R such that [3 −5 2] = [6 −10 5]. Comparing first coordinates shows that 3 = 6, or  = 2. But this value of  must work in all coordinates. However, it does not work in the third coordinate, since (2)(2) 6= 5. (e) True. Multiply both sides of x = 0 by 1 .

(f) False. Parallel vectors can be in opposite directions. For example, [1 0] and [−2 0] are parallel, but are not in the same direction. (g) False. The properties of vectors in Theorem 1.3 are independent of the location of the initial points of the vectors. p √ √ (h) False. ||[9 −8 −4]|| = 92 + (−8)2 + (−4)2 = 81 + 64 + 16 = 161  1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

4

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.2

Section 1.2 x·y (1) In each part, we use the formula cos  = kxkkyk . [−43]·[6−1] x·y √2 = k[−43]kk[6−1]k = √ (−4)(6)+(3)(−1) (a) cos  = kxkkyk 2 2

6 +(−1)2

(−4) +3

√ . And so, using a calculator yields = 5−27 37

27 ), or about 1526◦ , or 266 radians.  = arccos(− 5√ 37 [7−42]·[−6−101] x·y √ = k[7−42]kk[−6−101]k = √ 2 (7)(−6)+(−4)(−10)+(2)(1) (c) cos  = kxkkyk 2 2 2 2 7 +(−4) +2

(−6) +(−10) +12

0 = √69√ = 0. And so, 137

 = arccos(0), which is 90◦ , or 2 radians.

(4) When a force f moves an object by a displacement d, the work performed is f · d.

1 [−2 4 5] = (b) First, we need to compute f . A unit vector in the direction of f is u = k[−245]k 1 1 √ [−2 4 5] = √ [−2 4 5] = 3√ [−2 4 5]. Thus, since kf k = 26, f = 26u = 326 5 5 (−2)2 +42 +52 h √ √ √ i −52 5 104 5 26 5 , where we have rationalized the denominators.

15  15  3

1 [−1 2 2] = Next, we need to compute d. A unit vector in the direction of d is v = k[−122]k ¤ £ ¤ £ 1 20 20 √ [−1 2 2] = − 13  23  23 . Thus, since kdk = 10, d = 10v = − 10 3  3  3 . (−1)2 +22 +22 h √ √ √ i £ ¤ 5 104 5 26 5 20 20 Hence, the work performed is f · d = −52 · − 10 15  15  3 3  3  3 ³ √ ´¡ √ √ ¢ ¢ ³ 104√5 ´ ¡ 20 ¢ ³ 26√5 ´ ¡ 20 ¢ ¡ √ 5 10 − = 19 104 5 + 416 5 + 520 5 = −52 15 3 + 15 3 + 3 3 √

= 10409 5 , or 2584 joules.

(6) In all parts, let x = [1  2       ]  y = [1  2       ]  and z = [1  2       ]  (a) x · y = [1  2       ] · [1  2       ] = 1 1 + · · · +   = 1 1 + · · · +   = [1  2       ] · [1  2       ] = y · x

(b) x · x = [1  2       ] · [1  2       ] = 1 1 + · · · +   = 21 + · · · + 2 . Now 21 + · · · + 2 is a sum of squares, each of which must be nonnegative. Hence, the sum is also nonnegative, and ³p ´2 21 + · · · + 2 = kxk2 . so its square root is defined. Thus, 0 ≤ x · x = 21 + · · · + 2 = (c) Suppose x · x = 0. From part (b), 0 = x · x = 21 + · · · + 2 ≥ 2 , for each , since all terms in the sum are nonnegative. Hence, 0 ≥ 2 for each . But 2 ≥ 0, because it is a square. Hence each  = 0. Therefore, x = 0.

(d) (x · y) =  ([1  2       ] · [1  2       ]) =  (1 1 + · · · +   ) = 1 1 + · · · +   = [1  2       ] · [1  2       ] = (x) · y.

Next, (x · y) = (y · x) (by part (a)) = (y) · x (by the above) = x · (y), by part (a) (e) (x + y) · z = ([1  2       ] + [1  2       ]) · [1  2       ] = [1 + 1  2 + 2       +  ] · [1  2       ] = (1 + 1 )1 + (2 + 2 )2 + · · · + ( +  ) = (1 1 + 2 2 + · · · +   ) + (1 1 + 2 2 + · · · +   ) Also, (x · z) + (y · z) = ([1  2       ] · [1  2       ]) + ([1  2       ] · [1  2       ]) = (1 1 + 2 2 + · · · +   ) + (1 1 + 2 2 + · · · +   ). Hence, (x + y) · z = (x · z) + (y · z).

(7) No. Consider x = [1 0], y = [0 1], and z = [1 1]. Then x · z = [1 0] · [1 1] = (1)(1) + (0)(1) = 1 and y · z = [0 1] · [1 1] = (0) (1) + (1)(1) = 1, so x · z = y · z. But x 6= y. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

5

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.2

2

(8) By part (2) of Theorem 1.5, (a − b) · (a − b) = ka − bk ≥ 0. Hence, by parts (5) and (6) of Theorem 1.5, (a · a) − (b · a) − (a · b) + (b · b) ≥ 0. By parts (1) and (2) of Theorem 1.5, we have 2 2 kak −2 (a · b)+kbk ≥ 0. Thus, 1−2 (a · b)+1 ≥ 0 because a and b are unit vectors, and so a·b ≤ 1 (16) Position the cube so that one of its corners is at the origin and the three edges from that corner are along the positive -, -, and -axes. Then the corner diagonally opposite from the origin is at the point (  ). The vector representing this diagonal is d = [  ]. √ √ √ (a) The length of the diagonal = kdk = 2 + 2 + 2 = 32 = 3.

(b) The angle d makes with a side µ of the cube is equal between d and [ 0 0], which ¶ to the angle ³ ³ 2 ´ ´ √ d·[00] 3 ◦ √ √ is arccos kdkk[00]k = arccos = arccos √()+(0)+(0) = arccos( 2 3 ) ≈ 547 , or 3 ( 3)( 2 +02 +02 ) 0955 radians. ³ ´ ³ ´ ³ ´ (2)(1)+(1)(4)+(5)(−3) a·b (17) (a) proja b = kak a = [215]·[14−3] [2 1 5] = [2 1 5] = −9 2 2 2 2 2 2 +1 +5 30 [2 1 5] k[215]k ¤ ¤ £ ¤ £ 3 £ 3 3 3 = − 5  − 10  − 32 . b − proja b = [1 4 −3] − − 35  − 10  − 32 = 85  43 10  − 2 . proja b is orthogonal ¤ £ ¤ £ 3 3 3  − 32 · 85  43 to (b − proja b) because proja b · (b − proja b) = − 5  − 10 10  − 2 = ¡ 3 ¢ ¡ 8 ¢ ¡ 3 ¢ ¡ 43 ¢ ¡ 3 ¢ ¡ 3 ¢ − 5 5 + − 10 10 + − 2 − 2 = 0. ³ ´ ³ ´ a·b (c) proja b = kak a = [10−12]·[3−10−1] [1 0 −1 2] = 2 2 k[10−12]k ³ ´ ¤ £ (1)(3)+(0)(−1)+(−1)(0)+(2)(−1) [1 0 −1 2] = 16 [1 0 −1 2] = 16  0 − 16  13 .

12 +02 +(−1)2 +22 ¤ £ ¤ £ 1 4 b−proja b = [3 −1 0 −1]− 16  0 − 16  13 = 17 6  −1 6  − 3 . proja b is orthogonal to (b − proja b) ¤ £ ¤ £ 1 4 because proja b · (b − proja b) = 16  0 − 16  13 · 17 6  −1 6  − 3 = ¡ 1 ¢ ¡ 17 ¢ ¡ 1¢ ¡1¢ ¡1¢ ¡ 4¢ − 3 = 0.

6 6 + (0) (−1) + − 6 6 + 3 ³ ´ ³ ´ ³ ´ i·a 1+0+0 (19) Let a = [  ]. Then proji a = kik i = [100]·[] [1 0 0] = 2 2 12 +02 +02 [1 0 0] = i. Similarly, k[100]k ³ ³ ´ ³ ´ ³ ´ ´ j·a 0+1+0 k·a a = projj a = kjk j = [010]·[] [0 1 0] = [0 1 0] = j, and proj k = 2 2 2 2 2 2 k 0 +1 +0 kkk ³ ´ ³ k[010]k ´ [001]·[] [0 0 1] = 00+0+1 [0 0 1] = k.

2 +02 +12 k[001]k2

(20) In each part, if v is the given vector, then by Theorem 1.11, x = projv x + (x − projv x), where projv x is parallel to v and (x − projv x) is orthogonal to v. ³ ´ v·x (a) In this case, x = [−6 2 7] and v = [2 −3 4]. Then projv x = kvk v= 2 ³ ´ ³ ´ ¢ ¡ [2−34]·[−627] 20 30 40 [2 −3 4] = (2)(−6)+(−3)(2)+(4)(7) [2 −3 4] = 10 22 +(−3)2 +42 29 [2 −3 4] = [ 29  − 29  29 ]. k[2−34]k2 ¡ ¢ And so projv x is clearly parallel to v since projv x = 10 29 v. Also, 30 40 194 88 163 (x − projv x) = [−6 2 7] − [ 20 29  − 29  29 ] = [− 29  29  29 ]. We can easily check that

30 40 194 88 163 projv x + (x − projv x) = [ 20 29  − 29  29 ] + [− 29  29  29 ] = [−6 2 7] = x, and that projv x

30 40 194 88 163 and (x − projv x) are orthogonal, since projv x · (x − projv x) = [ 20 29  − 29  29 ] · [− 29  29  29 ] ¡ 20 ¢ ¡ 194 ¢ ¡ 30 ¢ ¡ 88 ¢ ¡ 40 ¢ ¡ 163 ¢ = 29 − 29 + − 29 29 + 29 29 = 0.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

6

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.3

³ ´ v·x (c) In this case, x = [−6 2 7] and v = [3 −2 6]. Then projv x = kvk v= 2 ³ ´ ³ ´ ¢ £ 60 40 120 ¤ ¡ [3−26]·[−627] [3 −2 6] = (3)(−6)+(−2)(2)+(6)(7) [3 −2 6] = 20 32 +(−2)2 +62 49 [3 −2 6] = 49  − 49  49 . k[3−26]k2 ¡ ¢ And so, projv x is clearly parallel to v since projv x = 20 49 v. Also, £ 60 40 120 ¤ £ 354 138 223 ¤ (x − projv x) = [−6 2 7] − 49  − 49  49 = − 49  49  49 . We can easily check that ¤ £ 354 138 223 ¤ £ 40 120 = [−6 2 7] = x, and that projv x projv x + (x − projv x) = 60 49  − 49  49 + − 49  49  49 ¤ £ 354 138 223 ¤ £ 40 120 and (x − projv x) are orthogonal, since projv x · (x − projv x) = 60 49  − 49  49 · − 49  49  49 ¡ ¢ ¡ 354 ¢ ¡ 40 ¢ ¡ 138 ¢ ¡ 120 ¢ ¡ 223 ¢ − 49 + − 49 = 60 49 49 + 49 49 = 0.

(23) (a) Note that projx y = [ 85  − 65  2] = 25 x and y − projx y = [ 75  − 24 5  −4] is orthogonal to x. Let w = y − projx y Then since y = projx y + (y − projx y) we have y = 25 x + w where w is orthogonal to x.

(25) (a) True by part (4) of Theorem 1.5. (b) True. Theorem 1.7 (the Cauchy-Schwarz Inequality) states that |x · y| ≤ kxk kyk. Since x · y ≤ |x · y|, we have x · y ≤ kxk kyk. But then, because kxk  0, we can divide both sides by kxk to obtain the desired inequality (c) False. For example, if x = [1 0 0] and y = [0 1 0], then kx − yk = k[1 −1 0]k = p √ 12 + (−1)2 + 02 = 2, and kxk − kyk = k[1 0 0]k − k[0 1 0k = √ √ √ 12 + 02 + 02 − 02 + 12 + 02 = 1 − 1 = 0. However, 2 £ 0. (d) False. Theorem 1.9 shows that if   2 , then x · y  0. (Remember,  is defined to be ≤ .) (e) True. If  6= , then e · e = 0(0) + · · · + 1(0) + · · · + 0(1) + 0(0) + · · · + 0(0) = 0 + · · · + 0 = 0. |{z} |{z} (f) False. proja b =

³

a·b kak2

´

th terms

th terms

a, and so proja b is parallel to a, since it is a scalar multiple of a. Thus, if

proja b = b, a and b are parallel, example, suppose a = [1 0] ³ not´perpendicular. ³ ´For a particular ³ ´

and b = [2 0]. Then proja b =

a·b kak2

a=

[10]·[20] k[10]k

[1 0] =

1(2)+0(0) √ 12 +02

[1 0] = 2[1 0] = [2 0] =

b. However, a = [1 0] and b = [2 0] are parallel (since b = 2a) and not perpendicular (since a · b = [1 0] · [2 0] = 1(2) + 0(0) = 2 6= 0.

Section 1.3 (1) (b) Let  = max{|| ||}. Then kx ± yk ≤ (kxk + kyk).

The proof is as follows: kx ± yk = ≤ = ≤ =

kx + (±y)k kxk + k±yk by the Triangle Inequality || kxk + | ± | kyk by Theorem 1.1  kxk +  kyk because || ≤  and | ± | ≤   (kxk + kyk)

(2) (b) The converse is: If an integer has the form 3 + 1, then it also has the form 6 − 5, where  and  are integers. For a counterexample, consider that the number 4 = 3 + 1 with  = 1, but there is no integer  such that 4 = 6 − 5, since this equation implies  = 32 , which is not an integer. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

7

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.3

2

2

(6) (a) Consider x = [1 0 0] and y = [1 1 0]. Then x · y = [1 0 0] · [1 1 0] = 1 = k[1 0 0]k = kxk , but x 6= y. (b) If x 6= y, then x · y 6= kxk2 .

(c) Yes. Using the counterexample from part (a), x 6= y, but x · y = kxk2 .

(7) Assuming x and y are vectors in R , the contrapositive of the given statement is: “If kx + yk2 = kxk2 + kyk2 , then x · y = 0.” To prove the contrapositive, we assume the premise kx + yk2 = kxk2 + kyk2 . Then by applying part (2) of Theorem 1.5 to both sides of this equation, we have (x + y) · (x + y) = x · x + y · y Expanding the left side (using the distributive laws from Theorem 1.5), we find x · x + 2(x · y)+ y· y = x · x + y · y, which reduces to x · y = 0 the desired conclusion. Since the contrapositive of the original statement has been proven, the original statement is proven as well.

(9) (a) Contrapositive: If x = 0, then x is not a unit vector.

Converse: If x is nonzero, then x is a unit vector.

Inverse: If x is not a unit vector, then x = 0. (c) (Let x, y be nonzero vectors.)

Contrapositive: If projy x 6= 0, then projx y 6= 0.

Converse: If projy x = 0, then projx y = 0.

Inverse: If projx y 6= 0, then projy x 6= 0.

(11) (b) Converse: Let x and y be vectors in R . If kx + yk ≥ kyk, then x · y = 0. The original statement is true, but the converse is false in general.

Proof of the original statement: kx + yk2

= = = ≥

(x + y) · (x + y) kxk2 + 2(x · y) + kyk2 kxk2 + kyk2 kyk2

by part (2) of Theorem 1.5 using Theorem 1.5 since x · y = 0 by part (2) of Theorem 1.5

Taking the square root of both sides yields kx + yk ≥ kyk. √ Counterexample √ to converse: let x = [1 0] and y = [1 1]. Then kx + yk = k[2 1]k = 5 and kyk = k[1 1]k = 2. Hence kx + yk ≥ kyk. However, x · y = [1 0] · [1 1] = 1 6= 0.

(12) Assume each coordinate of both x and y is equal to either 1 or −1. For a proof by contradiction of the statement “If x is orthogonal to y, then  is even,” we assume the premise and we assume the negation of the desired conclusion, and then show that this leads to a logical contradiction.

That is, we assume P that x ⊥ y and we also assume that  is odd. Then x ·y = 0. Now x ·y = =1   . But each product   equals either 1 or −1. If exactly  of these products equal 1, then x · y =  − ( − ) = − + 2.

Hence − + 2 = 0, and so  = 2, contradicting the fact that  is odd. Since we have reached a logical contradiction by assuming that  is odd, it follows that  is even.

(17) Base Step ( = 1): kx1 k2 = kx1 k2 .

Inductive Step: Assume kx1 + · · · + x k2 = kx1 k2 + · · · + kx k2 . Prove kx1 + · · · + x + x+1 k2 = kx1 k2 + · · · + kx k2 + kx+1 k2 . We have k(x1 + · · · + x ) + x+1 k2 = kx1 + · · · + x k2 + 2((x1 + · · · + x )·x+1 ) + kx+1 k2 = kx1 + · · · + x k2 + kx+1 k2 (since x+1 is orthogonal to all of x1      x ) = kx1 k2 + · · · + kx k2 + kx+1 k2  by the inductive hypothesis.

(19) Step 1 cannot be reversed, because  could equal −(2 + 2) and then the converse would not hold.

Step 2 cannot be reversed, because 2 could equal 4 +42 + with   4. For a specific counterexample c 2016 Elsevier Ltd. All rights reserved.

Copyright °

8

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.4

√ to the converse, set  = 4 + 42 + 5.

3  +8 Step 3 is reversible. To prove the converse, multiply both sides of  = 4 2 by 2.

Step 4 cannot be reversed, because in general  does not have to equal 2 + 2. In particular,  could equal 2 + 5.

2 +2) Step 5 can be reversed by multiplying 2 by 2( 2(2 +2) .  Step 6 cannot be reversed, since  could equal 2 +  for some  6= 0.

(20) (a) Negation: For every unit vector x in R3 , x · [1 −2 3] 6= 0. (Note that the existential quantifier has changed to a universal quantifier.) (c) Negation: x = 0 or kx + yk 6= kyk, for all vectors x and y in R . (The “and” has changed to an “or,” and the existential quantifier has changed to a universal quantifier.) (e) Negation: There is an x ∈ R3 such that for every nonzero y ∈ R3 , x · y 6= 0. (Note that the universal quantifier on x has changed to an existential quantifier, and the existential quantifier on y has changed to a universal quantifier.)

(21) (a) When negating the conclusion “x = 0 or kx − yk  kyk” of the original statement, we must change the “or” to “and.”

Contrapositive: If x 6= 0 and kx − yk ≤ kyk, then x · y 6= 0.

Converse: If x = 0 or kx − yk  kyk, then x · y = 0.

Inverse: If x · y 6= 0, then x 6= 0 and kx − yk ≤ kyk.

(25) (a) False. We must also verify that the reversed steps are truly reversible. That is, we must also supply valid reasons for the forward steps thus generated, as in the proof of Result 3. (b) True. “If not  then not ” is the contrapositive of “If  then .” A statement and its contrapositive are always logically equivalent. (c) True. We saw that “ only if ” is an equivalent form for “If  then ,” whose converse is “If  then .” (d) False. “ is a necessary condition for ” is an equivalent form for “If  then .” As such, it does not include the statement “If  then ” that is included in the “ if and only if ” statement. (e) False. “ is a necessary condition for ” and “ is a sufficient condition for ” are both equivalent forms for “If  then .” As such, neither includes the statement “If  then ” that is included in the “ if and only if ” statement. (f) False. The inverse of a statement is the contrapositive of the converse. Hence, the converse and the inverse are logically equivalent, and always have the same truth values. (g) False. The problem only describes the inductive step. The base step of the proof is also required. (h) True. This rule is given in the section for negating statements with quantifiers. (i) False. The “and” must change to an “or.” Thus, the negation of “ and ” is “not  or not .”

Section 1.4 ⎡

−4 (1) (a) ⎣ 0 6 ⎡ 2 =⎣ 2 9

⎤ ⎡ ⎤ ⎡ ⎤ 2 3 6 −1 0 (−4 + 6) (2 + (−1)) (3 + 0) 5 −1 ⎦ + ⎣ 2 2 −4 ⎦ = ⎣ (0 + 2) (5 + 2) ((−1) + (−4)) ⎦ 1 −2 3 −1 1 (6 + 3) (1 + (−1)) ((−2) + 1) ⎤ 1 3 7 −5 ⎦ 0 −1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

9

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.4

⎡

⎤ ⎡ ⎤ ⎡ ⎤ −4 2 3 4(−4) 4(2) 4(3) −16 8 12 0 20 −4 ⎦ (c) 4 ⎣ 0 5 −1 ⎦ = ⎣ 4(0) 4(5) 4(−1) ⎦ = ⎣ 6 1 −2 4(6) 4(1) 4(−2) 24 4 −8

(e) Impossible. C is a 2 × 2 matrix, F is a 3 × 2 matrix, and E is a 3 × 3 matrix. Hence, 3F is 3 × 2 and −E is 3 × 3. Thus, C, 3F, and −E have different sizes. However, in order to add matrices, they must be the same size. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −4 2 3 3 −3 5 6 −1 0 0 −2 ⎦ − ⎣ 2 2 −4 ⎦ (g) 2 ⎣ 0 5 −1 ⎦ − 3 ⎣ 1 6 1 −2 6 7 −2 3 −1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2(−4) 2(2) 2(3) −3(3) −3(−3) −3(5) −6 −(−1) −0 −3(0) −3(−2) ⎦ + ⎣ −2 −2 −(−4) ⎦ = ⎣ 2(0) 2(5) 2(−1) ⎦ + ⎣ −3(1) 2(6) 2(1) 2(−2) −3(6) −3(7) −3(−2) −3 −(−1) −1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −8 4 6 −9 9 −15 −6 1 0 0 6 ⎦ + ⎣ −2 −2 4 ⎦ = ⎣ 0 10 −2 ⎦ + ⎣ −3 12 2 −4 −18 −21 6 −3 1 −1 ⎡ ⎤ ⎡ ⎤ (−8) + (−9) + (−6) 4+9+1 6 + (−15) + 0 −23 14 −9 10 + 0 + (−2) (−2) + 6 + 4 ⎦ = ⎣ −5 8 8 ⎦ = ⎣ 0 + (−3) + (−2) 12 + (−18) + (−3) 2 + (−21) + 1 (−4) + 6 + (−1) −9 −18 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −4 2 3 3 −3 5 −4 0 6 3 1 6 0 −2 ⎦ = ⎣ 2 5 1 ⎦ + ⎣ −3 0 7 ⎦ (i) ⎣ 0 5 −1 ⎦ + ⎣ 1 6 1 −2 6 7 −2 3 −1 −2 5 −2 −2 ⎡ ⎤ ⎡ ⎤ ((−4) + 3) (0 + 1) (6 + 6) −1 1 12 ⎦ = ⎣ −1 (5 + 0) (1 + 7) 5 8 ⎦ = ⎣ (2 + (−3)) (3 + 5) ((−1) + (−2)) ((−2) + (−2)) 8 −3 −4

(l) Impossible. Since C is a 2 × 2 matrix, both C and 2C are 2 × 2. Also, since F is a 3 × 2 matrix, −3F is 3 × 2. Thus, 2C and −3F have different sizes. However, in order to add matrices, they must be the same size. ⎡ ⎤ ⎡ ⎤ 10 2 −3 3 1 6 0 7 ⎦. (n) (B − A) = ⎣ −3 −3 −2 ⎦ and E = ⎣ −3 −3 −3 3 5 −2 −2 ⎡ ⎤ ⎡ ⎤ 13 3 3 13 −6 2 ¢ ¡  Thus (B − A) + E = ⎣ −6 −3 5 ⎦, and so (B − A) + E = ⎣ 3 −3 −5 ⎦.

2 −5 1 3 5 1

(2) For a matrix to be square, it must have the same number of rows as columns. A matrix X is diagonal if and only if it is square and  = 0 for  6= . A matrix X is upper triangular if and only if it is square and  = 0 for   . A matrix X is lower triangular if and only if it is square and  = 0 for   . A matrix X is symmetric if and only if it is square and  =  for all  . A matrix X is skew-symmetric if and only if it is square and  = − for all  . Notice that this implies that if X is skew-symmetric, then  = 0 for all . Finally, the transpose of an  ×  matrix X is the  ×  matrix whose ( )th entry is  . We now check each matrix in turn.

A is a 3 × 2 matrix, so is not square. Therefore it can not be diagonal, upper triangular, lower ∙ ¸ −1 0 6  . triangular, symmetric, or skew-symmetric. Finally, A = 4 1 0 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

10

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.4

B is a 2 × 2 matrix, and so is square. Because 12 = 21 = 0, B is diagonal, upper triangular, lower triangular, and symmetric. B is not skew-symmetric because 11 6= 0. Finally, B = B. C is a 2 × 2 matrix, and so is square. Because 12 6= 0, C is neither diagonal, nor lower triangular.

Because 21 6= 0, C is not upper triangular. C is not symmetric because 12 6= 21 . C is not skew∙ ¸ −1 −1 . symmetric because 11 6= 0. Finally, C = 1 1

D is a 3 × 1 matrix, so is not square. Therefore it can not be diagonal, upper triangular, lower £ ¤ triangular, symmetric, or skew-symmetric. Finally, D = −1 4 2 .

E is a 3 × 3 matrix, and so is square. Because 13 6= 0, E is neither diagonal, nor lower triangular.

Because 31 6= 0, E is not upper triangular. E is not symmetric because 31 6= 13 . E is not skew⎡ ⎤ 0 0 −6 0 ⎦. symmetric because 22 6= 0. Finally, E = ⎣ 0 −6 6 0 0 F is a 4 × 4 matrix, and so is square. Because 14 6= 0, F is neither diagonal, nor lower triangular.

Because 41 6= 0, F is not upper triangular. F is symmetric because  =  for every  . F is not skew-symmetric because 32 6= −23 . Finally, F = F.

G is a 3 × 3 matrix, and so is square. Because  = 0 whenever  6= , G is diagonal, lower triangular, upper triangular, and symmetric. G is not skew-symmetric because 11 6= 0. Finally, G = G.

H is a 4 × 4 matrix, and so is square. Because 14 6= 0, H is neither diagonal, nor lower triangular.

Because 41 6= 0, H is not upper triangular. H is not symmetric because 12 6= 21 . H is skewsymmetric because  = − for every  . Finally, H = −H.

J is a 4 × 4 matrix, and so is square. Because 12 6= 0, J is neither diagonal, nor lower triangular.

Because 21 6= 0, J is not upper triangular. J is symmetric because  =  for every  . J is not skew-symmetric because 32 6= −23 . Finally, J = J.

K is a 4 × 4 matrix, and so is square. Because 14 6= 0, K is neither diagonal, nor lower triangular.

Because 41 6= 0, K is not upper triangular. K is not symmetric because 12 6= 21 . K is not skew⎡ ⎤ 1 −2 −3 −4 ⎢ 2 1 −5 −6 ⎥ ⎥. symmetric because 11 6= 0. Finally, K = ⎢ ⎣ 3 5 1 −7 ⎦ 4 6 7 1

L is a 3 × 3 matrix, and so is square. Because 13 6= 0, L is neither diagonal, nor lower triangular.

Because  = 0 whenever ̇  , L is upper triangular. L is not symmetric because 31 6= 13 . L is not ⎡ ⎤ 1 0 0 skew-symmetric because 11 6= 0 (or because 21 6= −12 ). Finally, L = ⎣ 1 1 0 ⎦.

1 1 1 M is a 3 × 3 matrix, and so is square. Because 31 6= 0, M is neither diagonal, nor upper triangular.

Because  = 0 whenever ̇  , M is lower triangular. M is not symmetric because 31 6= 13 . ⎡ ⎤ 0 1 1 M is not skew-symmetric because 31 6= −13 . Finally, M = ⎣ 0 0 1 ⎦.

0 0 0

N is a 3 × 3 matrix, and so is square. Because  = 0 whenever  6= , N is diagonal, lower triangular, upper triangular, and symmetric. N is not skew-symmetric because 11 6= 0. Finally, N = N = I3 .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

11

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.4

P is a 2 × 2 matrix, and so is square. Because 12 6= 0, P is neither diagonal, nor lower triangular.

Because 21 6= 0, P is not upper triangular. P is symmetric because 12 = 21 . P is not skew-symmetric because 12 6= −21 . Finally, P = P.

Q is a 3 × 3 matrix, and so is square. Because 31 6= 0, Q is neither diagonal, nor upper triangular.

Because  = 0 whenever ̇  , Q is lower triangular. Q is not symmetric because 31 6= 13 . Q is ⎡ ⎤ −2 4 −1 2 ⎦. not skew-symmetric because 11 6= 0 (or because 31 6= −13 ). Finally, Q = ⎣ 0 0 0 0 3 R is a 3 × 2 matrix, so is not square. Therefore it can not be diagonal, upper triangular, lower ∙ ¸ 6 3 −1 triangular, symmetric, or skew-symmetric. Finally, R = .

2 −2 0

(3) Just after Theorem 1.15 in the textbook, the desired decomposition is described as A = S + V, where S = 12 (A + A ) is symmetric and V = 12 (A − A ) is skew-symmetric. ⎛⎡ ⎤ ⎡ ⎤⎞ 3 −1 4 3 0 1 2 5 ⎦ + ⎣ −1 2 −3 ⎦⎠ (a) S = 12 (A + A ) = 12 ⎝⎣ 0 1 −3 2 4 5 2 ⎛⎡ ⎤⎞ (3 + 3) ((−1) + 0) (4 + 1) (2 + 2) (5 + (−3)) ⎦⎠ = 12 ⎝⎣ (0 + (−1)) (1 + 4) ((−3) + 5) (2 + 2) ⎡ ⎤ ⎡ ⎤ ⎡ 1 (6) 1 (−1) 1 (5) ⎤ 3 − 12 25 2 2 2 6 −1 5 ⎥ ⎥ ⎢ ⎢ 1 1 1 ⎢ 4 2 ⎦ = ⎣ 12 (−1) 2 1 ⎥ = 12 ⎣ −1 2 (4) 2 (2) ⎦ = ⎣ − 2 ⎦, 1 1 1 5 2 4 5 1 2 2 (5) 2 (2) 2 (4) 2 which is symmetric, since  =  for all  . ⎛⎡ ⎤ ⎡ ⎤⎞ 3 −1 4 3 0 1 2 5 ⎦ − ⎣ −1 2 −3 ⎦⎠ V = 12 (A − A ) = 12 ⎝⎣ 0 1 −3 2 4 5 2 ⎛⎡ ⎤⎞ (3 − 3) ((−1) − 0) (4 − 1) (2 − 2) (5 − (−3)) ⎦⎠ = 12 ⎝⎣ (0 − (−1)) (1 − 4) ((−3) − 5) (2 − 2) ⎡ ⎤ ⎡ ⎤ ⎡ 1 (0) 1 (−1) 1 (3) ⎤ 0 − 12 23 2 2 2 0 −1 3 ⎥ ⎥ ⎢ ⎢ 1 1 1 ⎢ 0 8 ⎦ = ⎣ 12 (1) 0 4 ⎥ = 12 ⎣ 1 2 (0) 2 (8) ⎦ = ⎣ 2 ⎦, 1 1 1 −3 −8 0 − 32 −4 0 2 (−3) 2 (−8) 2 (0) which is skew-symmetric, since  = − for all  . ⎡ ⎤ ⎡ ⎤ 3 − 12 25 0 − 12 23 ⎢ ⎥ ⎢ ⎥ 1 1 ⎢ 2 1 ⎥ 0 4 ⎥ Note that S + V = ⎢ 2 ⎣ −2 ⎦+⎣ ⎦ 5 2

1

2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

− 32

12

−4

0

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

(3 + 0)

¢ ⎢ ¡ = ⎣ − 12 + 12 ¡ 5 ¡ 3 ¢¢ 2 + −2

Section 1.5

¡ 1 ¡ 1 ¢¢ ¡ 5 3 ¢ ⎤ ⎡ ⎤ −2 + −2 2 + 2 3 −1 4 ⎥ 2 5 ⎦ = A. (2 + 0) (1 + 4) ⎦ = ⎣ 0 1 −3 2 (1 + (−4)) (2 + 0)

(5) (d) The matrix (call it A) must be a square zero matrix; that is, O , for some . First, since A is diagonal, it must be square (by definition of a diagonal matrix). To prove that  = 0 for all  , we consider two cases: First, if  6= , then  = 0 since A is diagonal. Second, if  = , then  =  = 0 since A is skew-symmetric ( = − for all , implying  = 0 for all ). Hence, every entry  = 0, and so A is a zero matrix.

(10) (a) Part (1): Let B = A and C = (A ) . Then  =  =  . Since this is true for all  and , we have C = A; that is, (A ) = A. (b) Part (2), Subtraction Case: Let D = A − B, F = (A − B) and G = A − B . Then  =  =  −  =  . Since this is true for all  and , we have F = G; that is, (A − B) = A − B .

(c) Part (3): Let B = (A ), D = A and F = (A) . Then  =  =  =  . Since this is true for all  and , we have F = B; that is, (A) = (A ).

(13) (a) Trace (B) = 11 + 22 = 2 + (−1) = 1, trace (C) = 11 + 22 = (−1) + 1 = 0, trace (E) = 11 + 22 + 33 = 0 + (−6) + 0 = −6, trace (F) = 11 + 22 + 33 + 44 = 1+0+0+1 = 2, trace (G) = 11 +22 +33 = 6+6+6 = 18, trace (H) = 11 +22 +33 +44 = 0+0+0+0 = 0, trace (J) = 11 +22 +33 +44 = 0+0+1+0 = 1, trace (K) = 11 +22 +33 +44 = 1 + 1 + 1 + 1 = 4, trace (L) = 11 + 22 + 33 = 1 + 1 + 1 = 3, trace (M) = 11 + 22 + 33 = 0 + 0 + 0 = 0, trace (N) = 11 + 22 + 33 = 1 + 1 + 1 = 3, trace (P) = 11 + 22 = 0 + 0 = 0, trace (Q) = 11 + 22 + 33 = (−2) + 0 + 3 = 1. (c) Not necessarily true when   1. Consider matrices L and N in Exercise 2. From part (a) of this exercise, trace(L) = 3 = trace(N). However, L 6= N. (14) (a) False. The main diagonal entries of a 5 × 6 matrix A are 11 , 22 , 33 , 44 , and 55 . Thus, there are 5 entries on the main diagonal. (There is no entry 66 because there is no 6th row.) (b) True. Suppose A is lower triangular. Then  = 0 whenever   . If B = A , then  =  = 0, if   , and so B is upper triangular. (c) False. The square zero matrix O is both skew-symmetric and diagonal. (d) True. This follows from the definition of a skew-symmetric matrix. ³ ¡ ³ ³¡ ¢ ´´ ¡ ¡ ¢¢ ¢ ´  (e) True.  A + B =  A + B (by part (3) of Theorem 1.13) =  A + B ¢ ¡ (by part (2) of Theorem 1.13) =  A + B (by part (1) of Theorem 1.13) = A + B (by part (5) of Theorem 1.12) = B + A (by part (1) of Theorem 1.12).

Section 1.5

⎡

⎤⎡ ⎤ −5 3 6 −2 3 5 ⎦ (1) (b) BA = ⎣ 3 8 0 ⎦ ⎣ 6 −2 0 4 1 −4 ⎡ ⎤ ⎡ ⎤ ((−5) (−2) + (3) (6) + (6) (1)) ((−5) (3) + (3) (5) + (6) (−4)) 34 −24 ((3) (3) + (8) (5) + (0) (−4)) ⎦ = ⎣ 42 49 ⎦ = ⎣ ((3) (−2) + (8) (6) + (0) (1)) ((−2) (−2) + (0) (6) + (4) (1)) ((−2) (3) + (0) (5) + (4) (−4)) 8 −22

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

13

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.5

(c) Impossible; (number of columns of J) = 1 6= 2 = (number of rows of M) ⎡ ⎤ 8 £ ¤ (e) RJ = −3 6 −2 ⎣ −1 ⎦ = [(−3) (8) + (6) (−1) + (−2) (4)] = [−38] 4 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ (8) (−3) (8) (6) (8) (−2) −24 48 −16 8 £ ¤ 3 −6 2 ⎦ (f) JR = ⎣ −1 ⎦ −3 6 −2 = ⎣ (−1) (−3) (−1) (6) (−1) (−2) ⎦ = ⎣ (4) (−3) (4) (6) (4) (−2) −12 24 −8 4

(g) Impossible; (number of columns of R) = 3 6= 1 = (number of rows of T) (j) Impossible; (number of columns of F) = 2 6= 4 = (number of rows of F) ⎡ ⎤2 1 1 0 1 ⎢ 1 0 1 0 ⎥ ⎥ (l) E3 = E(E2 ). Let W = E2 = ⎢ ⎣ 0 0 0 1 ⎦ . Then 1 0 1 0 11 = (1) (1) + (1) (1) + (0) (0) + (1) (1) = 3 12 = (1) (1) + (1) (0) + (0) (0) + (1) (0) = 1 13 = (1) (0) + (1) (1) + (0) (0) + (1) (1) = 2 14 = (1) (1) + (1) (0) + (0) (1) + (1) (0) = 1 21 = (1) (1) + (0) (1) + (1) (0) + (0) (1) = 1 22 = (1) (1) + (0) (0) + (1) (0) + (0) (0) = 1 23 = (1) (0) + (0) (1) + (1) (0) + (0) (1) = 0 24 = (1) (1) + (0) (0) + (1) (1) + (0) (0) = 2 31 = (0) (1) + (0) (1) + (0) (0) + (1) (1) = 1 32 = (0) (1) + (0) (0) + (0) (0) + (1) (0) = 0 33 = (0) (0) + (0) (1) + (0) (0) + (1) (1) = 1 34 = (0) (1) + (0) (0) + (0) (1) + (1) (0) = 0 41 = (1) (1) + (0) (1) + (1) (0) + (0) (1) = 1 42 = (1) (1) + (0) (0) + (1) (0) + (0) (0) = 1 43 = (1) (0) + (0) (1) + (1) (0) + (0) (1) = 0 44 = (1) (1) + (0) (0) + (1) (1) + (0) (0) = 2 ⎡ ⎤⎡ 1 1 0 1 3 ⎢ 1 0 1 0 ⎥⎢ 1 2 3 ⎥⎢ Let Y = E = EE = EW = ⎢ ⎣ 0 0 0 1 ⎦⎣ 1 1 0 1 0 1 11 = (1) (3) + (1) (1) + (0) (1) + (1) (1) = 5 12 = (1) (1) + (1) (1) + (0) (0) + (1) (1) = 3 13 = (1) (2) + (1) (0) + (0) (1) + (1) (0) = 2 14 = (1) (1) + (1) (2) + (0) (0) + (1) (2) = 5 21 = (1) (3) + (0) (1) + (1) (1) + (0) (1) = 4 22 = (1) (1) + (0) (1) + (1) (0) + (0) (1) = 1 23 = (1) (2) + (0) (0) + (1) (1) + (0) (0) = 3 24 = (1) (1) + (0) (2) + (1) (0) + (0) (2) = 1 31 = (0) (3) + (0) (1) + (0) (1) + (1) (1) = 1 32 = (0) (1) + (0) (1) + (0) (0) + (1) (1) = 1 33 = (0) (2) + (0) (0) + (0) (1) + (1) (0) = 0 34 = (0) (1) + (0) (2) + (0) (0) + (1) (2) = 2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

14

1 1 0 1

2 0 1 0

⎤ 1 2 ⎥ ⎥. So, 0 ⎦ 2

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.5

41 = (1) (3) + (0) (1) + (1) (1) + (0) (1) = 4 42 = (1) (1) + (0) (1) + (1) (0) + (0) (1) = 1 43 = (1) (2) + (0) (0) + (1) (1) + (0) (0) = 3 44 = (1) (1) + (0) (2) + (1) (0) + (0) (2) = 1 ⎡ ⎤ 5 3 2 5 ⎢ 4 1 3 1 ⎥ ⎥ Hence, E3 = Y = ⎢ ⎣ 1 1 0 2 ⎦.

4 1 3 1 ⎛⎡ ⎤ ⎞ 9 −3 ∙ ¸ ⎜⎢ 5 −4 ⎥ 2 1 −5 ⎟ ⎢ ⎥ ⎟ (n) D(FK) = D ⎜ ⎝⎣ 2 0 ⎦ 0 2 7 ⎠ 8 −3 ⎡ ⎤ ((9) (2) + (−3) (0)) ((9) (1) + (−3) (2)) ((9) (−5) + (−3) (7)) ⎢ ((5) (2) + (−4) (0)) ((5) (1) + (−4) (2)) ((5) (−5) + (−4) (7)) ⎥ ⎥ = D⎢ ⎣ ((2) (2) + (0) (0)) ((2) (1) + (0) (2)) ((2) (−5) + (0) (7)) ⎦ ((8) (2) + (−3) (0)) ((8) (1) + (−3) (2)) ((8) (−5) + (−3) (7)) ⎤ ⎡ ⎡ ⎤ 18 3 −66 −1 4 3 7 ⎢ 10 −3 −53 ⎥ ⎥ = Z. Then, 5 ⎦⎢ =⎣ 2 1 7 ⎣ 4 2 −10 ⎦ 0 5 5 −2 16 2 −61

11 = (−1) (18) + (4) (10) + (3) (4) + (7) (16) = 146 12 = (−1) (3) + (4) (−3) + (3) (2) + (7) (2) = 5 13 = (−1) (−66) + (4) (−53) + (3) (−10) + (7) (−61) = −603 21 = (2) (18) + (1) (10) + (7) (4) + (5) (16) = 154 22 = (2) (3) + (1) (−3) + (7) (2) + (5) (2) = 27 23 = (2) (−66) + (1) (−53) + (7) (−10) + (5) (−61) = −560 31 = (0) (18) + (5) (10) + (5) (4) + (−2) (16) = 38 32 = (0) (3) + (5) (−3) + (5) (2) + (−2) (2) = −9 33 = (0) (−66) + (5) (−53) + (5) (−10) + (−2) (−61) = −193 ⎡ ⎤ 146 5 −603 Hence, D(FK) = Z = ⎣ 154 27 −560 ⎦.

38 −9 −193 ∙ ¸∙ ¸ ∙ ¸ 10 9 7 −1 ((10) (7) + (9) (11)) ((10) (−1) + (9) (3)) (2) (a) No. LM = = 8 7 11 3 ((8) (7) + (7) (11)) ((8) (−1) + (7) (3)) ∙ ¸ ∙ ¸∙ ¸ 169 17 7 −1 10 9 = , but ML = 133 13 11 3 8 7 ∙ ¸ ∙ ¸ ((7) (10) + (−1) (8)) ((7) (9) + (−1) (7)) 62 56 = = . Hence, LM 6= ML. ((11) (10) + (3) (8)) ((11) (9) + (3) (7)) 134 120

(c) No. Since A is a 3 × 2 matrix and K is a 2 × 3 matrix, AK is a 3 × 3 matrix, while KA is a 2 × 2 matrix. Therefore AK can not equal KA, since they are different sizes. ∙ ¸∙ ¸ ∙ ¸ ∙ ¸ 0 0 3 −1 ((0) (3) + (0) (4)) ((0) (−1) + (0) (7)) 0 0 (d) Yes. NP = = = .

0 0 4 7 ((0) (3) + (0) (4)) ((0) (−1) + (0) (7)) 0 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

15

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Also, PN =

∙

3 −1 4 7

¸∙

0 0 0 0

¸

=

∙

((3) (0) + (−1) (0)) ((3) (0) + (−1) (0)) ((4) (0) + (7) (0)) ((4) (0) + (7) (0))

Hence, NP = PN, and so N and P commute.

(3) (a) (2nd row of BG) = (2nd row of B)G =

£

3 8 0

¤

Section 1.5 ¸

=

∙

0 0 0 0

¸ .

⎡

⎤ 5 1 0 ⎣ 0 −2 −1 ⎦. The entries obtained from 1 0 3

this are: 1st column entry = ((3) (5) + (8) (0) + (0) (1)) = 15, 2nd column entry = ((3) (1) + (8) (−2) + (0) (0)) = −13, 3rd column entry = ((3) (0) + (8) (−1) + (0) (3)) = −8.

Hence, (2nd row of BG) = [15 −13 −8].

⎡

⎤ 1 £ ¤⎢ 1 ⎥ ⎥ (c) (1st column of SE) = S(1st column of E) = 6 −4 3 2 ⎢ ⎣ 0 ⎦ 1 = [(6) (1) + (−4) (1) + (3) (0) + (2) (1)] = [4] (4) (a) Valid, by Theorem 1.16, part (1). (b) Invalid. The equation claims that all pairs of matrices that can be multiplied in both orders commute. However, parts (a) and (c) of Exercise 2 illustrate two different pairs of matrices that do not commute (see above). (c) Valid, by Theorem 1.16, part (1). (d) Valid, by Theorem 1.16, part (2). (e) Valid, by Theorem 1.18. (f) Invalid. For a counterexample, consider the matrices L and M from∙ Exercises ¸ ∙1 through 3. ¸ 10 9 62 56 Using our computation of ML from Exercise 2(a), above, L(ML) = 8 7 134 120 ∙ ¸ ∙ ¸ ((10) (62) + (9) (134)) ((10) (56) + (9) (120)) 1826 1640 = = , ((8) (62) + (7) (134)) ((8) (56) + (7) (120)) 1434 1288 µ∙ ¸∙ ¸¶ 10 9 10 9 2 M but L M = (LL)M = 8 7 8 7 ∙ ¸ ∙ ¸∙ ¸ ((10) (10) + (9) (8)) ((10) (9) + (9) (7)) 172 153 7 −1 = M= ((8) (10) + (7) (8)) ((8) (9) + (7) (7)) 136 121 11 3 ∙ ¸ ∙ ¸ ((172) (7) + (153) (11)) ((172) (−1) + (153) (3)) 2887 287 = = . ((136) (7) + (121) (11)) ((136) (−1) + (121) (3)) 2283 227

(g) Valid, by Theorem 1.16, part (3).

(h) Valid, by Theorem 1.16, part (2). (i) Invalid. For a counterexample, consider the matrices A and K from Exercises 1 through 3. Note that A is a 3 × 2 matrix and K is a 2 × 3 matrix. Therefore, AK is a 3 × 3 matrix. Hence, (AK) is a 3 × 3 matrix. However, A is a 2 × 3 matrix and K is a 3 × 2 matrix. Thus, A K is a 2 × 2 matrix, and so can not equal (AK) .

The equation is also false in general for square matrices, where it is not the sizes of the matrices c 2016 Elsevier Ltd. All rights reserved.

Copyright °

16

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

that cause the problem. For example, let A =

1 1 0 0

¸

, and let K =

∙

1 −1 0 0

¸ . Then,

¸∙ ¸¶ ∙ ¸ 1 1 1 −1 ((1) (1) + (1) (0)) ((1) (−1) + (1) (0)) (AK) = = 0 0 0 0 ((0) (1) + (0) (0)) ((0) (−1) + (0) (0)) ∙ ¸ ∙ ¸ ∙ ¸∙ ¸ 1 −1 1 0 1 0 1 0 = = , while A K = 0 0 −1 0 1 0 −1 0 ∙ ¸ ∙ ¸ ((1) (1) + (0) (−1)) ((1) (0) + (0) (0)) 1 0 = = . ((1) (1) + (0) (−1)) ((1) (0) + (0) (0)) 1 0 

µ∙

∙

Section 1.5

However, there are some rare instances for which the equation is true, such as when A = K, or when either A or K equals I . Another example for which the equation holds is for A = G and K = H, where G and H are the matrices used in Exercises 1, 2, and 3. (j) Valid, by Theorem 1.16, part (3), and Theorem 1.18.

(5) To find the total in salaries paid by Outlet 1, we must compute the total amount paid to executives, the total amount paid to salespersons, and the total amount paid to others, and then add these amounts.

But each of these totals is found by multiplying the number of that type of employee working at Outlet 1 by the salary for that type of employee. Hence, the total salary paid at Outlet 1 is (3) (30000) + (7) (22500) + (8) (15000) = 367500 | {z } . | {z } | {z } | {z } Executives

Salesp ersons

Others

Salary Total

Note that this is the (1 1) entry obtained when multiplying the two given matrices. A similar analysis shows that multiplying the two given matrices gives all the desired salary and fringe benefit totals. In particular, if ⎡ ⎤ ⎤ 3 7 8 ⎡ ⎢ 2 4 5 ⎥ 30000 7500 ⎥⎣ ⎦ W=⎢ ⎣ 6 14 18 ⎦ 22500 4500 , then 15000 3000 3 6 9 11 = (3) (30000) + (7) (22500) + (8) (15000) = 367500 12 = (3) (7500) + (7) (4500) + (8) (3000) = 78000 21 = (2) (30000) + (4) (22500) + (5) (15000) = 225000 22 = (2) (7500) + (4) (4500) + (5) (3000) = 48000 31 = (6) (30000) + (14) (22500) + (18) (15000) = 765000 32 = (6) (7500) + (14) (4500) + (18) (3000) = 162000 41 = (3) (30000) + (6) (22500) + (9) (15000) = 360000 42 = (3) (7500) + (6) (4500) + (9) (3000) = 76500

Hence, we have: W =

Salary Fringe Benefits ⎡ ⎤ Outlet 1 $367500 $78000 Outlet 2 ⎢ $48000 ⎥ ⎢ $225000 ⎥. ⎣ Outlet 3 $765000 $162000 ⎦ Outlet 4 $360000 $76500

(7) To compute the tonnage of a particular chemical applied to a given field, for each type of fertilizer we must multiply the percent concentration of the chemical in the fertilizer by the number of tons of that fertilizer applied to the given field. After this is done for each fertilizer type, we add these results to c 2016 Elsevier Ltd. All rights reserved.

Copyright °

17

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.5

get the total tonnage of the chemical that was applied. For example, to compute the number of tons of potash applied to field 2, we compute 005 |{z}

% Potash in Fert. 1

·

2 |{z}

Tons of Fert. 1

+

005 |{z}

% Potash in Fert. 2

·

1 |{z}

Tons of Fert. 2

+

020 |{z}

% Potash in Fert. 3

·

1 |{z}

Tons of Fert. 3

=

035 |{z}



Tons of Potash

Note that this is the dot product (3rd column of A)·(2nd column of B). Since this is the dot product of two columns, it is not in the form of an entry of a matrix product. To turn this into a form that looks like an entry of a matrix product, we consider the transpose of A , and express it as (3rd row of A )·(2nd column of B). Thus, this is the (3 2) entry of A B. A similar analysis shows that computing all of the entries of A B produces all of the information requested in the problem. Hence, to solve the problem, we compute ⎡ ⎤⎡ ⎤ 010 025 000 5 2 4 Y = A B = ⎣ 010 005 010 ⎦ ⎣ 2 1 1 ⎦  005 005 020 3 1 3

Solving for each entry produces

11 = (010) (5) + (025) (2) + (000) (3) = 100 12 = (010) (2) + (025) (1) + (000) (1) = 045 13 = (010) (4) + (025) (1) + (000) (3) = 065 21 = (010) (5) + (005) (2) + (010) (3) = 090 22 = (010) (2) + (005) (1) + (010) (1) = 035 23 = (010) (4) + (005) (1) + (010) (3) = 075 31 = (005) (5) + (005) (2) + (020) (3) = 095 32 = (005) (2) + (005) (1) + (020) (1) = 035 33 = (005) (4) + (005) (1) + (020) (3) = 085 Field 1 Field 2 Field 3 ⎤ Nitrogen 100 045 065 090 035 075 ⎦(in tons).

Hence, Y = Phosphate ⎣ Potash 095 035 085 ∙ ¸ ∙ ¸∙ ¸ 1 1 1 1 1 1 (9) (a) An example: A = . Note that A2 = 0 −1 0 −1 0 −1 ∙ ¸ ((1) (1) + (1) (0)) ((1) (1) + (1) (−1)) = = I2 . ((0) (1) + (−1) (0)) ((0) (1) + (−1) (−1)) ⎡ ⎤ ⎡ ⎤⎡ ⎤ 1 1 0 1 1 0 1 1 0 (b) One example: Suppose A = ⎣ 0 −1 0 ⎦. Let W = A2 = ⎣ 0 −1 0 ⎦ ⎣ 0 −1 0 ⎦.

0 0 1 0 0 1 0 0 1 Computing each entry yields ⎡

11 = (1) (1) + (1) (0) + (0) (0) = 1 12 = (1) (1) + (1) (−1) + (0) (0) = 0 13 = (1) (0) + (1) (0) + (0) (1) = 0 21 = (0) (1) + (−1) (0) + (0) (0) = 0 22 = (0) (1) + (−1) (−1) + (0) (0) = 1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

18

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.5

23 = (0) (0) + (−1) (0) + (0) (1) = 0 31 = (0) (1) + (0) (0) + (1) (0) = 0 32 = (0) (1) + (0) (−1) + (1) (0) = 0 33 = (0) (0) + (0) (0) + (1) (1) = 1 Hence, A2 = W = I3 . ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 0 1 0 0 1 0 0 1 (c) Consider A = ⎣ 1 0 0 ⎦. Let Y = A2 = ⎣ 1 0 0 ⎦ ⎣ 1 0 0 ⎦. First, we compute each 0 1 0 0 1 0 0 1 0 entry of Y.

11 = (0) (0) + (0) (1) + (1) (0) = 0 12 = (0) (0) + (0) (0) + (1) (1) = 1 13 = (0) (1) + (0) (0) + (1) (0) = 0 21 = (1) (0) + (0) (1) + (0) (0) = 0 22 = (1) (0) + (0) (0) + (0) (1) = 0 23 = (1) (1) + (0) (0) + (0) (0) = 1 31 = (0) (0) + (1) (1) + (0) (0) = 1 32 = (0) (0) + (1) (0) + (0) (1) = 0 33 = (0) (1) + (1) (0) + (0) (0) = 0 ⎡ ⎤ ⎡ ⎤⎡ ⎤ 0 1 0 0 1 0 0 0 1 And so Y = A2 = ⎣ 0 0 1 ⎦. Hence, A3 = A2 A = YA = ⎣ 0 0 1 ⎦ ⎣ 1 0 0 ⎦.

1 0 0 1 0 0 0 1 0 Letting this product equal Z, we compute each entry as follows: 11 = (0) (0) + (1) (1) + (0) (0) = 1 12 = (0) (0) + (1) (0) + (0) (1) = 0 13 = (0) (1) + (1) (0) + (0) (0) = 0 21 = (0) (0) + (0) (1) + (1) (0) = 0 22 = (0) (0) + (0) (0) + (1) (1) = 1 23 = (0) (1) + (0) (0) + (1) (0) = 0 31 = (1) (0) + (0) (1) + (0) (0) = 0 32 = (1) (0) + (0) (0) + (0) (1) = 0 33 = (1) (1) + (0) (0) + (0) (0) = 1 Therefore, A3 = Z = I3 .

(10) (a) This is the dot product of the 3rd row of A with the 4th column of B. Hence, the result is the 3rd row, 4th column entry of AB. (c) This is (2nd column of A)·(3rd row of B) = (3rd row of B)·(2nd column of A) = 3rd row, 2nd column entry of BA. P (11) (a) ((3 2) entry of AB) = (3rd row of A)·(2nd column of B) = =1 3 2 (12) (a) 3v1 − 2v2 + 5v3 = 3[4 7 −2] − 2[−3 −6 5] + 5[−9 2 −8] = [12 21 −6] + [6 12 −10] + [−45 10 −40] = [−27 43 −56] ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 4 7 −2 (b) 2w1 + 6w2 − 3w3 = 2 ⎣ −3 ⎦ + 6 ⎣ −6 ⎦ − 3 ⎣ 5 ⎦ −9 2 −8

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

19

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.5

⎡

⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 8 42 6 56 = ⎣ −6 ⎦ + ⎣ −36 ⎦ + ⎣ −15 ⎦ = ⎣ −57 ⎦ −18 12 24 18

(15) Several crucial steps in the following proofs rely on Theorem 1.5 for their validity. (a) Proof of Part (2): The ( ) entry of A(B + C) = (th row of A)·(th column of (B + C)) = (th row of A)·(th column of B + th column of C) = (th row of A)·(th column of B) + (th row of A)·(th column of C) = (( ) entry of AB) + (( ) entry of AC) = ( ) entry of (AB + AC). (b) Proof of Part (3): The ( ) entry of (A + B)C = (th row of (A + B))·(th column of C) = ((th row of A) + (th row of B)) · (th column of C) = (th row of A)·(th column of C) + (th row of B)·(th column of C) = (( ) entry of AC) + (( ) entry of BC) = ( ) entry of (AC + BC). (c) For the first equation in Part (4), the ( ) entry of (AB) = ((th row of A)·(th column of B)) = ((th row of A))·(th column of B) = (th row of A)·(th column of B) = ( ) entry of (A)B. Similarly, the ( ) entry of (AB) = ((th row of A)·(th column of B)) = (th row of A)·((th column of B)) = (th row of A)·(th column of B) = ( ) entry of A(B).

(20) (a) Proof of Part (1): We use induction on the variable . Base Step: A+0 = A = A I = A A0 .

Inductive Step: Assume A+ = A A for some  ≥ 0. We must prove A+(+1) = A A+1 . But A+(+1) = A(+)+1 = A+ A = (A A )A (by the inductive hypothesis) = A (A A) = A A+1 . (b) Proof of Part (2): Again, we use induction on the variable . Base Step: (A )0 = I = A0 = A0 .

Inductive Step: Assume (A ) = A for some integer  ≥ 0. We must prove (A )+1 = A(+1) .

But (A )+1 = (A ) A (by definition) = A A (by the inductive hypothesis) = A+ (by part (1)) = A(+1) . Finally, reversing the roles of  and  in the proof above shows that (A ) = A , which equals A . ∙ ¸∙ ¸ ∙ ¸ 1 0 1 0 1 0 (29) (a) Consider any matrix A of the form . Then A2 =  0  0  0 ∙ ¸ ∙ ¸ ((1) (1) + (0) ()) ((1) (0) + (0) (0)) 1 0 = = A. So, for example, is idempotent. (() (1) + (0) ()) (() (0) + (0) (0)) 1 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

20

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 1.5

⎤ ⎡ ⎤ ∙ ¸ 1 −2 1 −2 1 2 −1 1 2 −1 ⎣ 1 ⎦ . Then AB = 0 1 ⎦ (30) (b) Consider A = and B = ⎣ 0 2 4 −2 2 4 −2 1 0 1 0 ∙ ¸ ((1) (1) + (2) (0) + (−1) (1)) ((1) (−2) + (2) (1) + (−1) (0)) = = O2 . ((2) (1) + (4) (0) + (−2) (1)) ((2) (−2) + (4) (1) + (−2) (0)) ∙

¸

⎡

(31) A 2 × 2 matrix A that commutes with every other 2 × 2 matrix must have the form A = I2 . To prove ∙ ¸ 0 1 this, note that if A commutes with every other 2×2 matrix, then AB = BA, where B = . But 0 0 ¸∙ ¸ ∙ ¸ ¸ ∙ ∙ 0 1 ((11 ) (0) + (12 ) (0)) ((11 ) (1) + (12 ) (0)) 0 11 11 12 = and = AB = 0 0 21 22 0 21 ((21 ) (0) + (22 ) (0)) ((21 ) (1) + (22 ) (0)) ¸ ∙ ¸ ∙ ¸ ∙ ¸∙ ((0) (11 ) + (1) (21 )) ((0) (12 ) + (1) (22 )) 21 22 0 1 11 12 = = .

BA = 21 22 0 0 0 0 ((0) (11 ) + (0) (21 )) ((0) (12 ) + (0) (22 )) ¸ ∙  12 .

Setting these equal shows that 21 = 0 and 11 = 22 . Let  = 11 = 22 . Then A = 0  ¸∙ ¸ ∙ ¸ ∙ 0 0 0 0  12 Let D = . Then AD = 0  1 0 1 0 ¸ ∙ ¸ ∙ ¸ ∙ ¸∙ (() (0) + (12 ) (1)) (() (0) + (12 ) (0)) 12 0 0 0  12 = = and DA = 0  ((0) (0) + () (1)) ((0) (0) + () (0))  0 1 0 ¸ ¸ ∙ ∙ 0 0 ((0) () + (0) (0)) ((0) (12 ) + (0) ()) . = =  12 ((1) () + (0) (0)) ((1) (12 ) + (0) ()) Setting AD = DA shows that 12 = 0, and so A = I2 .

Finally, we must show that I2 actually does commute with every 2 × 2 matrix. If M is a 2 × 2 matrix, then (I2 )M = (I2 M) = M. Similarly, M(I2 ) = (MI2 ) = M. Hence I2 commutes with M. (32) (a) True. This is a “boxed” result given in the section. (b) True. D (A + B) = DA+DB (by part (2) of Theorem 1.16) = DB+DA (by part (1) of Theorem 1.12) (c) True. This is part (4) of Theorem 1.16. ∙ ¸ ∙ ¸ ∙ ¸∙ ¸ 1 1 1 −2 1 1 1 −2 (d) False. Let D = , and let E = . Now, DE = 0 −1 0 1 0 −1 0 1 ∙ ¸ ∙ ¸ ((1) (1) + (1) (0)) ((1) (−2) + (1) (1)) 1 −1 = = . Hence, ((0) (1) + (−1) (0)) ((0) (−2) + (−1) (1)) 0 −1 ∙ ¸∙ ¸ ∙ ¸ 1 −1 1 −1 ((1) (1) + (−1) (0)) ((1) (−1) + (−1) (−1)) 2 = = I2 .

(DE) = 0 −1 0 −1 ((0) (1) + (−1) (0)) ((0) (−1) + (−1) (−1)) ∙ ¸∙ ¸ ∙ ¸ 1 1 1 1 ((1) (1) + (1) (0)) ((1) (1) + (1) (−1)) = = I2 . Also, But, D2 = 0 −1 0 −1 ((0) (1) + (−1) (0)) ((0) (1) + (−1) (−1)) ∙ ¸∙ ¸ ∙ ¸ ∙ ¸ 1 −2 1 −2 ((1) (1) + (−2) (0)) ((1) (−2) + (−2) (1)) 1 −4 = = .

E2 = 0 1 0 1 ((0) (1) + (1) (0)) ((0) (−2) + (1) (1)) 0 1

Hence, D2 E2 = I2 E2 = E2 6= I2 . Thus, (DE)2 6= D2 E2 . The problem here is that D and E do

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

21

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chapter 1 Review

not commute. For, if they did, we would have (DE)2 = (DE) (DE) = D(ED)E = D(DE)E = (DD)(EE) = D2 E2 . (e) False. See the answer for part (i) of Exercise 4 for a counterexample. We know from Theorem 1.18 that (DE) = E D . So if D and E do not commute, then (DE) 6= D E . (f) False. See the answer for part (b) of Exercise 30 for a counterexample. DE = O whenever every row of D is orthogonal to every column of E, but neither matrix is forced to be zero for DE = O to be true. ∙ ¸ ∙ ¸ ∙ ¸ 1 0 2 −1 1 −1  and AA =  while (g) False. Consider A =  Then A = −1 1 −1 1 0 1 ∙ ¸ 1 −1 A A =  Thus, AA 6= A A in this case. −1 2

Chapter 1 Review Exercises q¡ ¢ 1 2

q √ ¡ ¢2 ¡ ¢2 q 1 394 9 9 394 + − 35 + 34 = 16 + 25 + 16 = 16·25 = 20 ≈ 099247. h i x 5 Thus, u = kxk = √394  − √12  √15 ≈ [02481 −05955 07444]. This is slightly longer than x 394 394 because we have divided x by a scalar with absolute value less than 1, which amounts to multiplying x by a scalar having absolute value greater than 1.

(2) kxk =

4

1 f . For the force (4) We will use Newton’s Second Law which states that¶ f = a, or equivalently, a =  µ ³ ³ ³ ´´ ´ [617−6] √ f1 , a1 = 17 133 k[617−6]k = 19 [617−6] = 19 √ 2[617−6] = [6 17 −6]. Similarly, for the 361 6 +172 +(−6)2 ¶ µ ³ ³ ³ ´´ ´ [−8−48] √ = 24 [−8−48] force f2 , a2 = 17 168 k[−8−48]k = 24 √ [−8−48] = [−16 −8 16]. Hence, 2 2 2 144 (−8) +(−4) +8

a = a1 + a2 = [6 17 −6] + [−16 −8 16] = [−10 9 10], in m sec2 . ³ ´ ³ ´ ¡ −69 ¢ x·y −69 ≈ cos−1 (−07237) ≈ 136◦ . ≈ cos−1 (1005)(949) ≈ cos−1 9534 (6) The angle  = cos−1 kxkkyk ¶¶ µ µ ¶¶ µ µ ³ ³ ´´ ³ ³ ´´ [15−8] [−724] [15−8] [−724] √ √ (8) Work = f · d = 34 k[15−8]k · 75 = · 75 k[−724]k = 34 152 +(−8)2 (−7)2 +242 ³ ³ ´´ ³ ³ ´´ 34 [15−8] · 75 [−724] = (2 [15 −8])·(3 [−7 24]) = [30 −16]·[−21 72] = (30)(−21)+(−16)(72) 17 25 = −1782 joules.

(10) First, x 6= 0, or else projx y is not defined. Also, y 6= 0, since that would imply³ proj´x y = ³y. Now, ´ x·y x·x assume x k y. Then, there is a scalar  6= 0 such that y = x. Hence, projx y = kxk x = kxk x 2 2 ³ ´ 2 = kxk x = x = y, contradicting the assumption that y 6= projx y. kxk2 ⎤ ∙ ¸ ∙ ¸ 3 5 5 −2 −1 5 −2 −1 3 −2 −4 − 4 ⎣ −2 4 ⎦ = 3 −4 (11) (a) 3A − 4C = 3 3 −1 4 3 −1 4 5 4 3 −4 ∙ ¸ ∙ ¸ 3∙ ¸ 15 −6 −3 12 −8 −16 3 2 13 = − = ; 9 −3 12 20 16 12 −11 −19 0 ⎡ ⎤ ∙ ¸ 2 −3 −1 5 −2 −1 ⎣ −4 5 −2 ⎦ AB = 3 −1 4 3 −4 3 ∙

¸

⎡

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

22

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

=

=

∙

∙

Chapter 1 Review

(5)(2) + (−2)(−4) + (−1)(3) (5)(−3) + (−2)(5) + (−1)(−4) (5)(−1) + (−2)(−2) + (−1)(3) (3)(2) + (−1)(−4) + (4)(3) (3)(−3) + (−1)(5) + (4)(−4) (3)(−1) + (−1)(−2) + (4)(3) 15 −21 −4 22 −30 11

¸

;

BA is not defined because B has 3 columns, which does not match the number of rows of A, which is 2; ⎡ ⎤ ∙ ¸ 3 5 5 −2 −1 ⎣ −2 4 ⎦ AC = 3 −1 4 −4 3 ∙

(5)(3) + (−2)(−2) + (−1)(−4) (5)(5) + (−2)(4) + (−1)(3) = (3)(5) + (−1)(4) + (4)(3) (3)(3) + (−1)(−2) + (4)(−4) ⎡ ⎤ ¸ 3 5 ∙ 5 −2 −1 CA = ⎣ −2 4 ⎦ 3 −1 4 −4 3

¸

=

∙

23 14 −5 23

⎡

¸ ;

⎤ ⎡ ⎤ (3)(5) + (5)(3) (3)(−2) + (5)(−1) (3)(−1) + (5)(4) 30 −11 17 2 0 18 ⎦; = ⎣ (−2)(5) + (4)(3) (−2)(−2) + (4)(−1) (−2)(−1) + (4)(4) ⎦ = ⎣ (−4)(5) + (3)(3) (−4)(−2) + (3)(−1) (−4)(−1) + (3)(4) −11 5 16

A3 is not defined. (Powers of matrices are only defined for square matrices.); ⎛⎡

⎤⎡ ⎤⎞ 2 −3 −1 2 −3 −1 5 −2 ⎦ ⎣ −4 5 −2 ⎦⎠ . Thus, B3 = (B2 )(B). Now B2 = ⎝⎣ −4 3 −4 3 3 −4 3 (1 1) entry of B2 = (2)(2) + (−3)(−4) + (−1)(3) = 13; (1 2) entry of B2 = (2)(−3) + (−3)(5) + (−1)(−4) = −17; (1 3) entry of B2 = (2)(−1) + (−3)(−2) + (−1)(3) = 1; (2 1) entry of B2 = (−4)(2) + (5)(−4) + (−2)(3) = −34; (2 2) entry of B2 = (−4)(−3) + (5)(5) + (−2)(−4) = 45; (2 3) entry of B2 = (−4)(−1) + (5)(−2) + (−2)(3) = −12; (3 1) entry of B2 = (3)(2) + (−4)(−4) + (3)(3) = 31; (3 2) entry of B2 = (3)(−3) + (−4)(5) + (3)(−4) = −41; (3 3) entry of B2 = (3)(−1) + (−4)(−2) + (3)(3) = 14. ⎡

⎤ 13 −17 1 45 −12 ⎦ . Hence, Thus, B2 = ⎣ −34 31 −41 14 ⎡

⎤⎡ ⎤ 13 −17 1 2 −3 −1 45 −12 ⎦ ⎣ −4 5 −2 ⎦ . Thus, B3 = B2 (B) = ⎣ −34 31 −41 14 3 −4 3

(1 1) entry of B3 = (13)(2) + (−17)(−4) + (1)(3) = 97; (1 2) entry of B3 = (13)(−3) + (−17)(5) + (1)(−4) = −128;

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

23

¸

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chapter 1 Review

(1 3) entry of B3 = (13)(−1) + (−17)(−2) + (1)(3) = 24; (2 1) entry of B3 = (−34)(2) + (45)(−4) + (−12)(3) = −284; (2 2) entry of B3 = (−34)(−3) + (45)(5) + (−12)(−4) = 375; (2 3) entry of B3 = (−34)(−1) + (45)(−2) + (−12)(3) = −92; (3 1) entry of B3 = (31)(2) + (−41)(−4) + (14)(3) = 268; (3 2) entry of B3 = (31)(−3) + (−41)(5) + (14)(−4) = −354; (3 3) entry of B3 = (31)(−1) + (−41)(−2) + (14)(3) = 93. ⎡

⎤ 97 −128 24 375 −92 ⎦ .

Hence, B3 = ⎣ −284 268 −354 93 (b) Third row of BC = (3rd row of B)C =

£

3 −4 3

¤

⎡

⎤ 3 5 ⎣ −2 4 ⎦ −4 3

= [(3)(3) + (−4)(−2) + (3)(−4) (3)(5) + (−4)(4) + (3)(3)] = [5 8]. ¢ ¡ (13) Now, (3(A − B) ) = 3 (A − B) (by part (3) of Theorem 1.13) = 3(A − B) (by part (1) of Theorem 1.13). Also, −(3(A − B) ) = 3(−1)(A − B ) (parts (2) and (3) of Theorem 1.13) = 3(−1)((−A)−(−B)) (definition of skew-symmetric) = 3(A−B). Hence, (3(A−B) ) = −(3(A−B) ), and so 3(A − B) is skew-symmetric. ⎡ ⎤ ∙ ¸ 5200 4300 20 15 (15) Let A = , the matrix of costs vs. commodities (steel and iron), and B = ⎣ 6300 5100 ⎦, 3 2 4600 4200 the matrix of companies vs. commodities. In order for the units to line up properly for matrix multiplication, the “commodities” label, the shared unit, must represent the columns of the first matrix in the product as well as the rows of the second matrix. One way to do this is to compute the product BA . Now, ⎡ ⎤ ⎤ ¸ (5200)(20) + (4300)(15) (5200)(3) + (4300)(2) 5200 4300 ∙ 20 3 = ⎣ (6300)(20) + (5100)(15) (6300)(3) + (5100)(2) ⎦ BA = ⎣ 6300 5100 ⎦ 15 2 (4600)(20) + (4200)(15) (4600)(3) + (4200)(2) 4600 4200 ⎡

Price Shipping Cost ⎡ ⎤ Company I $168500 $24200 $29100 ⎦ . = Company II ⎣ $202500 Company III $155000 $22200

¢ ¢ ¡ ¡ (16) Take the transpose of both sides of A B = B A to get A B = B A , which by Theorem 1.18 and part (1) of Theorem 1.13 implies that BA = AB. Then, (AB)2 = (AB)(AB) = A(BA)B = A(AB)B = A2 B2 .

(18) If A 6= O22 , then some row of A, say the th row, is nonzero. Let x = (th row of A). Then,∙ by ¸ Result 1 5 in Section 1.3, either x · [1 0] 6= 0 or x · [0 1] 6= 0. But x · [1 0] is the ( 1) entry of A , and 0 ∙ ¸ ∙ ¸ 0 1 x · [0 1] is the ( 1) entry of A . Thus, either at least one of the entries of A is nonzero or 1 0 ∙ ¸ 0 at least one of the entries of A is nonzero.

1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

24

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chapter 1 Review

(19) (a) Let A and B be  ×  matrices having the properties given in the exercise. Let C = AB. Then we know that  = 0 for all   ,  = 0 for all   ,  = 0 for all  6= , and that  6= 0 and  6= 0 for all . We need to prove that  = 0 for all   .PWe use induction on . P   Base Step P ( = 1): Let    = 1. Then 0 = 1 = =1  1 = 1 11 + =2  1  = 1 11 + =2  · 0 = 1 11 . Since 11 6= 0, this implies that 1 = 0, completing the Base Step.

Inductive Step: Assume for  = 1 2      − 1 that  = 0 for all    That is, assume we have already proved that, for some  ≥ 2, that the first  − 1 columns of A have all zeros below the main diagonal. To complete the inductive step, we need to prove that the th column of A has all zeros below the main P diagonal. That P−1is, we must prove thatP = 0 for all   +   + =+1     . Let   . Then, 0 =  = =1   = =1 P−1 P = =1 0 ·  +   + =+1  · 0 =   . But, since  6= 0, we must have  = 0, completing the proof.

(20) (a) False. If A is a nonzero matrix, there must exist some  and  such that  6= 0. For this  and , the ( ) entry of A, which equals  , must be nonzero.

1 x. (b) True. A nonzero vector x is parallel to the unit vector u = kxk

(c) False. For example, [1 4 3] − [2 5 4] = [−1 −1 −1] is a linear combination of [1 4 3] and [2 5 4] having negative entries.

(d) False. By definition, an angle between vectors has a measure from 0 to . But 3 2  . The measure of the angle between [1 0] and [0 −1] is 2 .

(e) False. projx y could have the opposite direction from x. For example, if x = [1 0] and y = [−1 1], then projx y = [−1 0]. (f) True. Apply the Triangle Inequality twice: kx + y + zk = kx + (y + z)k + zk ≤ °P≤ kxk° + ky °  ° P kxk+kyk+kzk. Note that a proof by induction can be used to prove that ° =1 x ° ≤ =1 kx k (see Exercise 16 in Section 1.3).

(g) False. Recall that a set {v1      v } of vectors is mutually orthogonal if and only if for every pair of distinct vectors v , v in the set, v · v = 0. When negating statements involving a universal quantifier, such as this, the universal quantifier must change to an existential quantifier. The correct negation is: There is a pair of distinct vectors v , v in {v1      v } such that v · v 6= 0.

(h) False. Disproving a statement involving a universal quantifier typically involves finding a counterexample. To disprove a statement involving an existential quantifier, we must prove its negation, which is a universally quantified statement. Thus, the negation must be proven in all cases.

For example, to disprove the statement “There is a 2 × 2 matrix A such that AO22 6= O22 ” we must show that all 2 × 2 matrices A have the property AO22 = O22  It is not enough to provide just one specific matrix A for which AO22 6= O22 . ∙ ¸ 1 1 As a further illustration, consider the following statement: “Let A = . Then there is 1 1 a nonzero 2×2 ∙ matrix B¸such that AB = O22 .” This statement is actually true, since the nonzero 1 1 matrix B = does satisfy AB = O22  Now it is possible to find other matrices, such −1 −1 as B = I2  for which AB 6= O22 . However, the existence of such possibilities for B does not prove the original existential statement false. The moral here is that finding a “counterexample” to an existential statement is not enough to ensure that the statement is false. ∙ ¸ ∙ ¸ ∙ ¸ 1 1 2 0 3 1 (i) False. For a counterexample, note that + = (not symmetric).

0 1 2 2 2 3 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

25

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chapter 1 Review

(j) True. Let A be a skew-symmetric matrix. Then  = − , and so it must be true that  = − .

Hence, all of the main diagonal entries of A are zero. Since the trace of a matrix is the sum of its main diagonal entries, trace(A) = 0. (k) True. If A ∈ U ∩ L , then A is both upper and lower triangular. Because A is upper triangular,  = 0 for all ̇  . Because A is lower triangular,  = 0 for all   . Hence, all of the entries of A off the main diagonal equal zero. Therefore, A ∈ D . This proves that U ∩ L ⊆ D . It is also clear that every diagonal matrix is both upper and lower triangular, since all of its entries both below and above the main diagonal equal zero. Hence, D ⊆ U ∩ L . Thus, because we have subset inclusion in both directions, U ∩ L = D . (l) True. This is easily proven using parts (2) and (3) of Theorem 1.13 and an induction argument.

(m) True. First, because A is × and B is ×1, AB is an ×1 matrix. Also, if a = [1       ] represents the th column of A, then AB = 11 a1 + · · · + 1 a , which shows that AB is a linear combination of the columns of A. (To verify this, note that, by definition, the ( 1) entry of AB equals 1 11 + · · · +  1 , while the ( 1) entry of 11 a1 + · · · + 1 a = 11 1 + · · · + 1  , and so AB = 11 a1 + · · · + 1 a .)

(n) False. In fact, using an argument similar to the one used in part∙(m), AD ¸ ∙is the matrix ¸ ∙ whose th ¸ 1 2 3 0 3 10 column is the th column of A multiplied by  . For example, = .

3 4 0 5 9 20

(o) False. If A is  ×  and B is  × , then both AB and BA are defined, with AB being  ×  and BA being  × . If  6= , then, clearly, AB 6= BA. For a specific counterexample, choose A = O23 and B = O32 .

23 (p) False. (A + B)2 = A2 + 2AB + B2 if∙and only ¸ if A and ∙B commute ¸ (see part (b) of Exercise ∙ ¸ 1 1 1 2 7 6 2 in Section 1.5). In particular, if A = and B = , then (A + B) = , 1 1 0 −1 2 3 ∙ ¸ 5 4 but A2 + 2AB + B2 = .

4 5 ∙ ¸∙ ¸ ∙ ¸ 0 2 0 −3 6 0 (q) False. For a counterexample, note that = . −2 0 3 0 0 6 (r) True. This follows immediately from part (2) of Theorem 1.17.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

26

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

Chapter 2 Section 2.1 (1) In each part, we first set up the augmented matrix corresponding to the given system of linear equations.

Then we perform the row operations designated by Gaussian Elimination. Finally, we use the final augmented matrix obtained to give the solution set for the original system. (a) The given augmented matrix is ¯ ⎤ −5 −2 2 ¯¯ 16 ⎣ 3 1 −1 ¯¯ −9 ⎦  2 2 −1 ¯ −4 ⎡

The first pivot is the (1 1) entry. We make this a “1” using the following Type (I) row operation: Row Operation

Resulting Matrix ¯ ⎤ 1 25 − 25 ¯ − 16 ° 5 ¯ ⎣ 3 1 −1 ¯ −9 ⎦  ¯ 2 2 −1 ¯ −4 ⎡

(I): h1i ← 15 h1i

We now target the (2 1) entry, and make it “0.” We do this using the Type (II) row operation: h2i ← (−3) × h1i + h2i  To perform this operation, we first do the following side calculation: (−3)× (row 1) (row 2)

−3

− 65

0

− 15

3

(sum)

1

Row Operation

6 5

48 5

−1

−9

1 5

3 5

Matrix ¯ ⎡ Resulting ⎤ 2 1 ° − 25 ¯¯ − 16 5 5 ⎢ 3 ⎥  1 ¯ ⎣ 0 − 15 5 ¯¯ 5 ⎦ ¯ 2 2 −1 −4

(II): h2i ← (−3) × h1i + h2i

Next, we target the (3 1) entry using the Type (II) row operation: h3i ← (−2) × h1i + h3i  We first do the following side calculation: (−2)× (row 1) (row 3)

−2

− 45

0

6 5

2

(sum) Row Operation

32 5

−1

−4

− 15

12 5

Matrix ¯ ⎡ Resulting 2 2 ¯ 16 ⎤ 1 ° − 5 5 ¯ − 5 ⎢ 3 ⎥  1 ¯ ⎣ 0 − 15 5 ¯¯ 5 ⎦ 12 6 1 ¯ 0 −5 5 5

(II): h3i ← (−2) × h1i + h3i

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

2

4 5

27

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

Now we move the pivot to the (2 2) entry. We make this entry a “1” using the following Type (I) row operation: Row Operation

Matrix ¯ ⎡Resulting ⎤ 1 25 − 25 ¯¯ − 16 5 ⎢ ⎥  ¯ 1 −1 ¯ −3 ⎦ ⎣ 0 ° ¯ 12 0 65 − 15 ¯ 5

(I): h2i ← −5 h2i

Now we target the (3 2) entry using the Type (II) row operation: h3i ← (− 65 ) × h2i + h3i: (− 65 )× (row 2)

0

(row 3)

0

(sum)

0

− 65 6 5

0

Row Operation

6 5 − 15

1

18 5 12 5

6

Matrix ¯ ⎡ Resulting ⎤ 1 25 − 25 ¯¯ − 16 5 ⎢ ⎥  ¯ 1 −1 ¯ −3 ⎦ ⎣ 0 ° ¯ 0 0 1 ¯ 6

(II): h3i ← (− 65 ) × h2i + h3i

The pivot moves to the (3 3) entry. However, the (3 3) entry already equals 1. Since there are no rows below the third row, we are finished performing row operations. The final augmented matrix corresponds to the following linear system: ⎧ 2 2 16 ⎨ 1 + 5 2 − 5 3 = − 5 2 − 3 = −3 ⎩ 6 3 =

The last equation clearly gives 3 = 6. Substituting this value into the second equation yields 2 −6 = −3, or 2 = 3. Substituting both values into the first equation produces 1 + 25 (3)− 25 (6) = − 16 5 , which leads to 1 = −2. Hence, the complete solution set for the system is {(−2 3 6)}. The system is consistent. (c) The given augmented matrix is ¯ ⎡ ⎤ 3 −2 4 ¯¯ −54 ⎣ −1 1 −2 ¯¯ 20 ⎦  ¯ 5 −4 8 −83 The first pivot is the (1 1) entry. We make this a “1” by performing the following Type (I) row operation: Row Operation Resulting Matrix ⎡ ⎤ 4 ¯¯ 1 − 23 ° 3 ¯ −18  ⎣ −1 (I): h1i ← 13 h1i 1 −2 ¯¯ 20 ⎦ 5 −4 8 ¯ −83

We now target the (2 1) entry, and make it “0.” We do this using the Type (II) row operation: h2i ← (1) × h1i + h2i  To perform this operation, we first do the following side calculation: (1)× (row 1)

1

(row 2)

−1

(sum) c 2016 Elsevier Ltd. All rights reserved.

Copyright °

0

28

− 23

1

1 3

4 3

−2

− 23

−18

20 2

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Row Operation

Section 2.1

Matrix ¯ ⎡ Resulting ⎤ 4 ¯ 1 − 23 ° 3 ¯ −18 ⎢ ¯ ⎥  1 − 23 ¯ 2 ⎦ ⎣ 0 3 ¯ 5 −4 8 ¯ −83

(II): h2i ← (1) × h1i + h2i

Next, we target the (3 1) entry using the Type (II) row operation: h3i ← (−5) × h1i + h3i  We first do the following side calculation: (−5)× (row 1) (row 3)

−5

5

(sum)

0

10 3

− 20 3

−4

8

− 23

Row Operation

4 3

90 −83

7

Matrix ¯ ⎡ Resulting ⎤ 4 ¯ 1 − 23 ° 3 ¯ −18 ⎢ ¯ ⎥  1 − 23 ¯ 2 ⎦ ⎣ 0 3 ¯ 4 ¯ 7 0 − 23 3

(II): h3i ← (−5) × h1i + h3i

Now we move the pivot to the (2 2) entry. We make this entry a “1” using the following Type (I) row operation: Row Operation

Matrix ¯ ⎡ Resulting ⎤ 4 ¯ 1 − 23 3 ¯ −18 ¯ ⎢ ⎥  1 −2 ¯ 6 ⎦ ⎣ 0 ° ¯ 4 ¯ 0 − 23 7 3

(I): h2i ← 3 h2i

Now we target the (3 2) entry using the Type (II) row operation: h3i ← ( 23 ) × h2i + h3i  We first do the following side calculation: ( 23 )× (row 2)

0

(row 3)

0

(sum)

0

Row Operation

2 3 − 23

0

− 43

4

4 3

7

0

11

Resulting Matrix ⎤ 4 ¯¯ 1 − 23 3 ¯ −18  ⎣ 0 ° 1 −2 ¯¯ 6 ⎦ 0 0 0 ¯ 11 ⎡

(II): h3i ← ( 23 ) × h2i + h3i

The pivot moves to the (3 3) entry. However, the (3 3) entry equals 0. Because there are no rows below the third row, we can not switch with a row below the third row to produce a nonzero pivot.

Also, since there are no more columns before the augmentation bar, we are finished performing row operations. The final augmented matrix corresponds to the following linear system: ⎧ 2 4 ⎨ 1 − 3 2 + 3 3 = −18 6 2 − 23 = ⎩ 0 = 11

The last equation states that 0 = 11. Since this equation can not be satisfied, the original linear system has no solutions. The solution set is empty. The system is inconsistent. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

29

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

(e) The given augmented matrix is ⎡

¯ ⎤ 6 −12 −5 16 −2 ¯¯ −53 ⎣ −3 6 3 −9 1 ¯¯ 29 ⎦  ¯ −4 8 3 −10 1 33

The first pivot is the (1 1) entry. We make this a “1” by performing the following Type (I) row operation: Row Operation

Resulting Matrix ¯ ⎤ 8 1 −2 − 56 ° − 13 ¯ − 53 3 6 ¯  ⎣ −3 6 3 −9 1 ¯¯ 29 ⎦ −4 8 3 −10 1 ¯ 33 ⎡

(I): h1i ← 16 h1i

We now target the (2 1) entry, and make it “0.” We do this using the Type (II) row operation: h2i ← (3) × h1i + h2i  We first perform the following side calculation: (3) × (row 1) (row 2)

(sum)

3 −3

−6

− 52

0

1 2

6

0

Row Operation

3

8 −9

−1

−1

− 53 2

0

5 2

1

29

Resulting Matrix ¯ ⎤ 8 1 −2 − 56 ° − 13 ¯¯ − 53 3 6 ⎢ ¯ 5 ⎥  1 0 −1 0 ¯ ⎣ 0 2 2 ⎦ ¯ 33 −4 8 3 −10 1 ¯ ⎡

(II): h2i ← (3) × h1i + h2i

Next, we target the (3 1) entry. We do this using the Type (II) row operation: h3i ← (4)×h1i+h3i  We first perform the following side calculation: (4) × (row 1) (row 3)

(sum)

4 −4

0

−8

− 10 3

0

− 13

8

Row Operation

3

32 3

−10 2 3

− 43

− 106 3

− 13

− 73

1

33

Resulting Matrix¯ ⎤ 8 1 −2 − 56 ° − 13 ¯¯ − 53 3 6 ⎢ ¯ 1 5 ⎥  0 −1 0 ¯ ⎣ 0 2 2 ⎦ ¯ 1 1 ¯ 2 −3 − 73 0 0 −3 3 ⎡

(II): h3i ← (4) × h1i + h3i

We attempt to move the pivot to the (2 2) entry; however, there is a “0” there. Also, the entry below this is also 0, so we can not switch rows to make the (2 2) entry nonzero. Thus, we move the pivot horizontally to the next column — to the (2 3) entry. We make this pivot a “1” using the following Type (I) row operation: Row Operation (I): h2i ← 2 h2i

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

Resulting Matrix¯ ⎤ 8 − 13 ¯¯ − 53 1 −2 − 56 3 6 ⎥  ⎢ ¯ 1 −2 0 ° 0 ¯ 5 ⎦ ⎣ 0 ¯ 2 − 13 ¯ − 73 0 0 − 13 3 ⎡

30

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

Now we target the (3 3) entry using the Type (II) row operation: h3i ← ( 13 ) × h2i + h3i  We first perform the following side calculation: ( 13 ) × (row 2)

0

0

(row 3)

0

0

(sum)

0

0

Row Operation

1 3 − 13

− 23

0

2 3

0

0 − 13

− 13

5 3 − 73 − 23

Resulting Matrix¯ ⎤ 8 1 −2 − 56 − 13 ¯¯ − 53 3 6 ⎥  ⎢ ¯ 1 −2 0 ° 0 ¯ 5 ⎦ ⎣ 0 ¯ 0 0 0 0 − 13 ¯ − 23 ⎡

(II): h3i ← ( 13 ) × h2i + h3i

We attempt to move the pivot diagonally to the (3 4) entry; however, that entry is also 0. Since there are no rows below with which to swap, we must move the pivot horizontally to the (3 5) entry. We make this pivot a “1” by performing the following Type (I) row operation: Row Operation

Resulting Matrix¯ ⎤ 8 1 −2 − 56 − 13 ¯¯ − 53 3 6 ⎢ ⎥  ¯ 0 1 −2 0 ¯ 5 ⎦ ⎣ 0 ¯ 1 ¯ 0 0 0 0 ° 2 ⎡

(I): h3i ← −3 h3i

At this point, we are finished performing row operations. The linear system corresponding to the final augmented matrix is: ⎧ ⎨ 1 − 22 − 56 3 + 83 4 − 13 5 = − 53 6 3 − 24 = 5 ⎩ 2 5 =

The last equation clearly gives us 5 = 2. The variable 4 corresponds to a nonpivot column, and hence is an independent variable. Let 4 = . Then the second equation yields 3 − 2 = 5, or 3 = 2 + 5. The variable 2 also corresponds to a nonpivot column, making it an independent variable as well. We let 2 = . Finally, plugging the all these values into the first equation produces 1 − 2 − 56 (2 + 5) + 83  − 13 (2) = − 53 6 . Solving this equation for 1 gives 1 = 2 −  − 4.

Hence, the complete solution set for the original system is

{(2 −  − 4  2 + 5  2) |   ∈ R} The system is consistent.

To find three particular solutions, we plug in three different sets of values for  and . Using  = 0,  = 0 yields the solution (−4 0 5 0 2). Letting  = 1,  = 0 gives us (−2 1 5 0 2).

Finally, setting  = 0,  = 1 produces (−5 0 7 1 2). (g) The given augmented matrix is ¯ ⎤ 4 −2 −7 ¯¯ 5 ⎢ −6 5 10 ¯¯ −11 ⎥ ⎢ ⎥ ⎣ −2 3 4 ¯¯ −3 ⎦ −3 2 5 ¯ −5 ⎡

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

31

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

The first pivot is the (1 1) entry. We make this a “1” by performing the following Type (I) row operation: Row Operation Resulting Matrix ¯ ⎡ 5 ⎤ 1 − 12 − 74 ¯ ° 4 ¯ ⎢ −6 ¯ −11 ⎥  5 10 1 ⎢ ⎥ ¯ (I): h1i ← 4 h1i ⎣ −2 3 4 ¯¯ −3 ⎦ −3 2 5 ¯ −5

We now target the (2 1) entry, and make it “0.” We do this using the Type (II) row operation: h2i ← (6) × h1i + h2i  We first perform the following side calculation: (6)× (row 1)

6

(row 2)

−6

(sum)

0

−3

− 21 2

2

− 12

5

Row Operation

10

15 2

−11 − 72

Matrix ¯ ⎡ Resulting 5 ⎤ 1 − 12 − 74 ¯¯ ° 4 ⎢ ⎥ ¯ 2 − 12 ¯ − 72 ⎥  ⎢ 0 ⎢ ⎥ ¯ ⎣ −2 3 4 ¯¯ −3 ⎦ −3 2 5 ¯ −5

(II): h2i ← (6) × h1i + h2i

Next, we target the (3 1) entry. We do this using the Type (II) row operation: h3i ← (2)×h1i+h3i  We first perform the following side calculation: (2)× (row 1)

2

(row 3)

−2

(sum)

0

−1

− 72

2

1 2

3

Row Operation

4

5 2

−3

− 12

Matrix ¯ ⎡ Resulting ⎤ 5 1 − 12 − 74 ¯¯ ° 4 ⎢ ⎥ ¯ ⎢ 0 2 − 12 ¯ − 72 ⎥  ⎢ ⎥ ¯ ⎢ 0 1 ¯ 1 ⎥ 2 − ¯ ⎣ 2 2 ⎦ ¯ ¯ −3 2 5 −5

(II): h3i ← (2) × h1i + h3i

To finish the first column, we now target the (4 1) entry using the Type (II) row operation: h4i ← (3) × h1i + h4i  We first perform the following side calculation: (3)× (row 1)

3

(row 4)

−3

(sum)

0

Row Operation

(II): h4i ← (3) × h1i + h4i

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

32

− 32

− 21 4

1 2

− 14

2

5

15 4

−5

− 54

Matrix ¯ ⎡ Resulting ⎤ 5 1 − 12 − 74 ¯¯ ° 4 ⎢ ⎥ ¯ ⎢ 0 2 − 12 ¯ − 72 ⎥  ⎢ ⎥ ¯ ⎢ 0 1 ¯ 1 ⎥ 2 ⎣ 2 ¯ −2 ⎦ ¯ 1 0 − 14 ¯ − 54 2

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

Now we move the pivot to the (2 2) entry. We make this entry a “1” using the following Type (I) row operation: Row Operation

Matrix ¯ ⎡ Resulting ⎤ 5 1 − 12 − 74 ¯¯ 4 ⎢ ⎥ ¯ ⎢ 0 ° 1 − 14 ¯ − 74 ⎥  ⎢ ⎥ ¯ ⎢ 0 1 ¯ 1 ⎥ 2 − ¯ ⎣ 2 2 ⎦ ¯ 1 1 ¯ 0 −4 − 54 2

(I): h2i ← 12 h2i

Now we target the (3 2) entry using the Type (II) row operation: h3i ← (−2) × h2i + h3i  We first perform the following side calculation: (−2)× (row 2)

0

(row 3)

0

(sum)

0

−2 2

1 2 1 2

0

1

Row Operation

7 2 − 12

3

Matrix ¯ ⎡ Resulting ⎤ 5 1 − 12 − 74 ¯¯ 4 ⎢ ⎥ ¯ ⎢ 0 ° 1 − 14 ¯ − 74 ⎥  ⎢ ⎥ ¯ ⎢ 0 0 1 ¯¯ 3 ⎥ ⎣ ⎦ ¯ 1 1 ¯ 5 − 0 − 2 4 4

(II): h3i ← (−2) × h2i + h3i

To finish the second column, we target the (4 2) entry using the Type (II) row operation: h4i ← (− 12 ) × h2i + h4i  We first perform the following side calculation: (− 12 )× (row 2)

0

(row 4)

0

(sum)

0

− 12 1 2

0

Row Operation

1 8 − 14 − 18

7 8 − 54 − 38

Matrix ¯ ⎡ Resulting ⎤ 5 1 − 12 − 74 ¯¯ 4 ⎢ ⎥ ¯ ⎢ 0 ° 1 − 14 ¯ − 74 ⎥  ⎢ ⎥ ¯ ⎢ 0 0 1 ¯¯ 3 ⎥ ⎣ ⎦ ¯ 0 0 − 18 ¯ − 38

(II): h4i ← (− 12 ) × h2i + h4i

The pivot now moves to the (3 3) entry. Luckily, the pivot already equals 1. Therefore, we next target the (4 3) entry using the Type (II) row operation: h4i ← ( 18 ) × h3i + h4i  We first perform the following side calculation: ( 18 )× (row 3)

0

0

(row 4)

0

0

(sum)

0

0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

33

1 8 − 18

0

3 8 − 38

0

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Row Operation

Section 2.1

Matrix ¯ ⎡ Resulting ⎤ 5 1 − 12 − 74 ¯¯ 4 ⎢ ⎥ ¯ ⎢ 0 1 − 14 ¯ − 74 ⎥  ⎢ ⎥ ¯ ⎢ 0 ⎥ ¯ 1 0 ° 3 ¯ ⎣ ⎦ ¯ 0 0 0 ¯ 0

(II): h4i ← ( 18 ) × h3i + h4i

This finishes the third column. Since there are no more columns before the augmentation bar, we are finished performing row operations. The linear system corresponding to the final matrix is: ⎧ 5 1 − 12 2 − 74 3 = ⎪ 4 ⎪ ⎪ ⎨ 2 − 14 3 = − 74 ⎪ ⎪ 3 = 3 ⎪ ⎩ 0 = 0

The last equation, 0 = 0, provides no information regarding the solution set for the system because it is true for every value of 1 , 2 , and 3 . Thus, we can ignore it. The third equation clearly gives 3 = 3. Plugging this value into the second equation yields 2 − 14 (3) = − 74 , or, 2 = −1. Substituting the values we have for 2 and 3 into the first equation produces 1 − 12 (−1) − 74 (3) = 54 , or, 1 = 6. Hence, the original linear system has a unique solution. The complete solution set is {(6 −1 3)}. The system is consistent.

(2) (a) The system of linear equations corresponding to the given augmented matrix is ⎧ 1 − 52 + 23 + 34 − 25 = −4 ⎪ ⎪ ⎨ 2 − 3 − 34 − 75 = −2 5 4 + 25 = ⎪ ⎪ ⎩ 0 = 0

The last equation, 0 = 0, provides no information regarding the solution set for the system because it is true for every value of 1 , 2 , 3 , 4 , and 5 . So, we can ignore it. The column corresponding to the variable 5 is not a pivot column, and so 5 is an independent variable. Let 5 = .

Substituting  for 5 into the third equation yields 4 + 2 = 5, or 4 = −2 + 5. The column corresponding to the variable 3 is not a pivot column, and so 3 is another independent variable.

Let 3 = . Substituting the expressions we have for 3 , 4 , and 5 into the second equation produces 2 −  − 3(−2 + 5) − 7 = −2. Solving for 2 and simplifying gives 2 =  +  + 13.

Finally, we plug all these expressions into the first equation to get 1 − 5( +  + 13) + 2 + 3(−2 + 5) − 2 = −4 which simplifies to 1 = 3 + 13 + 46. Hence, the complete solution set is {(3 + 13 + 46  +  + 13  −2 + 5 ) |   ∈ R}

(c) The system of linear equations corresponding to the given augmented matrix is ⎧ 1 + 42 − 83 − 4 + 25 − 36 = −4 ⎪ ⎪ ⎨ 2 − 73 + 24 − 95 − 6 = −3  2 5 − 46 = ⎪ ⎪ ⎩ 0 = 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

34

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

The last equation, 0 = 0, provides no information regarding the solution set for the system because it is true for every value of 1 , 2 , 3 , 4 , 5 and 6 . So, we can ignore it. The column corresponding to the variable 6 is not a pivot column, and so 6 is an independent variable. Let 6 =  . Substituting  for 6 into the third equation yields 5 − 4 = 2, or 5 = 4 + 2. The columns corresponding to the variables 3 and 4 are not pivot columns, and so 3 and 4 are also independent variables. Let 3 =  and 4 = . Substituting the expressions we have for 3 , 4 , 5 and 6 into the second equation produces 2 − 7 + 2 − 9(4 + 2) −  = −3. Solving for 2 and simplifying gives 2 = 7 − 2 + 37 + 15. Finally, we plug all these expressions into the first equation to get 1 + 4(7 − 2 + 37 + 15) − 8 −  + 2(4 + 2) − 3 = −4 which simplifies to 1 = −20 + 9 − 153 − 68. Hence, the complete solution set is {(−20 + 9 − 153 − 68 7 − 2 + 37 + 15   4 + 2  ) |    ∈ R} (3) Let  represent the number of nickels,  represent the number of dimes, and  represent the number of quarters. The fact that the total value of all of the coins is $17 gives us the following equation: 005 + 010 + 025 = 17, or, 5 + 10 + 25 = 1700.

Next, using the fact that there are twice as many dimes as quarters gives us  = 2, or,  − 2 = 0.

Finally, using that the total number of nickels and quarters is 20 more than the number of dimes produces  +  =  + 20, or  −  +  = 20. Putting these three equations into a single system yields ⎧  +  = 20 ⎨  −  − 2 = 0  ⎩ 5 + 10 + 25 = 1700

where we have arranged the equations in an order that will make the row operations easier. The augmented matrix for this system is ¯ ⎡ ⎤ 1 −1 1 ¯¯ 20 ⎣ 0 1 −2 ¯¯ 0 ⎦ 5 10 25 ¯ 1700

The (1 1) entry is the first pivot. Since this already equals 1, we continue by targeting nonzero entries in the first column. We get a zero in the (3 1) entry by performing the Type (II) row operation: h3i ← (−5) × h1i + h3i. We first perform the following side calculation: (−5)× (row 1) (row 3) (sum)

−5 5 0

5 10 15

Row Operation

−5 25 20

−100 1700 1600

¯ ⎡ Resulting Matrix ⎤ 1 −1 ° 1 ¯¯ 20  ⎣ 0 1 −2 ¯¯ 0 ⎦ 0 15 20 ¯ 1600

(II): h3i ← (−5) × h1i + h3i

Next, the pivot moves to the (2 2) entry. Since this already equals 1, we continue with the second column by targeting the (3 2) entry using the Type (II) row operation: h3i ← (−15) × h2i + h3i. We first perform the following side calculation: (−15)× (row 2) (row 3) (sum)

0 0 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

35

−15 15 0

30 20 50

0 1600 1600

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Row Operation

Section 2.1

¯ ⎡ Resulting Matrix ⎤ 1 −1 1 ¯¯ 20  ⎣ 0 ° 1 −2 ¯¯ 0 ⎦ ¯ 0 0 50 1600

(II): h3i ← (−15) × h2i + h3i

The pivot now moves to the (3 3) entry. We must perform the following Type (I) operation to turn the pivot into a “1”: Row Operation

¯ ⎡Resulting Matrix ⎤ 1 −1 1 ¯¯ 20  ⎣ 0 1 −2 ¯¯ 0 ⎦ 1 ¯ 32 0 0 °

1 h3i (I): h3i ← 50

This ends the row operations. The final matrix corresponds to the linear system ⎧ ⎨  −  +  = 20  − 2 = 0  ⎩  = 32

The third equation in this system yields  = 32. Plugging this value into the second equation produces −2(32) = 0, or  = 64. Finally, we substitute these values into the first equation to get −64+32 = 20, or  = 52. Hence, there are 52 nickels, 64 dimes, and 32 quarters.

(4) First, plug each of the three given points into the equation  = 2 +  + . This yields the following linear system: ⎧ ⎨ 20 = 9 + 3 +  ←− using (3 20) 11 = 4 + 2 +  ←− using (2 11)  ⎩ 15 = 4 − 2 +  ←− using (−2 15) The augmented matrix for this system is: ⎡

¯ ⎤ 9 3 1 ¯¯ 20 ⎣ 4 2 1 ¯¯ 11 ⎦  4 −2 1 ¯ 15

The first pivot is the (1 1) entry. We make this a “1” by performing the following Type (I) row operation: Row Operation Resulting Matrix ⎡ 1 1 ¯¯ 20 ⎤ 1 ° 3 9 ¯ 9  ⎣ 4 (I): h1i ← 19 h1i 2 1 ¯¯ 11 ⎦ 4 −2 1 ¯ 15

We now target the (2 1) entry, and make it “0.” We do this using the Type (II) row operation: h2i ← (−4) × h1i + h2i. We first perform the following side calculation: (−4)× (row 1) (row 2) (sum)

−4

− 43

− 49

− 80 9

0

2 3

5 9

19 9

4

Row Operation (II): h2i ← (−4) × h1i + h2i

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

36

2

1

11

Matrix ¯ ⎡Resulting 1 1 ¯ 20 ⎤ 1 ° 3 9 ¯ 9 ⎢ 2 5 ¯ 19 ⎥  0 ¯ ⎣ 3 9 ¯ 9 ⎦ 4 −2 1 ¯ 15

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

Next, we target the (3 1) entry using the Type (II) row operation: h3i ← (−4) × h1i + h3i. We first perform the following side calculation: (−4)× (row 1) (row 3) (sum)

−4

− 43

− 49

− 80 9

0

− 10 3

5 9

55 9

4

−2

Row Operation

1

15

Matrix ¯ ⎡ Resulting 1 ¯ 20 ⎤ 1 1 ° 3 9 ¯ 9 ⎢ 5 ¯ 19 ⎥  2 ⎣ 0 3 9 ¯¯ 9 ⎦ 10 5 ¯ 55 0 −3 9 9

(II): h3i ← (−4) × h1i + h3i

Now we move the pivot to the (2 2) entry. We make this entry a “1” using the following Type (I) row operation: Row Operation Matrix ¯ ⎡Resulting 1 1 ¯ 20 ⎤ 1 3 9 ¯ 9 ⎥  ⎢ ¯ 1 56 ¯ 19 ° (I): h2i ← 32 h2i ⎦ ⎣ 0 ¯ 6 10 5 ¯ 55 0 −3 9 9

Now we target the (3 2) entry using the Type (II) operation: h3i ← ( 10 3 ) × h2i + h3i. We first perform the following side calculation: ( 10 3 )× (row 2)

0

(row 3)

0

(sum)

0

Row Operation

10 3 − 10 3

95 9 55 9 50 3

25 9 5 9 10 3

0

Matrix ¯ ⎡Resulting 1 ¯ 20 ⎤ 1 13 9 ¯ 9 ⎢ 5 ¯ 19 ⎥  1 0 ° ¯ ⎣ 6 ¯ 6 ⎦ ¯ 50 0 0 10 3 3

(II): h3i ← ( 10 3 ) × h2i + h3i

Finally, we move the pivot to the (3 3) entry and perform the following Type (I) row operation to turn that entry into a “1.”

Row Operation

Resulting Matrix ¯ ⎡ 1 ¯ 20 ⎤ 1 13 9 ¯ 9 ⎥  ⎢ ¯ ⎦ ⎣ 0 1 56 ¯ 19 ¯ 6 ¯ 1 0 0 ° 5

3 (I): h3i ← 10 h3i

This matrix corresponds to the following linear system: ⎧ 1 1 ⎪ ⎨  + 3 + 9 =  + 56  = ⎪ ⎩  =

20 9 19 6



5

The last equation clearly gives  = 5. Plugging  = 5 into the second equation yields  + 56 (5) = 19 6 , or, , or  = 2.  = −1. Substituting these values into the first equation produces  + 13 (−1) + 19 (5) = 20 9 Hence the desired quadratic equation is  = 22 −  + 5. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

37

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

(6) First, plug each of the three given points into the equation 2 +  2 +  +  = . This yields the following linear system: ⎧ 2 ⎨ (6 + 82 ) + 6 + 8 =  ←− using (6 8) (82 + 42 ) + 8 + 4 =  ←− using (8 4)  ⎩ 2 (3 + 92 ) + 3 + 9 =  ←− using (3 9) Rearranging terms leads to the following augmented matrix: ¯ ⎡ ⎤ 6 8 −1 ¯¯ −100 ⎣ 8 4 −1 ¯ −80 ⎦  ¯ 3 9 −1 ¯ −90

The first pivot is the (1 1) entry. We make this a “1” by performing the following Type (I) row operation: Row Operation Resulting Matrix ¯ ⎡ ⎤ 1 43 − 16 ¯ − 50 ° 3 ¯ ⎣ 8 4 −1 ¯ −80 ⎦  (I): h1i ← 16 h1i ¯ 3 9 −1 ¯ −90

We now must target the (2 1) entry, and make it “0.” We do this using the Type (II) row operation: h2i ← (−8) × h1i + h2i. We first perform the following side calculation: (−8)× (row 1) (row 2) (sum)

−8

− 32 3

0

− 20 3

8

4

Row Operation

4 3

400 3

−1

−80

1 3

160 3

Matrix ¯ ⎡ Resulting 4 1 ¯ 50 ⎤ 1 ° − 3 6 ¯ − 3 ⎢ 160 ⎥  1 ¯ ⎣ 0 − 20 3 3 ¯¯ 3 ⎦ ¯ 3 9 −1 −90

(II): h2i ← (−8) × h1i + h2i

Next, we target the (3 1) entry using the Type (II) row operation: h3i ← (−3) × h1i + h3i. We first perform the following side calculation: (−3)× (row 1) (row 3) (sum)

−3

−4

0

5

3

Row Operation

9

1 2

50

−1

−90

− 12

−40

Matrix ¯ ⎡ Resulting ⎤ 4 1 ° − 16 ¯¯ − 50 3 3 ⎢ 160 ⎥  1 ¯ ⎣ 0 − 20 3 3 ¯¯ 3 ⎦ 0 5 − 12 ¯ −40

(II): h3i ← (−3) × h1i + h3i

Now we move the pivot to the (2 2) entry. We make this entry a “1” using the following Type (I) row operation: Row Operation Matrix ¯ ⎡ Resulting ⎤ 1 43 − 16 ¯¯ − 50 3 ⎢ ⎥  1 ¯ 3 1 − 20 (I): h2i ← − 20 h2i ¯ −8 ⎦ ⎣ 0 ° ¯ 0 5 − 12 ¯ −40

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

38

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.1

Now we target the (3 2) entry using the Type (II) row operation: h3i ← (−5) × h2i + h3i. We first perform the following side calculation: (−5)× (row 2)

0

(row 3)

0

(sum)

0

−5

5 0

Row Operation

1 4 − 12 − 14

40 −40

0

Matrix ¯ ⎡ Resulting ⎤ 1 ¯ 1 43 − 6 ¯ − 50 3 ⎢ ⎥  1 ¯ 1 − 20 ¯ −8 ⎦ ⎣ 0 ° ¯ 0 0 − 14 ¯ 0

(II): h3i ← (−5) × h2i + h3i

Finally, we move the pivot to the (3 3) entry and perform the following Type (I) row operation to turn that entry into a “1”: Row Operation

Matrix ¯ ⎡ Resulting ⎤ 1 43 − 16 ¯¯ − 50 3 ⎢ ⎥  1 ¯ ¯ −8 ⎦ ⎣ 0 1 − 20 ¯ 1 ¯ 0 0 ° 0

(I): h3i ← (−4) h3i

This matrix corresponds to the following linear system: ⎧ 4 1 50 ⎪ 6 = − 3 ⎨  + 3 − 1  − 20  = −8  ⎪ ⎩  = 0

1 The last equation clearly gives  = 0. Plugging  = 0 into the second equation yields  − 20 (0) = −8, 4 1 or,  = −8. Substituting these values into the first equation produces  + 3 (−8) − 6 (0) = − 50 3 , or,  = −6. Hence the desired equation of the circle is 2 +  2 − 6 − 8 = 0, or, ( − 3)2 + ( − 4)2 = 25. ⎡ ⎤ 26 15 −6 ⎢ 6 4 1 ⎥ ⎥. To find (AB) we perform the Type (II) row (7) (a) First, we compute AB = ⎢ ⎣ 18 6 15 ⎦ 10 4 −14 operation  : h3i ← (−3) × h2i + h3i  We first perform the following side calculation:

(−3)× (row 2) (row 3) (sum) Row Operation

−12 6 −6

−3 15 12

Resulting Matrix ⎡ ⎤ 26 15 −6 ⎢ 6 4 1 ⎥ ⎥  (AB) = ⎢ ⎣ 0 −6 12 ⎦ 10 4 −14

(II): h3i ← (−3) × h2i + h3i

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

−18 18 0

39

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

⎡

2 ⎢ 0 Next we compute (A). Using A = ⎢ ⎣ −2 3

3 1 1 0

perform the following side calculation:

(−3)× (row 2) (row 3) (sum) Row Operation

Section 2.1

⎤ 4 1 ⎥ ⎥ and  : h3i ← (−3) × h2i + h3i. We first 5 ⎦ 1 0 −2 −2

−3 1 −2

−3 5 2

Resulting Matrix ⎤ ⎡ 2 3 4 ⎢ 0 1 1 ⎥ ⎥  (A) = ⎢ ⎣ −2 −2 2 ⎦ 3 0 1

(II): h3i ← (−3) × h2i + h3i

Multiplying (A) by the given matrix B produces the same result computed above for (AB).

(8) (a) To save space, we use the notation hCi for the th row of a matrix C. Also, recall the hint from the textbook, which we can rewrite as hABi = hAi B in the notation we have just defined.

For the Type (I) operation  : hi ← hi: Now, h(AB)i = hABi = hAi B (by the hint) = h(A)i B = h(A)Bi . But, if  6= , h(AB)i = hABi = hAi B (by the hint) = h(A)i B = h(A)Bi .

For the Type (II) operation  : hi ← hi + hi: Now, h(AB)i = hABi + hABi = hAi B + hAi B (by the hint) = (hAi + hAi )B = h(A)i B = h(A)Bi . But, if  6= , h(AB)i = hABi = hAi B (by the hint) = h(A)i B = h(A)Bi .

For the Type (III) operation  : hi ↔ hi: Now, h(AB)i = hABi = hAi B (by the hint) = h(A)i B = h(A)Bi . Similarly, h(AB)i = hABi = hAi B (by the hint) = h(A)i B = h(A)Bi . And, if  6=  and  6= , h(AB)i = hABi = hAi B (by the hint) = h(A)i B = h(A)Bi .

(11) (a) True. The augmented matrix contains all of the information necessary to solve the system completely. (b) False. A linear system has either no solutions, one solution, or infinitely many solutions. Exactly three solutions is not one of the possibilities. (c) False. A linear system is consistent if it has at least one solution. It could have either just one, or infinitely many solutions. (d) False. Type (I) operations are used to convert nonzero pivot entries to 1. When following the standard algorithm for Gaussian Elimination, Type (II) row operations are never used for this purpose. Type (II) row operations are only used to convert a nonzero target to a zero. (e) True. In the standard algorithm for Gaussian Elimination, this is the only instance in which a Type (III) row operation is used. (f) True. This is the statement of part (1) of Theorem 2.1 expressed in words rather than symbols. (g) False. The row of all zeroes in the augmented matrix corresponds to the equation 0 = 0, which is certainly true. The system could be consistent or inconsistent depending upon the other equations

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

40

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

in the system. For example, if the final augmented matrix is consistent. One particular solution is  = 1,  = 1.

∙

Section 2.2

¯ ¸ 1 1 ¯¯ 2 , then the system is 0 0 ¯ 0

Section 2.2 (1) The matrix in (e) is in reduced row echelon form, as it satisfies all four conditions. The matrices in (a), (b), (c), (d), and (f) are not in reduced row echelon form.

The matrix in (a) fails condition 2 of the definition, since the first nonzero entry in row 2 is in the 3rd column, while the first nonzero entry in row 3 is only in the second column.

The matrix in (b) fails condition 4 of the definition, as the second row contains all zeroes, but later rows do not.

The matrix in (c) fails condition 1 of the definition because the first nonzero entry in the second row equals 2, not 1.

The matrix in (d) fails conditions 1, 2, and 3 of the definition. Condition 1 fails because the first nonzero entry in row 3 equals 2, not 1. Condition 2 fails because the first nonzero entries in rows 2 and 3 appear in the same column. Condition 3 fails because the (3 3) entry below the first nonzero entry of the second column is nonzero.

The matrix in (f) fails condition 3 of the definition since the entries in the 5th column above the “1” in the third row do not equal zero.

(2) (a) First convert the pivot (the (1 1) entry) to “1.” ⎡ ⎣

(I): h1i ← ( 15 ) h1i

1 °

4

3 −4

12 −16

⎡

1 °

− 18 5

−14 13

Next, target the (2 1) and (3 1) entries.

(II): h2i ← (−3) × h1i + h2i (II): h3i ← (4) × h1i + h3i

⎢ ⎣

4

0

0

0

0

− 18 5

− 16 5 − 75

¯ 11 ⎤ ¯ −5 ¯ ¯ 3 ⎦ ¯ ¯ 13 ¯ 11 ⎤ ¯ −5 ¯ ¯ 48 ⎥ ¯ 5 ⎦ ¯ ¯ 21 5

We try to move the pivot to the second column, but can not, due to the zeroes in the (2 2) and (3 2) entries. Hence, the pivot moves to the (2 3) entry. We set that equal to 1. ¯ 11 ⎤ ⎡ ¯ −5 1 4 − 18 5 ¯ ⎥ ⎢ ¯ 5 1 ¯ −3 ⎦ ) h2i ° (I): h2i ← (− 16 ⎣ 0 0 ¯ 0 0 − 7 ¯ 21 5

5

Finally, we target the (1 3) and (3 3) entries.

(II): h1i ← ( 18 5 ) × h2i + h1i (II): h3i ← ( 75 ) × h2i + h3i This is the desired reduced row echelon form matrix.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

41

⎡

1 ⎣ 0 0

4 0 0

0 1 ° 0

¯ ⎤ ¯ −13 ¯ ¯ −3 ⎦ ¯ ¯ 0

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

(b) First convert the pivot (the (1 1) entry) to “1.” ⎡

(I): h1i ← (− 12 ) h1i Next, target the (2 1), (3 1) and (4 1) entries.

(I):

− 12

6 1 −5

−1 −1 −5

⎡

1 °

− 12

− 12

1

⎢ ⎢ 0 ⎢ ⎢ 0 ⎣

1 2 − 12 − 15 2

1 °

− 12

⎡

(II): h1i ← ( 12 ) × h2i + h1i (II): h3i ← ( 12 ) × h2i + h3i (II): h4i ← ( 15 2 ) × h2i + h4i

1

0

⎡

1 0

⎢ ⎢ 0 1 ⎢ ⎢ 0 0 ⎣

0

1 2 − 14 − 15 4

− 14 1 2

1 °

0 0 − 15 4

Next, target the (1 3), (2 3) and (4 3) entries.

− 15 2

(II): h1i ← ( 14 ) × h3i + h1i

(II): h2i ← (− 12 ) × h3i + h2i (II): h4i ← ( 15 4 ) × h3i + h4i

⎡

1 ⎢ 0 ⎢ ⎣ 0 0

− 15 2

⎤

− 21 4

⎤

9 2 − 72 − 103 2

− 14

0

⎢ ⎢ 0 ° 1 ⎢ ⎢ 0 0 ⎣

The pivot now moves to the (3 3) entry.

(I): h3i ← (−4) h3i

⎤

−36 ⎥ ⎥ −11 ⎦ −14

−2 −1 −5

− 12

0 − 15 2

Next, target the (1 2), (3 2) and (4 2) entries.

− 15 2

⎤

⎥ ⎢ ⎢ 0 2 1 9 ⎥ ⎥ ⎢ ⎢ 0 −1 1 7 ⎥ − − ⎣ 2 2 2 ⎦ 15 15 103 0 −2 −2 − 2 ⎡

h2i ← ( 12 ) h2i

− 12

⎢ ⎢ ⎣

(II): h2i ← (−6) × h1i + h2i (II): h3i ← (−1) × h1i + h3i (II): h4i ← (5) × h1i + h4i The pivot now moves to the (2 2) entry.

1 − 12 °

⎥ ⎥ ⎥ ⎥ ⎦

9 2 − 54 − 71 4

− 21 4

⎥ ⎥ ⎥ ⎥ ⎦

⎤

⎥ ⎥ ⎥ 5 ⎥ ⎦

9 2

− 71 4

⎤ 0 0 −4 1 0 2 ⎥ ⎥ 1 0 ° 5 ⎦ 0 0 1

The pivot moves to the (4 4) entry. Since this already equals 1, we now only need to target the (1 4), (2 4) and (3 4) entries. ⎤ ⎡ 1 0 0 0 (II): h1i ← (4) × h4i + h1i ⎢ 0 1 0 0 ⎥ ⎥ ⎢ (II): h2i ← (−2) × h4i + h2i ⎣ 0 0 1 0 ⎦ (II): h3i ← (−5) × h4i + h3i 1 0 0 0 °

Thus, I4 is the desired reduced row echelon form matrix. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

42

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

(c) First convert the pivot (the (1 1) entry) to “1.” ⎡

(I): h1i ← (− 15 ) h1i Next, target the (2 1), (3 1) and (4 1) entries.

1 °

−2

19 5

17 5

−4

⎤

⎡

1 −2 °

19 5 2 5 3 5 − 15

17 5 − 45 − 65 2 5

−4

⎤

⎢ −3 ⎢ ⎣ −7

6 −11 −11 14 ⎥ ⎥ 14 −26 −25 31 ⎦ 9 −18 34 31 −37

⎢ ⎢ 0 ⎢ ⎢ 0 ⎣ 0

(II): h2i ← (3) × h1i + h2i (II): h3i ← (7) × h1i + h3i (II): h4i ← (−9) × h1i + h4i

0 0 0

⎥ 2 ⎥ ⎥ 3 ⎥ ⎦ −1

We try to move the pivot to the second column, but can not, due to the zeroes in the (2 2), (3 2) and (4 2) entries. Hence, the pivot moves to the (2 3) entry. We set that equal to 1. ⎡ ⎤ 17 −4 1 −2 19 5 5 ⎢ ⎥ ⎢ 0 1 −2 0 ° 5 ⎥ 5 ⎢ ⎥ (I): h2i ← ( 2 ) h2i ⎢ 0 ⎥ 3 6 − 3 0 ⎣ ⎦ 5 5 2 −1 0 0 − 15 5 Next, target the (1 3), (3 3) and (4 3) entries.

⎡

1 ⎢ 0 ⎢ ⎣ 0 0

(II): h1i ← (− 19 5 ) × h2i + h1i (II): h3i ← (− 35 ) × h2i + h3i (II): h4i ← ( 15 ) × h2i + h4i

This is the desired reduced row echelon form matrix. (e) First convert the pivot (the (1 1) entry) to “1.” " (I):

h1i ← (− 13 ) h1i

−2 0 0 0

1 −2 °

−1

2

Next, target the (2 1) entry. "

(II): h2i ← (1) × h1i + h2i

1 −2 ° 0

0

1 3 10 3

0 1 ° 0 0

11 −2 0 0

⎤ −23 5 ⎥ ⎥ 0 ⎦ 0

¯ # 0 ¯¯ 53 ¯ 3 −5 10 ¯ 5

1 3

5 3

5 3 − 10 3

¯ # 0 ¯¯ 53 ¯ 10 ¯ 20 3

We try to move the pivot to the (2 2) entry, but it equals zero. There are no rows below it to provide a nonzero pivot to be swapped up, and so the pivot moves to the (2 3) entry. We set that equal to 1. ¯ " # 5 1 −2 13 0 ¯¯ 53 3 3 (I): h2i ← ( 10 ) h2i ¯ 1 −1 3 ¯ 2 0 0 ° Finally, we target the (1 3) entry to obtain the desired reduced row echelon form matrix. ¯ ¸ ∙ 1 −2 0 2 −1 ¯¯ 1 1 (II): h1i ← (− 3 ) × h2i + h1i 1 −1 0 0 ° 3 ¯ 2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

43

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

(3) (a) The final matrix obtained in Exercise 1(a) in Section 2.1 is ¯ ⎡ ⎤ 1 25 − 25 ¯¯ − 16 5 ⎢ ⎥ ¯ ⎣ 0 1 −1 ¯ −3 ⎦  ¯ 0 0 1 ¯ 6

We target the entries above the pivots to obtain the reduced row echelon form matrix. First, we use the (2 2) entry as the pivot and target the (1 2) entry. ¯ ⎡ ⎤ 1 0 0 ¯¯ −2 ⎣ 0 ° 1 −1 ¯¯ −3 ⎦ (II): h1i ← (− 25 ) × h2i + h1i 0 0 1 ¯ 6

Next, pivot at the (3 3) entry, and target the (2 3) entry. (Note that the (1 3) entry is already zero.) ¯ ⎡ ⎤ 1 0 0 ¯¯ −2 ⎣ 0 1 0 ¯ 3 ⎦ (II): h2i ← (1) × h3i + h2i ¯ 1 ¯ 6 0 0 °

This matrix corresponds to the linear system ⎧ ⎨ 1 2 ⎩

3

= −2 = 3  = 6

Hence, the unique solution to the system is (−2 3 6).

(e) The final matrix obtained in Exercise 1(e) in Section 2.1 is ⎡

¯ ⎤ − 13 ¯ − 53 6 ¯ 1 −2 0 ¯¯ 5 ⎦ ¯ 0 0 1 2

1 −2 − 56

⎣ 0 0

0 0

8 3

We target the entries above the pivots to obtain the reduced row echelon form matrix. First, we use the (2 3) entry as the pivot and target the (1 3) entry. ⎡

⎣ 0

(II): h1i ← ( 56 ) × h2i + h1i

¯ ⎤ 1 − 13 ¯ − 14 3 ¯ 1 −2 0 ° 0 ¯¯ 5 ⎦ ¯ 0 0 0 1 2

1 −2

0

0

Next, pivot at the (3 5) entry, and target the (1 5) entry. (Note that the (2 5) entry is already zero.) ¯ ⎡ ⎤ 1 −2 0 1 0 ¯¯ −4 ⎣ 0 (II): h1i ← ( 13 ) × h3i + h1i 0 1 −2 0 ¯¯ 5 ⎦ ¯ 1 0 0 0 0 ° 2 This matrix corresponds to the linear system ⎧ ⎨ 1 − 22 3 ⎩

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

44

+ 4 − 24

5

= −4 = 5  = 2

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

Columns 2 and 4 are not pivot columns, so 2 and 4 are independent variables. Let 2 =  and 4 = . Then the first equation gives 1 − 2 +  = −4, or, 1 = 2 −  − 4. The second equation yields 3 − 2 = 5, or, 3 = 2 + 5. The last equation states that 5 = 2. Hence, the complete solution set is {(2 −  − 4  2 + 5  2) |   ∈ R}.

(g) The final matrix obtained in Exercise 1(g) in Section 2.1 is ¯ ⎡ 5 ⎤ 1 − 12 − 74 ¯¯ 4 ⎥ ⎢ ¯ 1 − 14 ¯ − 74 ⎥ ⎢ 0 ⎥ ⎢ ¯ ⎣ 0 0 1 ¯¯ 3 ⎦ 0 0 0 ¯ 0

We target the entries above the pivots to obtain the reduced row echelon form matrix. First, we use the (2 2) entry as the pivot and target the (1 2) entry. ¯ ⎡ 3 ⎤ ¯ 1 0 − 15 8 ¯ 8 ⎢ ⎥ ¯ 1 − 14 ¯ − 74 ⎥ ⎢ 0 ° 1 (II): h1i ← ( 2 ) × h2i + h1i ⎢ ⎥ ¯ ⎣ 0 0 1 ¯¯ 3 ⎦ 0 0 0 ¯ 0

Next, we use the (3 3) entry as a pivot and target the (1 3) and (2 3) entries. ¯ ⎡ ⎤ 1 0 0 ¯¯ 6 15 ⎢ 0 1 0 ¯ −1 ⎥ (II): h1i ← ( 8 ) × h3i + h1i ⎢ ¯ ⎥ ⎣ 0 0 ° 1 1 ¯¯ 3 ⎦ (II): h2i ← ( 4 ) × h3i + h2i 0 0 0 ¯ 0

Ignoring the last row, which gives the equation 0 = 0, this matrix yields the linear system ⎧ = 6 ⎨ 1 = −1  2 ⎩ 3 3 =

producing the unique solution (6 −1 3).

(4) (a) The augmented matrix for the system is ¯ ⎡ ⎤ −2 −3 2 −13 ¯¯ 0 ⎣ −4 −7 4 −29 ¯¯ 0 ⎦  1 2 −1 8 ¯ 0

The (1 1) entry is the first pivot. We convert it to 1. ⎡ ⎤ 3 13 ¯¯ 1 ° −1 2 2 ¯ 0 ⎣ −4 −7 (I): h1i ← (− 12 ) h1i 4 −29 ¯¯ 0 ⎦ 1 2 −1 8 ¯ 0 Next, we target the (2 1) and (3 1) entries.

(II): h2i ← (4) × h1i + h2i (II): h3i ← (−1) × h1i + h3i c 2016 Elsevier Ltd. All rights reserved.

Copyright °

45

⎡

1 °

3 2

⎢ ⎣ 0 −1 1 0 2

¯ ⎤ ¯ 0 ¯ ¯ ⎥ 0 −3 ¯ 0 ⎦ ¯ 3 ¯ 0 0 2

−1

13 2

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

The (2 2) entry is the second pivot. We convert it to 1. ⎡ 1 32 −1 ⎢ 1 0 (I): h2i ← (−1) h2i ⎣ 0 ° 1 2

0

Next, we target the (1 2) and (3 2) entries.

(II): h1i ← (− 32 ) × h2i + h1i

(II): h3i ← (− 12 ) × h2i + h3i

0

Section 2.2

¯ ⎤ ¯ 0 ¯ ¯ ⎥ 3 ¯ 0 ⎦ ¯ 3 ¯ 0

13 2

2

¯ ⎤ 1 0 −1 2 ¯¯ 0 ⎣ 0 ° 1 0 3 ¯¯ 0 ⎦ 0 0 0 0 ¯ 0 ⎡

This matrix is in reduced row echelon form. It corresponds to the linear system ⎧ − 3 + 24 = 0 ⎨ 1 + 34 = 0  2 ⎩ 0 = 0

We can ignore the last equation, since it provides us with no information regarding the values of the variables. Columns 3 and 4 of the matrix are nonpivot columns, and so 3 and 4 are independent variables. Let 3 =  and 4 = . Substituting these into the first equation yields 1 −  + 2 = 0, or, 1 =  − 2. The second equation gives us 2 + 3 = 0, or, 2 = −3.

Therefore. the complete solution set is {( − 2 −3  ) |   ∈ R}. Letting  = 1 and  = 2 produces the particular nontrivial solution (−3 −6 1 2). (c) The augmented matrix for the system is ¯ ⎡ ⎤ 7 28 4 −2 10 19 ¯¯ 0 ⎢ −9 −36 −5 3 −15 −29 ¯¯ 0 ⎥ ⎢ ⎥ ⎣ 3 12 2 0 6 11 ¯¯ 0 ⎦ 6 24 3 −3 10 20 ¯ 0 The (1 1) entry is the first pivot. We convert it to 1. ⎡ ⎤ 10 19 ¯¯ 4 1 − 27 ° 4 7 7 7 ¯ 0 ⎢ −9 −36 −5 3 −15 −29 ¯¯ 0 ⎥ ⎢ ⎥ (I): h1i ← ( 17 ) h1i ⎣ 3 12 2 0 6 11 ¯¯ 0 ⎦ 6 24 3 −3 10 20 ¯ 0 Next, we target the (2 1), (3 1) and (4 1) entries. ⎡ 1 ° ⎢ (II): h2i ← (9) × h1i + h2i ⎢ 0 ⎢ (II): h3i ← (−3) × h1i + h3i ⎢ 0 ⎣ (II): h4i ← (−6) × h1i + h4i 0

4 0 0 0

4 7 1 7 2 7 − 37

− 27

3 7 6 7 − 97

10 7 − 15 7 12 7 10 7

19 7 − 32 7 20 7 26 7

¯ ⎤ ¯ 0 ¯ ¯ ⎥ ¯ 0 ⎥ ¯ ⎥ ¯ 0 ⎥ ¯ ⎦ ¯ ¯ 0

We try to move the pivot to the second column, but can not, due to the zeroes in the (2 2), (3 2) and (4 2) entries. Hence, the pivot moves to the (2 3) entry. We convert that to 1. ¯ ⎤ ⎡ 4 10 19 ¯ − 27 1 4 7 7 7 ¯ 0 ¯ ⎥ ⎢ ⎢ 0 0 ° 1 3 −15 −32 ¯ 0 ⎥ ¯ ⎥ ⎢ (I): h2i ← (7) h2i ⎥ ⎢ 0 0 2 6 12 20 ¯ ⎣ 7 7 7 7 ¯ 0 ⎦ ¯ 10 26 ¯ 0 0 0 − 37 − 97 7 7

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

46

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Next, we target the (1 3), (3 3) and (4 3) entries. ⎡ 1 (II): h1i ← (− 47 ) × h2i + h1i ⎢ 0 2 ⎢ (II): h3i ← (− 7 ) × h2i + h3i ⎣ 0 3 (II): h4i ← ( 7 ) × h2i + h4i 0

Section 2.2

¯ ⎤ 4 0 −2 10 21 ¯¯ 0 1 0 ° 3 −15 −32 ¯¯ 0 ⎥ ⎥ 0 0 0 6 12 ¯¯ 0 ⎦ 0 0 0 −5 −10 ¯ 0

We try to move the pivot to the third column, but can not, due to the zeroes in the (3 4) and (4 4) entries. Hence, the pivot moves to the (3 5) entry. We convert that to 1. ¯ ⎡ ⎤ 1 4 0 −2 10 21 ¯¯ 0 ⎢ 0 0 1 3 −15 −32 ¯¯ 0 ⎥ ⎢ ⎥ (I): h3i ← ( 16 ) h3i ⎣ 0 0 0 1 0 ° 2 ¯¯ 0 ⎦ 0 0 0 0 −5 −10 ¯ 0 Next, we target the (1 5), (2 5) and (4 5) entries.

⎡

1 ⎢ 0 ⎢ ⎣ 0 0

(II): h1i ← (−10) × h3i + h1i (II): h2i ← (15) × h3i + h2i (II): h4i ← (5) × h3i + h4i

4 0 0 0

¯ ⎤ 0 −2 0 1 ¯¯ 0 1 3 0 −2 ¯¯ 0 ⎥ ⎥ 1 0 ° 2 ¯¯ 0 ⎦ 0 0 0 0 0 ¯ 0

This matrix is in reduced row echelon form. It corresponds to the linear system ⎧ − 24 + 6 = 0 1 + 42 ⎪ ⎪ ⎨ − 26 = 0 3 + 34  + 26 = 0  ⎪ 5 ⎪ ⎩ 0 = 0

We can ignore the last equation, since it provides us with no information regarding the values of the variables. Columns 2, 4 and 6 of the matrix are nonpivot columns, and so 2 , 4 and 6 are independent variables. Let 2 = , 4 =  and 6 =  . Substituting these into the first equation yields 1 + 4 − 2 +  = 0, or, 1 = −4 + 2 −  . The second equation gives us 3 + 3 − 2 = 0, or, 3 = −3 + 2 . The third equation produces 5 + 2 = 0, or, 5 = −2 . Therefore. the complete solution set is {(−4 + 2 −   −3 + 2  −2  ) |    ∈ R} Letting  = 1,  = 2, and  = 3 produces the particular nontrivial solution (−3 1 0 2 −6 3).

(5) (a) The augmented matrix for the system is ¯ ⎡ ⎤ −2 1 8 ¯¯ 0 ⎣ 7 −2 −22 ¯ 0 ⎦  ¯ 3 −1 −10 ¯ 0 The (1 1) entry is the first pivot. We convert it to 1. ⎡

1 (I): h1i ← (− ) h1i 2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

⎣ 7 3 47

¯ ⎤ −4 ¯ 0 ¯ −2 −22 ¯¯ 0 ⎦ −1 −10 ¯ 0

1 − 12 °

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

Next, we target the (2 1) and (3 1) entries. ⎡

⎢ ⎣ 0 0

(II): h2i ← (−7) × h1i + h2i (II): h3i ← (−3) × h1i + h3i

¯ ⎤ −4 ¯¯ 0 ¯ ⎥ 6 ¯ 0 ⎦ ¯ 2 ¯ 0

1 − 12 ° 3 2 1 2

The (2 2) entry is the second pivot. We convert it to 1. ¯ ⎡ ⎤ 1 − 12 −4 ¯¯ 0 ⎢ ¯ ⎥ 1 4 ¯ 0 ⎦ (I): h2i ← ( 23 ) h2i ⎣ 0 ° ¯ 1 2 ¯ 0 0 2

Next, we target the (1 2) and (3 2) entries.

¯ ⎤ 1 0 −2 ¯¯ 0 ⎣ 0 ° 1 4 ¯¯ 0 ⎦ 0 0 0 ¯ 0 ⎡

(II): h1i ← ( 12 ) × h2i + h1i

(II): h3i ← (− 12 ) × h2i + h3i

This matrix is in reduced row echelon form. It corresponds to the linear system ⎧ − 23 = 0 ⎨ 1 2 + 43 = 0  ⎩ 0 = 0

We can ignore the last equation, since it provides us with no information regarding the values of the variables. Column 3 of the matrix is a nonpivot column, and so 3 is an independent variable. Let 3 = . Substituting this into the first equation yields 1 − 2 = 0, or, 1 = 2.

The second equation gives us 2 + 4 = 0, or, 2 = −4. Therefore. the complete solution set is {(2 −4 ) |  ∈ R} = {(2 −4 1) |  ∈ R}

(c) The augmented matrix for the system is ⎡ 2 6 ⎢ 1 4 ⎢ ⎣ 2 8 3 10

13 10 20 21

¯ ⎤ 1 ¯¯ 0 1 ¯¯ 0 ⎥ ⎥ 1 ¯¯ 0 ⎦ 2 ¯ 0

The (1 1) entry is the first pivot. We convert it to 1. ⎡ 1 ° ⎢ ⎢ 1 (I): h1i ← ( 12 ) h1i ⎣ 2

3

13 2

4 8 3 10

10 20 21

Next, we target the (2 1), (3 1) and (4 1) entries.

(II): h2i ← (−1) × h1i + h2i (II): h3i ← (−2) × h1i + h3i (II): h4i ← (−3) × h1i + h4i c 2016 Elsevier Ltd. All rights reserved.

Copyright °

48

⎡

1 3 °

⎢ ⎢ 0 1 ⎢ ⎢ 0 2 ⎣ 0 1

¯ ⎤ ¯ 0 ¯ 1 ¯¯ 0 ⎥ ⎥ 1 ¯¯ 0 ⎦ 2 ¯ 0

1 2

13 2 7 2

7 3 2

¯ ⎤ ¯ 0 ¯ ¯ ⎥ ¯ 0 ⎥ ¯ ⎥ 0 ¯¯ 0 ⎥ ⎦ ¯ 1 ¯ 0 2 1 2 1 2

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

The (2 2) entry is the second pivot. It already equals 1, so now we target the (1 2), (3 2) and (4 2) entries. ¯ ⎡ ⎤ 1 0 −4 −1 ¯¯ 0 (II): h1i ← (−3) × h2i + h1i ⎢ ⎥ 1 ¯ 7 1 ⎢ 0 ° 2 2 ¯¯ 0 ⎥ (II): h3i ← (−2) × h2i + h3i ⎢ ⎥ ⎣ 0 0 0 −1 ¯¯ 0 ⎦ (II): h4i ← (−1) × h2i + h4i 0 0 −2 0 ¯ 0

The (3 3) entry equals 0. However, we can switch the 3rd and 4th rows to bring a nonzero number into the (3 3) position. ¯ ⎤ ⎡ 1 0 −4 −1 ¯¯ 0 ⎥ ⎢ 7 1 ¯ ⎢ 0 1 2 2 ¯¯ 0 ⎥ (III): h3i ↔ h4i ⎥ ⎢ ⎣ 0 0 −2 0 ¯¯ 0 ⎦ 0 0 0 −1 ¯ 0 We now convert the pivot (the (3 3) entry) to 1.

¯ ⎤ 1 0 −4 −1 ¯¯ 0 ⎢ ⎥ 7 1 ¯ ⎢ 0 1 2 2 ¯¯ 0 ⎥ ⎢ ⎥ ⎣ 0 0 ° 1 0 ¯¯ 0 ⎦ 0 0 0 −1 ¯ 0 ⎡

(I): h3i ← (− 12 ) h3i

Next, we target the (1 3) and (2 3) entries. The (4 3) entry already equals 0. ¯ ⎡ ⎤ 1 0 0 −1 ¯¯ 0 ⎢ ⎥ 1 ¯ (II): h1i ← (4) × h3i + h1i ⎢ 0 1 0 2 ¯¯ 0 ⎥ ⎢ ⎥ ⎣ 0 0 ° (II): h2i ← (− 72 ) × h3i + h2i 1 0 ¯¯ 0 ⎦ 0 0 0 −1 ¯ 0 The (4 4) entry is the last pivot. We convert it to 1. ⎡ 1 ⎢ ⎢ 0 (I): h4i ← (−1) h4i ⎢ ⎣ 0 0

¯ ⎤ 0 0 −1 ¯¯ 0 ⎥ 1 ¯ 1 0 2 ¯¯ 0 ⎥ ⎥ 0 1 0 ¯¯ 0 ⎦ 1 ¯ 0 0 0 °

Finally, we target the (1 4) and (2 4) entries. (The (3 4) entry is already 0.) ¯ ⎡ ⎤ 1 0 0 0 ¯¯ 0 ⎢ 0 1 0 0 ¯ 0 ⎥ (II): h1i ← (1) × h4i + h1i ⎢ ¯ ⎥ 1 ⎣ 0 0 1 0 ¯ 0 ⎦ (II): h2i ← (− 2 ) × h4i + h2i ¯ 1 ¯ 0 0 0 0 °

This matrix is in reduced row echelon form. It corresponds to the linear system ⎧ = 0 1 ⎪ ⎪ ⎨ = 0 2  = 0  ⎪ 3 ⎪ ⎩ 4 = 0

Clearly, this system has only the trivial solution. The solution set is {(0 0 0 0)}. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

49

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

(6) (a) First we find the system of linear equations corresponding to the given chemical equation by considering each element separately. This produces ⎧ =  ← Carbon equation ⎨ 6 6 = 2 ← Hydrogen equation  ⎩ 2 = 2 +  ← Oxygen equation Moving all variables to the left side, and then creating an augmented matrix yields ¯ ⎡ ⎤ 6 0 −1 0 ¯¯ 0 ⎣ 6 0 0 −2 ¯¯ 0 ⎦  0 2 −2 −1 ¯ 0 The (1 1) entry is the first pivot. We convert it to 1. ¯ ⎡ ⎤ 1 0 − 16 ° 0 ¯ 0 ¯ ⎣ 6 0 (I): h1i ← ( 16 ) h1i 0 −2 ¯¯ 0 ⎦ 0 2 −2 −1 ¯ 0

Next, we target the (2 1) entry. (The (3 1) entry is already 0.) ¯ ⎡ ⎤ 1 0 − 16 0 ¯ 0 ° ¯ ⎣ 0 0 (II): h2i ← (−6) × h1i + h2i 1 −2 ¯¯ 0 ⎦ 0 2 −2 −1 ¯ 0

The pivot moves to the (2 2) entry, which is zero. So, we switch the 2nd and 3rd rows to move a nonzero number into the pivot position. ¯ ⎤ ⎡ 0 ¯ 0 1 0 − 16 ¯ ⎣ 0 ° (III): h2i ↔ h3i 2 −2 −1 ¯¯ 0 ⎦ 0 0 1 −2 ¯ 0 Now we convert the (2 2) entry to 1.

⎡

1

⎢ 1 ⎣ 0 ° 0 0

(I): h2i ← ( 12 ) h2i

¯ ⎤ 0 ¯¯ 0 ¯ ⎥ −1 − 12 ¯ 0 ⎦ ¯ 1 −2 ¯ 0

0 − 16

The pivot moves to the (3 3) entry. Since this already equals 1, we target the (1 3) and (2 3) entries. ¯ ⎡ ⎤ 1 0 0 − 13 ¯¯ 0 1 (II): h1i ← ( 6 ) × h3i + h1i ⎢ ¯ ⎥ ⎣ 0 1 0 − 52 ¯ 0 ⎦ ¯ (II): h2i ← (1) × h3i + h2i 1 −2 ¯ 0 0 0 ° The matrix is now in reduced row echelon form. It corresponds to the following linear system: ⎧ − 13  = 0 ⎪ ⎨   − 52  = 0  ⎪ ⎩  − 2 = 0

or  = 13 ,  = 52 , and  = 2. Setting  = 6 eliminates all fractions, and provides the smallest solution containing all positive integers. Hence, the desired solution is  = 2,  = 15,  = 12,  = 6.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

50

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

(c) First we find the system of linear equations corresponding to the given chemical equation by considering each element separately. This produces ⎧  =  ← Silver equation ⎪ ⎪ ⎨  =  ← Nitrogen equation  3 +  = 2 + 3 ← Oxygen equation ⎪ ⎪ ⎩ 2 =  ← Hydrogen equation Moving all variables to the left side, and then creating an augmented matrix yields ¯ ⎡ ⎤ 1 0 −1 0 0 ¯¯ 0 ⎢ 1 0 0 0 −1 ¯¯ 0 ⎥ ⎢ ⎥ ⎣ 3 1 0 −2 −3 ¯¯ 0 ⎦ 0 2 0 0 −1 ¯ 0

The (1 1) entry is the first pivot. Since it already equals 1, we target the (2 1) and (3 1) entries.

(The (4 1) entry is already 0.) ¯ ⎡ ⎤ 1 0 −1 ° 0 0 ¯¯ 0 ⎢ 0 0 1 0 −1 ¯¯ 0 ⎥ (II): h2i ← (−1) × h1i + h2i ⎢ ⎥ ⎣ 0 1 3 −2 −3 ¯¯ 0 ⎦ (II): h3i ← (−3) × h1i + h3i 0 2 0 0 −1 ¯ 0

The pivot moves to the (2 2) entry. However, it equals zero. So, we switch the 2nd and 3rd rows to move a nonzero number into the pivot position. ¯ ⎡ ⎤ 1 0 −1 0 0 ¯¯ 0 ⎢ 0 ° 1 3 −2 −3 ¯¯ 0 ⎥ ⎢ ⎥ (III): h2i ↔ h3i ⎣ 0 0 1 0 −1 ¯¯ 0 ⎦ 0 2 0 0 −1 ¯ 0

The (2 2) entry already equals 1, so we target the (4 2) entry. (The (1 2) and (3 2) entries already equal zero, and so do not need to be targeted.) ¯ ⎡ ⎤ 1 0 −1 0 0 ¯¯ 0 ⎢ 0 ° 1 3 −2 −3 ¯¯ 0 ⎥ ⎢ ⎥ (II): h4i ← (−2) × h2i + h4i ⎣ 0 0 1 0 −1 ¯¯ 0 ⎦ 0 0 −6 4 5 ¯ 0

The pivot moves to the (3 3) entry. Since this already equals 1, we target the (1 3), (2 3) and (4 3) entries. ¯ ⎡ ⎤ 1 0 0 0 −1 ¯¯ 0 (II): h1i ← (1) × h3i + h1i ⎢ 0 1 0 −2 0 ¯¯ 0 ⎥ ⎢ ⎥ (II): h2i ← (−3) × h3i + h2i ⎣ 0 0 ° 1 0 −1 ¯¯ 0 ⎦ (II): h4i ← (6) × h3i + h4i 0 0 0 4 −1 ¯ 0 The pivot moves to the (4 4) entry. We convert this to 1. ⎡ 1 0 0 0 ⎢ 0 1 0 −2 ⎢ (I): h4i ← ( 14 ) h4i ⎣ 0 0 1 0 0 0 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

51

¯ ⎤ −1 ¯ 0 ¯ 0 ¯ 0 ⎥ ¯ ⎥ −1 ¯ 0 ⎦ ¯ 1 − 14 ¯ 0 °

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

Finally, we target the (2 4) entry. ¯ ⎤ −1 ¯¯ 0 ⎢ ¯ ⎥ ⎢ 0 1 0 0 − 12 ¯ 0 ⎥ ⎢ ¯ ⎥ ⎢ 0 0 1 0 −1 ¯ 0 ⎥ ¯ ⎣ ⎦ ¯ 1 − 14 ¯ 0 0 0 0 ° ⎡

(II): h2i ← (2) × h4i + h2i

1 0 0

0

The matrix is now in reduced row echelon form. It corresponds to the following linear system: ⎧  −  = 0 ⎪ ⎪ ⎨  − 12  = 0   −  = 0 ⎪ ⎪ ⎩ 1  − 4 = 0

or  = ,  = 12 ,  = , and  = 14 . Setting  = 4 eliminates all fractions, and provides the smallest solution containing all positive integers. Hence, the desired solution is  = 4,  = 2,  = 4,  = 1,  = 4.

(7) (a) Combining the fractions on the right side of the given equation over a common denominator produces   ( − 3)( + 4) + ( − 1)( + 4) + ( − 1)( − 3)  + + = ( − 1) ( − 3) ( + 4) ( − 1)( − 3)( + 4) 2 ( +  + )  + ( + 3 − 4)  + (−12 − 4 + 3)  = ( − 1)( − 3)( + 4) Now we set the coefficients of 2 and , and the constant term equal to the corresponding coefficients in the numerator of the left side of the equation given in the problem. Doing this yields the following linear system: ⎧  +  +  = 5 ← coefficients of 2 ⎨  + 3 − 4 = 23 ← coefficients of   ⎩ −12 − 4 + 3 = −58 ← constant terms

The augmented matrix for the system is ¯ ⎡ ⎤ 1 1 1 ¯¯ 5 ⎣ 1 3 −4 ¯¯ 23 ⎦  −12 −4 3 ¯ −58

The (1 1) entry is the first pivot. Since it already equals 1, we target the (2 1) and (3 1) entries. ¯ ⎤ ⎡ 1 1 ° 1 ¯¯ 5 (II): h2i ← (−1) × h1i + h2i ⎣ 0 2 −5 ¯ 18 ⎦ ¯ (II): h3i ← (12) × h1i + h3i 0 8 15 ¯ 2 The (2 2) entry is the second pivot. We convert it to 1. ¯ ⎡ ⎤ 1 1 1 ¯¯ 5 ⎢ ¯ ⎥ 1 − 52 ¯ 9 ⎦ (I): h2i ← ( 12 ) h2i ⎣ 0 ° ¯ 0 8 15 ¯ 2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

52

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

Next, we target the (1 2) and (3 2) entries. ⎡

1 0

⎢ ⎣ 0 1 0 0

1 ) h3i (I): h3i ← ( 35

Finally, we target the (1 3) and (2 3) entries.

0

⎢ 1 ⎣ 0 ° 0 0

(II): h1i ← (−1) × h2i + h1i (II): h3i ← (−8) × h2i + h3i The (3 3) entry is the last pivot. We convert it to 1. ⎡

1

¯ ⎤ ¯ −4 ¯ ¯ ⎥ 9 ⎦ ¯ ¯ 35 ¯ −70

7 2 − 52

¯ ⎤ ¯ −4 ¯ ¯ ⎥ ¯ 9 ⎦ ¯ 1 ¯ −2 °

7 2 − 52

¯ ⎤ 1 0 0 ¯¯ 3 ⎣ 0 1 0 ¯ 4 ⎦ ¯ 1 ¯ −2 0 0 ° ⎡

(II): h1i ← (− 72 ) × h3i + h1i (II): h2i ← ( 52 ) × h3i + h2i

This matrix is in reduced row echelon form. It corresponds to the linear system ⎧ = 3 ⎨   = 4  ⎩  = −2

which gives the unique solution to the problem.

(8) We set up the augmented matrix having two columns to the right of the augmentation bar: ¯ ⎡ ⎤ 9 2 2 ¯¯ −6 −12 ⎣ 3 2 4 ¯ 0 −3 ⎦  ¯ 27 12 22 ¯ 12 8 The (1 1) entry is the first pivot. We convert it to 1.

(I): h1i ← ( 19 ) h1i We next target the (2 1) and (3 1) entries.

¯ 2 ⎤ ¯ − 3 − 43 ¯ ⎣ 3 2 4 ¯ 0 −3 ⎦ ¯ 27 12 22 ¯ 12 8 ⎡

1 °

⎡

2 9

2 9

1 °

2 9 4 3

⎢ ⎣ 0 0

(II): h2i ← (−3) × h1i + h2i (II): h3i ← (−27) × h1i + h3i

6

¯ 2 ⎤ ¯ − 3 − 43 ¯ ⎥ ¯ 1 ⎦ ¯ 2 ¯ 16 ¯ 30 44 2 9 10 3

The (2 2) entry is the second pivot. We convert it to 1. ¯ 2 ⎤ ⎡ 2 ¯ 1 29 − 43 9 ¯ −3 ⎢ 3 5 ¯ 3 ⎥ 1 (I): h2i ← ( 34 ) h2i ⎣ 0 ° 2 ¯¯ 2 4 ⎦ 0 6 16 ¯ 30 44

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

53

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.2

Next, we target the (1 2) and (3 2) entries.

(II):

¯ ⎤ 0 − 13 ¯¯ −1 − 32 ⎢ 3 5 ¯ 3 ⎥ 1 ⎣ 0 ° 2 ¯¯ 2 4 ⎦ 0 0 1 ¯ 21 79 2 ⎡

h1i ← (− 29 ) × h2i + h1i

(II): h3i ← (−6) × h2i + h3i

1

The (3 3) entry is the last pivot. Since it already equals 1, we target the (1 3) and (2 3) entries. ¯ ⎡ 35 ⎤ 1 0 0 ¯¯ 6 3 1 (II): h1i ← ( 3 ) × h3i + h1i ⎢ ⎥ ¯ 0 1 0 −51 −98 ¯ ⎣ ⎦ 5 ¯ (II): h2i ← (− 2 ) × h3i + h2i 79 ¯ 1 0 0 ° 21 2

This matrix is in reduced row echelon form. It gives the unique solution for each of the two original linear systems. In particular, the solution for the system AX = B1 is (6 −51 21), and the solution for

79 the system AX = B2 is ( 35 3  −98 2 ).

(11) (b) Any nonhomogeneous system with two equations and two unknowns that has a unique solution will serve as a counterexample. For instance, consider ½  +  = 1   −  = 1 This system has a unique solution: (1 0). Let (1  2 ) and (1  2 ) both equal (1 0). Then the sum of solutions is not a solution in this case. Also, if  6= 1, then the scalar multiple of a solution by  is not a solution.

(14) (a) True. Gaussian Elimination puts the augmented matrix for a system into row echelon form. The statement gives a verbal description of the definition of row echelon form given in the section. (b) True. Gaussian Elimination puts the augmented matrix for a system into row echelon form, while the Gauss-Jordan Method puts the matrix into reduced row echelon form. The statement describes the difference between those two forms for a matrix. (c) False. A column is skipped if the pivot entry that would usually be used is zero, and all entries below that position in that column are zero. For example, the reduced row echelon form matrix ∙ ¸ 1 2 0 has no pivot in the second column. Rows, however, are never skipped over, although 0 0 1 rows of zeroes may occur at the bottom of the matrix. (d) True. A homogeneous system always has the trivial solution. (e) False. This statement reverses the roles of the “independent (nonpivot column)” and “dependent (pivot column)” variables. Note, however, that the statement does correctly define the terms “independent” and “dependent.” (f) False. For a counterexample, consider the system ⎧ = 0 ⎨   = 0  ⎩  +  = 0 which clearly has only the trivial solution. The given statement in part (f) is related to Corollary 2.3, but with the number of variables and number of equations reversed, making it false, as the counterexample illustrates. (Of course, Corollary 2.3, as stated in the text, is true.)

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

54

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.3

Section 2.3 (1) Recall that a matrix A is row equivalent to a matrix B if a finite sequence of row operations will convert A into B. In parts (a) and (c), below, only one row operation is necessary, although longer finite sequences could be used. (a) A row operation of Type (I) converts A to B: h2i ← −5 h2i.

(c) A row operation of Type (II) converts A to B: h2i ← h3i + h2i.

(2) (b) We must first compute B by putting A into reduced row echelon form. Performing the following sequence of row operations converts A to B = I3 : (I): (II): (II): (III): (II):

h1i ← 14 h1i h2i ← 2 h1i + h2i h3i ← −3 h1i + h3i h2i ↔ h3i h1i ← 5 h3i + h1i

To convert B back to A we use the inverses of these row operations in the reverse order: (II): h1i ← −5 h3i + h1i (III): h2i ↔ h3i (II): h3i ← 3 h1i + h3i (II): h2i ← −2 h1i + h2i (I): h1i ← 4 h1i

which is the inverse of (II): h1i ← 5 h3i + h1i which is the inverse of (III): h2i ↔ h3i which is the inverse of (II): h3i ← −3 h1i + h3i which is the inverse of (II): h2i ← 2 h1i + h2i which is the inverse of (I): h1i ← 14 h1i

(3) (a) Let C be the common reduced row echelon form matrix for A and B. Then, A and B are both row equivalent to C. Also, by part (1) of Theorem 2.4, C is row equivalent to B. But, since A is row equivalent to C, and C is row equivalent to B, part (2) of Theorem 2.4 asserts that A is row equivalent to B. (b) Both A and B row reduce to I3 . To convert A to I3 , use the following sequence of row operations: (II): h3i ← 2 h2i + h3i (I): h3i ← −1 h3i (II): h1i ← −9 h3i + h1i (II): h2i ← 3 h3i + h2i To convert B to I3 , use these row operations: (I): h1i ← − 15 h1i (II): h2i ← 2 h1i + h2i

(II): h3i ← 3 h1i + h3i (I): h2i ← −5 h2i

(II): h1i ← 35 h2i + h1i

(II): h3i ← 95 h2i + h3i

(c) First we convert A to I3 using the sequence of row operations we computed in part (a): (II): h3i ← 2 h2i + h3i (I): h3i ← −1 h3i (II): h1i ← −9 h3i + h1i (II): h2i ← 3 h3i + h2i

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

55

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.3

Next, we continue by converting I3 to B. To do this, we use the inverses of the row operations that converted B to I3 (from part (a)), in the reverse order: (II): h3i ← − 95 h2i + h3i

which is the inverse of (II): h3i ← 95 h2i + h3i

(I): h2i ← − 15 h2i

which is the inverse of (I): h2i ← −5 h2i

(II): h1i ← − 35 h2i + h1i (II): h3i ← −3 h1i + h3i

(II): h2i ← −2 h1i + h2i (I): h1i ← −5 h1i

which is the inverse of (II): h1i ← 35 h2i + h1i which is the inverse of (II): h3i ← 3 h1i + h3i

which is the inverse of (II): h2i ← 2 h1i + h2i which is the inverse of (I): h1i ← − 15 h1i

(5) The rank of a matrix is the number of nonzero rows in its corresponding reduced row echelon form matrix. ⎡ ⎤ 1 0 2 (a) The given matrix reduces to ⎣ 0 1 −1 ⎦, which has 2 nonzero rows. Hence, its rank is 2.

0 0 0 ⎡ ⎤ 1 0 0 (c) The given matrix reduces to ⎣ 0 0 1 ⎦, which has 2 nonzero rows. Hence, its rank is 2.

0 0 0 ⎤ ⎡ 1 0 0 1 ⎢ 0 1 0 −1 ⎥ ⎥ ⎢ (e) The given matrix reduces to ⎢ 3 ⎥, which has 3 nonzero rows. Hence, its rank is 3. ⎣ 0 0 1 2 ⎦ 0 0 0 0

(6) (a) The augmented matrix for this system is ¯ ⎡ ⎤ −2 6 3 ¯¯ 0 ⎢ 5 −9 −4 ¯ 0 ⎥ ⎢ ¯ ⎥ ⎣ 4 −8 −3 ¯ 0 ⎦ , which reduces to ¯ 6 −11 −5 ¯ 0

⎡

1 ⎢ 0 ⎢ ⎣ 0 0

0 1 0 0

¯ ⎤ 0 ¯¯ 0 0 ¯¯ 0 ⎥ ⎥ 1 ¯¯ 0 ⎦ 0 ¯ 0

Since there are three nonzero rows in the reduced row echelon form matrix, its rank = 3. Theorem 2.7 predicts that the system has only the trivial solution. In fact, the reduced row echelon form matrix corresponds to the linear system ⎧ = 0 1 ⎪ ⎪ ⎨ = 0 2  = 0  ⎪ 3 ⎪ ⎩ 0 = 0 which clearly has the solution set {(0 0 0)}

(7) In each case, we give the smallest and largest rank possible, with a particular example of a linear system and the reduced row echelon form of its corresponding augmented matrix for each of the two ranks.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

56

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

(a)

Smallest rank = 1 ⎧ 1 + 22 + 33 = 4 ⎪ ⎪ ⎨ 1 + 22 + 33 = 4 ⎪ 1 + 22 + 33 = 4 ⎪ ⎩ 1 + 22 + 33 = 4 ⎡

1 ⎢ 0 ⎢ ⎣ 0 0

2 0 0 0

3 0 0 0

¯ ⎤ ¯ 4 ¯ ¯ 0 ⎥ ¯ ⎥ ¯ 0 ⎦ ¯ ¯ 0

Largest rank = 4

⎡

⎧ 1 ⎪ ⎪ ⎨ 2 ⎪ 3 ⎪ ⎩ 0

1 ⎢ 0 ⎢ ⎣ 0 0

0 1 0 0

= = = = 0 0 1 0

0 0 0 1 ¯ ⎤ ¯ 0 ¯ ¯ 0 ⎥ ¯ ⎥ ¯ 0 ⎦ ¯ ¯ 1

Smallest rank = 2

Largest rank = 3

⎧ ⎨ 1 + 2 + 3 + 4 = 0 1 + 2 + 3 + 4 = 1 ⎩ 1 + 2 + 3 + 4 = 0

⎧ ⎨ 1 + 3 + 4 = 0 2 + 3 + 4 = 0 ⎩ 1 + 3 + 4 = 1

(c)

⎡

1 1 1 1 ⎣ 0 0 0 0 0 0 0 0

¯ ⎤ ¯ 0 ¯ ¯ 1 ⎦ ¯ ¯ 0

Section 2.3

⎡

1 0 1 1 ⎣ 0 1 1 1 0 0 0 0

¯ ⎤ ¯ 0 ¯ ¯ 0 ⎦ ¯ ¯ 1

(8) In each part, to express the vector x as a linear combination of the vectors a1      a , we need to solve the linear system whose augmented matrix has the vectors a1      a as columns to the left of the augmentation bar, and has the vector x as a column to the right of the augmentation bar. A solution of this linear system gives the coefficients for the desired linear combination. (a) Following the procedure described above, we row reduce the matrix ¯ # " ¯ ∙ ¸ 1 0 ¯¯ − 21 1 −2 ¯¯ −3 11  to obtain ¯ 6 4 3 ¯ −6 0 1 ¯ 11

The unique solution to this system provides the coefficients for x as a linear combination of a1

6 and a2 , and so x = − 21 11 a1 + 11 a2 .

(c) Following the procedure described above, we row reduce the matrix ¯ ⎤ ⎡ ¯ ⎡ ⎤ 1 0 ¯¯ 11 9 3 2 ¯¯ 2 ⎥ ⎢ ¯ ⎣ 6 10 ¯ −1 ⎦ to obtain ⎣ 0 1 ¯ − 56 ⎦  ¯ ¯ 2 −4 ¯ 4 0 0 ¯ − 16 9

The last row of the reduced matrix shows that the system is inconsistent. Therefore, it is not possible to express x as a linear combination of a1 and a2 .

(e) Following the procedure described above, we row reduce the matrix ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 5 4 ¯¯ 7 1 0 −1 ¯¯ −3 ⎣ −2 −2 0 ¯ 2 ⎦ to obtain ⎣ 0 1 1 ¯¯ 2 ⎦  ¯ 3 6 3 ¯ 3 0 0 0 ¯ 0 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

57

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.3

The linear system has the infinite solution set {( − 3 − + 2 ) |  ∈ R}, and hence there are many sets of coefficients 1 , 2 , 3 for which x = 1 a1 + 2 a2 + 3 a3 . Choosing the particular solution when  = 0 yields x = −3a1 + 2a2 + 0a3 . Other choices for  produce alternate linear combinations. (g) Following the procedure described above, we row reduce the matrix ¯ ¯ ⎡ ⎤ ⎡ ⎤ 3 −2 6 ¯¯ 2 1 0 0 ¯¯ 2 ⎢ 2 ⎢ ¯ ⎥ 0 1 ¯¯ 3 ⎥ ⎢ ⎥ to obtain ⎢ 0 1 0 ¯ −1 ⎥  ¯ ¯ ⎣ −2 ⎦ ⎣ 1 2 ¯ −7 0 0 1 ¯ −1 ⎦ ¯ 4 −3 8 3 0 0 0 ¯ 0

The unique solution to this system provides the coefficients for x as a linear combination of a1 ,

a2 and a3 , and so x = 2a1 − a2 − a3 .

(9) To determine whether a given vector X is in the row space of matrix A, we row reduce the augmented ¤ £ matrix A |X . If the corresponding linear system is consistent, the solution set gives coefficients for expressing X as a linear combination of the rows of A. If the system is inconsistent, then X is not in the row space of A. (a) Following the procedure described above, we row reduce the matrix ¯ ¯ ⎡ ⎤ ⎡ ⎤ 3 2 2 ¯¯ 7 1 0 0 ¯¯ 5 ⎣ 6 10 −1 ¯ 1 ⎦ to obtain ⎣ 0 1 0 ¯ −3 ⎦  ¯ ¯ 2 −4 4 ¯ 18 0 0 1 ¯ −1

Since the system has the unique solution (5 −3 −1), X is in the row space of A, and X = 5(row 1) − 3(row 2) − 1(row 3).

(c) Following the procedure described above, we row reduce the matrix ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 0 2 ¯¯ 1 4 −2 6 ¯¯ 2 ⎣ −1 1 ⎦ 3 1 ¯¯ 2 ⎦ to obtain ⎣ 0 1 1 ¯¯ 0 0 0 ¯ −10 2 5 9 ¯ −3

The last row of the reduced matrix shows that the system is inconsistent. Hence, X is not in the row space of A.

(e) Following the procedure described above, we row reduce the matrix ¯ ¯ ⎡ ⎤ ⎡ ⎤ 2 7 3 ¯¯ 1 1 0 −2 ¯¯ −3 ⎢ −4 −1 ⎢ 7 ¯¯ 11 ⎥ 1 ¯¯ 1 ⎥ ⎢ ⎥ to obtain ⎢ 0 1 ⎥ ¯ ⎣ 1 −1 −3 −4 ⎦ ⎣ 0 0 0 ¯¯ 0 ⎦ ¯ −3 2 8 ¯ 11 0 0 0 ¯ 0

The linear system has the infinite solution set {(2 − 3 − + 1 ) |  ∈ R} There are many sets of coefficients 1 , 2 , 3 for which [1 11 −4 11] = 1 (row 1) + 2 (row 2) + 3 (row 3). Choosing the particular solution when  = 0 yields [1 11 −4 11] = −3(row 1) + 1(row 2) + 0(row 3). Other choices for  produce alternate linear combinations.

(10) (a) To express the vector [13 −23 60] as a linear combination of vectors q1  q2  and q3 , we need to solve the linear system whose augmented matrix has the vectors q1  q2 , and q3 as columns to c 2016 Elsevier Ltd. All rights reserved.

Copyright °

58

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.3

the left of the augmentation bar, and has the vector [13 −23 60] as a column to the right of the augmentation bar. Hence, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ −1 −10 7 ¯¯ 13 1 0 0 ¯¯ −2 ⎣ −5 3 −12 ¯¯ −23 ⎦ to obtain ⎣ 0 1 0 ¯¯ 1 ⎦  11 −8 30 ¯ 60 0 0 1 ¯ 3

The unique solution (−2 1 3) for the system corresponding to this matrix gives us the coefficients for the desired linear combination. Hence, [13 −23 60] = −2q1 + q2 + 3q3 .

(b) As in part (a), for each q , we need to solve a linear system whose augmented matrix of the form [r1 r2 r3 | q ], where the vectors listed represent columns of the matrix. But, using the method for solving several systems simultaneously, we can solve all three problems together by row reducing [r1 r2 r3 | q1 q2 q3 ]. Hence, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 3 2 4 ¯¯ −1 −10 7 1 0 0 ¯¯ 3 2 1 ⎣ −2 1 −1 ¯¯ −5 3 −12 ⎦ to get ⎣ 0 1 0 ¯¯ −1 2 −6 ⎦  ¯ ¯ 4 −3 2 11 −8 30 0 0 1 −2 −5 4 Reading the three solutions (3 −1 −2), (2 2 −5), and (1 −6 4) from the reduced matrix gives the coefficients needed for the desired linear combinations. Hence, q1 = 3r1 − r2 − 2r3 , q2 = 2r1 + 2r2 − 5r3 , and q3 = r1 − 6r2 + 4r3 .

(c) Combining the answers from parts (a) and (b) yields

[13 −23 60] = −2q1 + q2 + 3q3 = −2 (3r1 − r2 − 2r3 ) + (2r1 + 2r2 − 5r3 ) + 3 (r1 − 6r2 + 4r3 ) = −6r1 + 2r2 + 4r3 + 2r1 + 2r2 − 5r3 + 3r1 − 18r2 + 12r3 = −r1 − 14r2 + 11r3  ⎡

⎤ 1 0 −1 2 (11) (a) (i) A row reduces to B = ⎣ 0 1 3 2 ⎦.

0 0 0 0

(ii) To check that the th row, b , of B is in the row space of A, we must solve the system whose augmented matrix is [A |b ], where b is a column. But, we can solve this problem for the two nonzero rows of B simultaneously by row reducing the matrix [A |b1 b2 ]. Performing this row reduction produces the matrix ¯ ⎡ ⎤ ¯ 1 0 − 38 ¯ − 78 41 ⎢ ⎥ ¯ ⎢ 0 1 ⎥ 1 1 ¯ 0 ⎢ ⎥ ¯ 2 ⎢ ⎥ ¯ 2 ¯ ⎣ 0 0 0 ¯ 0 0 ⎦ 0 0 0 ¯ 0 0 Both systems have infinitely many solutions. Hence, there are many ways to express the two nonzero rows of B as linear combinations of the rows of A. The two solution sets are {( 38  − 78  − 12  + 12  ) |  ∈ R} for b1 and {( 38  + 14  − 12  ) |  ∈ R} for b2  If we find particular solutions by choosing  = 0, we get [1 0 −1 2] = − 78 [0 4 12 8] + 12 [2 7 19 18] + 0[1 2 5 6], and

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

59

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.3

[0 1 3 2] = 14 [0 4 12 8] + 0[2 7 19 18] + 0[1 2 5 6]. (iii) In this part, we perform the same operations as in part (ii), except that we reverse the roles of A and B. Hence, we row reduce the matrix [b1 b2 |A ], which yields ¯ ⎡ ⎤ 1 0 ¯¯ 0 2 1 ⎢ 0 1 ¯ 4 7 2 ⎥ ⎢ ¯ ⎥ ⎣ 0 0 ¯ 0 0 0 ⎦ ¯ 0 0 ¯ 0 0 0

In this case, each of the three linear systems has a unique solution. These three solutions are (0 4), (2 7), and (1 2). Each gives the coefficients for expressing a row of A as a linear combination of the two nonzero rows of B. Hence, we have [0 4 12 8] = 0[1 0 −1 2] + 4[0 1 3 2], [2 17 19 18] = 2[1 0 −1 2] + 7[0 1 3 2], and [1 2 5 6] = 1[1 0 −1 2] + 2[0 1 3 2].

(13) (a) Suppose we are performing row operations on an  ×  matrix A. Throughout this part, we use the notation hBi for the th row of a matrix B. For the Type (I) operation  : hi ← hi: Now −1 is hi ← 1 hi. Clearly,  and −1 change only the th row of A. We want to show that −1  leaves hAi unchanged. But h−1 ((A))i = 1 h(A)i = 1 (hAi ) = hAi .

For the Type (II) operation  : hi ← hi + hi: Now −1 is hi ← −hi + hi. Again,  and −1 change only the th row of A, and we need to show that −1  leaves hAi unchanged.

But h−1 ((A))i = −h(A)i + h(A)i = −hAi + h(A)i = −hAi + hAi + hAi = hAi .

For the Type (III) operation  : hi ↔ hi: Now, −1 = . Also,  changes only the th and th rows of A, and these get swapped. Obviously, a second application of  swaps them back to where they were, proving that  is indeed its own inverse. (b) Suppose C is row equivalent to D, and D is row equivalent to E. Then by the definition of row equivalence, there is a sequence 1  2    of row operations converting C to D, and a sequence 1  2    of row operations converting D to E. But then the combined sequence 1  2     1  2    of row operations converts C to E, and so C is row equivalent to E. (c) An approach similar to that used for Type (II) operations in the abridged proof of Theorem 2.5 in the text works just as easily for Type (I) and Type (III) operations.

For Type (I) operations: Suppose that the original system has the form ⎧ 11 1 + 12 2 + 13 3 + · · · + 1  = 1 ⎪ ⎪ ⎪ ⎨ 21 1 + 22 2 + 23 3 + · · · + 2  = 2 .. .. .. .. .. .. ⎪ . . . . . . ⎪ ⎪ ⎩ 1 1 + 2 2 + 3 3 + · · · +   = 

and that the row operation used is  : hi ←  hi. When  is applied to the corresponding augmented matrix, all rows except the th row remain unchanged. The new th equation then has the form (1 )1 +(2 )2 +· · ·+( ) =  . We must show that any solution (1  2       ) of the original system is a solution of the new one. Now, since (1  2       ) is a solution of the th equation in the original system, we have 1 1 + 2 2 + · · · +   =  . Multiplying this equation by  yields (1 )1 + (2 )2 + · · · + ( ) =  , and so (1  2       ) is also a c 2016 Elsevier Ltd. All rights reserved.

Copyright °

60

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.3

solution of the new th equation. Now, since none of the other equations in the system have changed, (1  2       ) is still a solution for each of them. Therefore, (1  2       ) is a solution to the new system formed.

For Type (III) operations: A Type (III) row operation merely changes the order of the rows of a matrix. Since the rows of the augmented matrix correspond to equations in the system, performing a Type (III) row operation only changes the order in which the equations are written, without making any changes to the actual equations at all. Hence, any solution to the first system of equations must also be a solution to the second system of equations, since none of the actual equations themselves have changed.

There is, however, an alternate approach to this problem that manages to prove the result for all three types of row operations at once: Suppose  is a row operation, and let X satisfy AX = B. Multiplying both sides of this matrix equation by the matrix (I) yields (I)AX = (I)B, implying (IA)X = (IB), by Theorem 2.1. Thus, (A)X = (B), showing that X is a solution to the new linear system obtained from AX = B after the row operation  is performed.

You might wonder what motivates the second method of proof used here. The technique is based on the topic of Elementary Matrices, which is covered in Section 8.6 of the textbook. You will have sufficient background to read that section after you finish studying Chapter 2.

(14) The zero vector is a solution to AX = 0, but it is not a solution for AX = B. Hence these two systems have different solution sets, and thus can not be equivalent systems.

(15) Consider the systems ½

 +   + 

= 1 = 0

and

½

 −   − 

= 1  = 2

The augmented matrices for these systems are, respectively, ¯ ¯ ∙ ¸ ∙ ¸ 1 1 ¯¯ 1 1 −1 ¯¯ 1 and  1 1 ¯ 0 1 −1 ¯ 2

The reduced row echelon matrices for these are, respectively, ¯ ¯ ∙ ¸ ∙ ¸ 1 1 ¯¯ 0 1 −1 ¯¯ 0 and  0 0 ¯ 1 0 0 ¯ 1

where we have continued the row reduction beyond the augmentation bar. Thus, the original augmented matrices are not row equivalent, since their reduced row echelon forms are different. However, the reduced row echelon form matrices reveal that both systems are inconsistent. Therefore, they both have the same solution set, namely, the empty set. Hence, the systems are equivalent.

(17) Because A and B are row equivalent, A =  (· · · (2 (1 (B))) · · · ) for some row operations 1       .

Now, if D is the unique reduced row echelon form matrix to which A is row equivalent, then for some additional row operations +1      + , D = + (· · · (+2 (+1 (A))) · · · ) = + (· · · (+2 (+1 ( (· · · (2 (1 (B))) · · · )))) · · · ) showing that B also has D as its reduced echelon form matrix. Therefore, by the definition of rank, rank(B) = the number of nonzero rows in D = rank(A). c 2016 Elsevier Ltd. All rights reserved.

Copyright °

61

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.4

(19) Let a1      a represent the rows of A, and let b1      b represent the rows of B. For the Type (I) operation  : hi ← hi: Now b = 0a1 + 0a2 + · · · + a + 0a+1 + · · · + 0a , and, for  6= , b = 0a1 + 0a2 + · · · + 1a + 0a+1 + · · · + 0a . Hence, each row of B is a linear combination of the rows of A, implying it is in the row space of A. For the Type (II) operation  : hi ← hi + hi: Now b = 0a1 + 0a2 + · · · + a + 0a+1 + · · · + a + 0a+1 +    + 0a , where our notation assumes   . (An analogous argument works for   .)

And, for  6= , b = 0a1 + 0a2 + · · · + 1a + 0a+1 + · · · + 0a . Hence, each row of B is a linear combination of the rows of A, implying it is in the row space of A. For the Type (III) operation  : hi ↔ hi: Now, b = 0a1 +0a2 +· · ·+1a +0a+1 +· · ·+0a , b = 0a1 +0a2 +· · ·+1a +0a+1 +· · ·+0a , and, for  6= ,  6= , b = 0a1 +0a2 +· · ·+1a +0a+1 +· · ·+0a .

Hence, each row of B is a linear combination of the rows of A, implying it is in the row space of A. (20) Let  be the number of matrices between A and B when performing row operations to get from A to B. Use a proof by induction on .

Base Step: If  = 0, then there are no intermediary matrices, and Exercise 19 shows that the row space of B is contained in the row space of A. Inductive Step: Given the chain A → D1 → D2 → · · · → D → D+1 → B we must show that the row space of B is contained in the row space of A. The inductive hypothesis shows that the row space of D+1 is in the row space of A, since there are only  matrices between A and D+1 in the chain. Thus, each row of D+1 can be expressed as a linear combination of the rows of A. But by Exercise 19, each row of B can be expressed as a linear combination of the rows of D+1 .

Hence, by Lemma 2.8, each row of B can be expressed as a linear combination of the rows of A, and therefore is in the row space of A. By Lemma 2.8 again, the row space of B is contained in the row space of A. (22) (a) True. This is the statement of Theorem 2.5. (b) True. In general, if two matrices are row equivalent, then they have the same rank. This is asserted in Exercise 17, whose proof appears above. (c) False. The inverse of the Type (I) row operation hi ←  hi is the Type (I) row operation hi ← 1 hi.

(d) False. The statement is true for homogeneous systems, but is false in general. If the system in question is not homogeneous, it could be inconsistent, and thus have no solutions at all, and ½ ++ =1 hence, no nontrivial solutions. For a particular example, consider the system . ++ =2 This system has three variables, but the rank of its augmented matrix is  3, since the matrix has only two rows. However, the system has no solutions because it is impossible for  +  +  to equal both 1 and 2.

(e) False. This statement directly contradicts part (2) of Theorem 2.7. (f) True. By the definition of row space, x is clearly in the row space of A. But Theorem 2.9 shows that A and B have the same row space. Hence x is also in the row space of B.

Section 2.4 (2) To find the rank of a matrix, find its corresponding reduced row echelon form matrix and count its nonzero rows. An  ×  matrix is nonsingular if and only if its rank equals . c 2016 Elsevier Ltd. All rights reserved.

Copyright °

62

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.4

(a) The given 2 × 2 matrix row reduces to I2 , which has two nonzero rows. Hence, the original matrix has rank = 2. Therefore, the matrix is nonsingular. (c) The given 3 × 3 matrix row reduces to I3 , which has three nonzero rows. Hence, the original matrix has rank = 3. Therefore, the matrix is nonsingular. ⎡ ⎤ 1 0 0 0 ⎢ 0 1 0 0 ⎥ ⎥ (e) The given 4 × 4 matrix row reduces to ⎢ ⎣ 0 0 1 −2 ⎦  which has three nonzero rows. Hence, 0 0 0 0 the rank of the original matrix is 3. But this is less than the number of rows in the matrix, and so the original matrix is singular.

(3) Use the formula for the inverse of a 2 × 2 matrix given in Theorem 2.14. (a) First we compute  = (4)(−3) − (2)(9) = −30. Since  6= 0, the given matrix is nonsingular, and ∙ ¸ " 1 1 # 10 15 −3 −2 1 = . its inverse is −30 3 −9 4 −2 10

15

(c) First we compute  = (−3)(−8) − (5)(−12) = 84. Since  6= 0, the given matrix is nonsingular, ∙ ¸ " −2 −5 # 21 84 −8 −5 1 = and its inverse is 84 .

1 1 12 −3 − 28 7 (e) First we compute  = (−6)(−8) − (12)(4) = 0. Since  = 0, the given matrix is singular; it has no inverse.

(4) To find the inverse of an  ×  matrix A, row reduce [A|I ] to obtain [I |A−1 ]. If row reduction does not produce I to the left of the augmentation bar, then A does not have an inverse. ¯ ¯ ⎡ ⎤ ⎡ ⎤ −4 7 6 ¯¯ 1 0 0 1 0 0 ¯¯ 1 3 2 2 ⎦. (a) We row reduce ⎣ 3 −5 −4 ¯¯ 0 1 0 ⎦ to obtain ⎣ 0 1 0 ¯¯ −1 0 −2 4 3 ¯ 0 0 1 0 0 1 ¯ 2 2 −1 ⎡ ⎤ 1 3 2 2 ⎦ is the inverse of the original matrix.

Therefore, ⎣ −1 0 2 2 −1 ¯ ⎡ ¯ ⎡ ⎤ 1 ⎤ 1 0 0 ¯¯ 32 0 2 2 −2 3 ¯¯ 1 0 0 ⎥ ⎢ ¯ 9 ¯¯ 0 1 0 ⎦ to obtain ⎣ 0 1 0 ¯ −3 12 − 12 ⎦. (c) We row reduce ⎣ 8 −4 ¯ −4 6 −9 ¯ 0 0 1 0 0 1 ¯ − 83 31 − 23 ⎡ 3 1 ⎤ 0 2 2 ⎢ ⎥ 1 1 ⎥ Therefore, ⎢ ⎣ − 3 2 − 2 ⎦ is the inverse of the original matrix. − 83 31 − 23 ¯ ⎡ 2 0 −1 3 ¯¯ 1 ⎢ 1 −2 3 1 ¯¯ 0 (e) We row reduce ⎢ ⎣ 4 1 0 −1 ¯¯ 0 1 3 −2 −5 ¯ 0

0 1 0 0

0 0 1 0

⎡ ⎤ 1 0 ⎢ ⎢ 0 0 ⎥ ⎥ to ⎢ 0 ⎦ ⎢ ⎣ 0 1 0

0 0

1 0 0 1 0 0

¯ 1 1 2 ¯ ¯ 5 15 15 ¯ 4 4 7 ¯ − 5 − 15 15 ¯ ¯ −3 2 4 ¯ 5 15 15 ¯ 0 ¯ 1 1 −1

8 15 − 47 15 − 29 15

0

⎤

⎥ 0 ⎥ ⎥.

0 ⎥ ⎦ 1

Therefore, since I4 is not obtained to the left of the augmentation bar, the original matrix does

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

63

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 2.4

not have an inverse. (Note: If you use a calculator to perform the row reduction above, it might continue to row reduce beyond the augmentation bar.)

(5) (c) To find the inverse of a diagonal matrix, merely find the diagonal matrix whose main diagonal entries are the reciprocals of the main diagonal entries of the original matrix. Hence, when this matrix is multiplied by the original, the product will be a diagonal matrix with all 1’s on the main diagonal — that is, the identity matrix. Therefore, the inverse matrix we need is ⎤ ⎡ 1 0 ··· 0 11 ⎥ ⎢ 1 ⎢ 0 ··· 0 ⎥ 22 ⎥ ⎢  ⎢ . .. ⎥ .. .. ⎥ ⎢ .. . . . ⎦ ⎣ 1 0 0 · · ·  (6) (a) Using Theorem 2.14,  = (cos )(cos ) − (sin )(− sin ) = cos2  + sin2  = 1. Therefore, the ∙ ¸ cos  sin  general inverse is . Plugging in the given values for  yields: − sin  cos    6

 4

 2

Original Matrix

Inverse Matrix

" √3

− 12

1 2

√ 3 2

#

"

" √2

− 22

#

"

2

2 √ 2 2

∙

√

√ 2 2

0 −1 1 0

¸

√ 3 2

− 12 √ 2 2 √ − 22

∙

1 2 √ 3 2

#

√ 2 2 √ 2 2

#

0 1 −1 0

¸

(b) It is easy to see that if A and B are 2 × 2 matrices with AB = C, then ⎤⎡ ⎤ ⎡ ⎤ ⎡ 11 12 0 11 12 0 11 12 0 ⎣ 21 22 0 ⎦ ⎣ 21 22 0 ⎦ = ⎣ 21 22 0 ⎦. (Check it out!) Therefore, to find the 0 0 1 0 0 1 0 0 1

inverse of the matrix given in part (b), we merely use the associated 2 × 2 inverses we found in part (a), glued into the upper corner of a 3 × 3 matrix, as illustrated. Hence, the general inverse ⎡ ⎤ cos  sin  0 is ⎣ − sin  cos  0 ⎦ . Substituting the given values of  produces: 0 0 1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

64

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition



Original Matrix

 6

⎡ √3

 4

 2

2

⎢ ⎢ ⎣

1 2

0

⎡ √2 ⎢ ⎢ ⎣

2 √ 2 2

0 ⎡

− 12 √ 3 2

0

⎤

⎥ 0 ⎥ ⎦

0 1 √

− 22 √ 2 2

0

⎤

⎥ 0 ⎥ ⎦

0 1

⎤ 0 −1 0 ⎣ 1 0 0 ⎦ 0 0 1

Section 2.4

Inverse Matrix ⎡

√ 3 2

1 2 √ 3 2

⎢ ⎢ −1 ⎣ 2

0

⎡ ⎢ ⎢ ⎣

0

⎤

⎥ 0 ⎥ ⎦

0 1

√ 2 2 √ − 22

0 ⎡

√ 2 2 √ 2 2

0

⎤

⎥ 0 ⎥ ⎦

0 1

⎤ 0 1 0 ⎣ −1 0 0 ⎦ 0 0 1

(7) By Theorem 2.16, if A is a nonsingular matrix, then the unique solution to AX = B is A−1 B. ¸ ∙ ¸ ∙ ¸∙ 20 5 −1 1 = . We (a) The given system corresponds to the matrix equation −31 2 −7 2 ∙ ¸ 5 −1 use Theorem 2.14 to compute the inverse of . First,  = (5)(2) − (−7)(−1) = 3. −7 2 " 2 1 # ∙ ¸ 3 3 2 1 Hence, the inverse = 13 = . The unique solution to the system is thus 7 5 7 5 3 3 " 2 1 #∙ ¸ ∙ ¸ 3 3 20 3 = . That is, the solution set for the system is {(3 −5)}.

7 5 −31 −5 3 3 ⎡ ⎤ ⎡ ⎤⎡ ⎤ 0 −2 5 1 25 1 ⎢ −7 −4 ⎥ ⎢ ⎢ ⎥ 5 22 ⎥ ⎥ ⎢ 2 ⎥ = ⎢ −15 ⎥. (c) The system corresponds to the matrix equation ⎢ ⎣ 5 ⎣ 3 −4 −16 ⎦ ⎣ 3 ⎦ 9 ⎦ −3 −1 0 9 4 −16 We find the inverse of the coefficient matrix by row reducing ¯ ⎡ ⎤ ⎡ 0 −2 5 1 ¯¯ 1 0 0 0 1 0 ⎢ −7 −4 ¯ 0 1 0 0 ⎥ ⎢ 0 1 5 22 ⎢ ¯ ⎥ to obtain ⎢ ⎣ 5 ⎣ 0 0 3 −4 −16 ¯¯ 0 0 1 0 ⎦ ¯ −3 −1 0 9 0 0 0 1 0 0

0 0 1 0

¯ ⎤ 0 ¯¯ 1 −13 −15 5 0 ¯¯ −3 3 0 −7 ⎥ ⎥ ¯ 0 ¯ −1 2 1 −3 ⎦ 1 ¯ 0 −4 −5 1

The inverse of the coefficient matrix is the 4 × 4 matrix to the right of the augmentation bar in the row reduced matrix. Hence, the unique solution to the system is ⎡ ⎤⎡ ⎤ ⎡ ⎤ 1 −13 −15 5 25 5 ⎢ −3 ⎢ ⎥ ⎢ ⎥ 3 0 −7 ⎥ ⎢ ⎥ ⎢ −15 ⎥ = ⎢ −8 ⎥  ⎣ −1 ⎦ ⎣ ⎦ ⎣ 2 1 −3 9 2 ⎦ 0 −4 −5 1 −16 −1 That is, the solution set for the linear system is {(5 −8 2 −1)}.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

65

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

(8) (a) Through trial and error, we find the involutory matrix

∙

0 1 1 0

Section 2.4

¸ .

(b) Using the answer to part (a), the comments at the beginning of the answer to Section 2.4, Exercise ⎡ ⎤ 0 1 0 6(b) in this manual suggests that we consider the matrix ⎣ 1 0 0 ⎦. Squaring this matrix 0 0 1 verifies that it is, in fact involutory. (c) If A is involutory, then A2 = AA = I . Hence, A itself satisfies the definition of an inverse for A, and so A−1 = A. (10) (a) Since A is nonsingular, A−1 exists. Hence AB = O =⇒ A−1 (AB) = A−1 O =⇒ (A−1 A)B = O =⇒ I B = O =⇒ B = O . Thus, B must be the zero matrix. (b) No. AB = I implies B = A−1 , and so A is nonsingular. We can now apply part (a), substituting C where B appears, proving that AC = O =⇒ C = O .

(11) Now A4 = I =⇒ A3 A = I =⇒ A−1 = A3 . Also, since I = I for every integer , we see that A4 = (A4 ) = I = I for every integer . Hence, for every integer , A4+3 = A4 A3 = I A3 = A3 = A−1 .

Thus, all  ×  matrices of the form A4+3 equal A−1 . These are     A−9  A−5  A−1  A3  A7  A11    .

Now, could any other powers of A also equal A−1 ? Suppose that A = A−1 . Then, A+1 = A A = A−1 A = I . Now, if  + 1 = 4 for some integer , then  = 4( − 1) + 3, and so  is already on our list of powers of A such that A = A−1 . On the other hand, if  + 1 is not a multiple of 4, then, dividing  + 1 by 4 yields an integer , with a remainder  of either 1, 2, or 3. Then  + 1 = 4 + , and so A4+ = I implying A4 A = I . But A4 = I from above, and so A = I . But, since  equals either 1, 2, or 3, this contradicts the given conditions of the problem, and so  + 1 can not equal 4 + . Therefore, the only powers of A that equal A−1 are of the form A4+3 .

(12) By parts (1) and (3) of Theorem 2.12, B−1 A is the inverse of A−1 B. (Proof: (A−1 B)−1 = B−1 (A−1 )−1 = B−1 A.) Thus, if A−1 B is known, simply compute its inverse to find B−1 A. (14) (a) Suppose first that the matrix contains a column of zeroes. No row operations can alter a column of zeroes, so the unique reduced row echelon form for the matrix must also contain a column of zeroes, and thus cannot equal I . Hence the matrix has rank less than  But by Theorem 2.15, an  ×  matrix is nonsingular if and only if its rank equals . Thus, the matrix must be singular.

Suppose a matrix contains a row of all zeroes. We assume that it is nonsingular and derive a contradiction. If the matrix is nonsingular, then part (4) of Theorem 2.12 says that its transpose is also nonsingular. But the transpose contains a column of all zeroes. This contradicts the result obtained in the first paragraph.

(15) (a) Part (1): Since AA−1 = I , we must have (A−1 )−1 = A. Part (2): For   0, to show (A )−1 = (A−1 ) , we must show that A (A−1 ) = I . We proceed by induction on .

Base Step: For  = 1, clearly AA−1 = I .

Inductive Step: Assume A (A−1 ) = I . Prove A+1 (A−1 )+1 = I .

Now, A+1 (A−1 )+1 = AA (A−1 ) A−1 = AIn A−1 = AA−1 = I . This concludes the proof for   0.

We now show A (A−1 ) = I for  ≤ 0.

For  = 0, clearly A0 (A−1 )0 = I I = I . The case  = −1 is covered by part (1) of the theorem.

For  ≤ −2, (A )−1 = ((A−1 )− )−1 (by definition) = ((A− )−1 )−1 (by the   0 case) = A− (by part (1)). c 2016 Elsevier Ltd. All rights reserved.

Copyright °

66

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 2 Review

(17) (a) Let  = −,  = −. Then    0. Now, A+ = A−(+) = (A−1 )+ = (A−1 ) (A−1 ) (by Theorem 1.17) = A− A− = A A .

(21) Suppose A is an  ×  matrix and B is a  ×  matrix. Suppose, further, that AB = I and   . By Corollary 2.3, the homogeneous system having B as its matrix of coefficients has a nontrivial solution X. That is, there is a nonzero vector X such that BX = 0. But then X = I X = (AB)X = A(BX) = A0 = 0, a contradiction.

(22) (a) False. Many  ×  matrices are singular, having no inverse. For example, the 2 × 2 matrix ∙ ¸ 6 3 has no inverse, by Theorem 2.14, because  = (6)(4) − (3)(8) = 0.

8 4 (b) True. This follows directly from Theorem 2.10.

(c) True. ((AB) )−1 = ((AB)−1 ) (by part (4) of Theorem 2.12) = (B−1 A−1 ) (by part (3) of Theorem 2.12) = (A−1 ) (B−1 ) (by Theorem 1.18).

¸ 11 12 is nonsingular (d) False. This statement contradicts Theorem 2.14, which states that 21 22 if and only if  = 11 22 − 12 21 6= 0. Consider O2 for a specific counterexample. ∙

(e) False. This statement contradicts the Inverse Method given in Section 2.4 as well as Theorem 2.15. To correct the statement, change the word “nonsingular” to “singular.” Consider O2 for a specific counterexample for the “if” half and I2 as a specific counterexample for the “only if.” (f) True. This follows directly from combining Theorems 2.15 and 2.16.

Chapter 2 Review Exercises (1) (a) First augmented matrix:

¯ ⎤ 2 5 −4 ¯¯ 48 ⎢ 1 −3 2 ¯¯ −40 ⎥ ⎢ ⎥ ⎣ −3 4 7 ¯¯ 15 ⎦ −2 3 −1 ¯ 41 ⎡

The first pivot is the (1 1) entry. We make this “1” using a Type (I) row operation: ¯ ⎤ 24 −2 ¯ ¯ ⎢ 1 −3 2 ¯¯ −40 ⎥ ⎢ ⎥ ⎣ −3 15 ⎦ 4 7 ¯¯ −2 3 −1 ¯ 41 ⎡

(I): h1i ← 12 h1i

1 °

5 2

We now target the (2 1), (3 1), and (4 1) entries to make them “0” using Type (II) row operations: ¯ ⎤ ⎡ 5 1 −2 ¯¯ 24 ° 2 ¯ ⎥ ⎢ (II): h2i ← (−1) × h1i + h2i ⎢ 0 − 11 4 ¯ −64 ⎥ 2 ¯ ⎥ ⎢ (II): h3i ← (3) × h1i + h3i ⎢ 0 23 1 ¯¯ 87 ⎥ ⎦ ⎣ 2 (II): h4i ← (2) × h1i + h4i ¯ 0 8 −5 ¯ 89

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

67

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 2 Review

Now we move the pivot to the (2 2) entry. We make this entry a “1” using a Type (I) row operation: ¯ ⎡ ⎤ 5 −2 ¯¯ 24 1 2 ⎢ 8 ¯¯ 128 ⎥ ⎢ 0 ° ⎥ 1 − 11 2 ⎢ ¯ 11 ⎥  h2i (I): h2i ← − 11 ⎢ 0 23 ¯ 1 ¯ 87 ⎥ ⎣ ⎦ 2 ¯ 0 8 −5 ¯ 89 Now we target the (3 2) and (4 2) entries using the following Type (II) row operations: ¯ ⎤ ⎡ −2 ¯¯ 1 52 24 ⎢ 128 ⎥ 8 ¯¯ ⎥ ⎢ 0 ° 1 − 11 (II): h3i ← (− 23 11 ⎥ 2 ) × h2i + h3i ⎢ ¯ ⎢ 0 0 103 ¯ − 515 ⎥  (II): h4i ← (−8) × h2i + h4i ⎣ 11 ¯ 11 ⎦ ¯ 9 ¯ 0 0 − 45 11 11

Now we move the pivot to the (3 3) entry. We make this entry a “1” using a Type (I) row operation: ¯ ⎡ ⎤ 24 −2 ¯¯ 1 52 ⎢ 8 ¯¯ 128 ⎥ ⎢ 0 1 − 11 ⎥ 11 ⎥ 11 ⎢ ¯ (I): h3i ← 103 h3i  ⎢ 0 0 ¯ 1 ¯ −5 ⎥ ° ⎣ ⎦ ¯ 9 ¯ 0 0 − 45 11 11 We target the (4 3) entry using the following Type (II) operation: ¯ ⎡ ⎤ −2 ¯¯ 24 1 52 ⎢ 8 ¯¯ 128 ⎥ ⎢ 0 1 − 11 ⎥ 9 ⎢ ¯ 11 ⎥  (II): h4i ← (− 11 ) × h3i + h4i ⎢ 0 0 ¯ 1 ¯ −5 ⎥ ° ⎣ ⎦ ¯ 0 0 0 0 ¯

This finishes the third column. Since there are no more columns before the augmentation bar, we are finished performing row operations. The linear system corresponding to the final matrix is: ⎧ 1 + 52 2 − 23 = 24 ⎪ ⎪ ⎪ ⎨ 8 2 − 11 3 = 128 11 ⎪ ⎪ 3 = −5 ⎪ ⎩ 0 = 0

The last equation, 0 = 0, provides no information regarding the solution set for the system because it is true for every value of 1 , 2 , and 3 . Thus, we can ignore it. The third equation 8 (−5) = 128 clearly gives 3 = −5. Plugging this value into the second equation yields 2 − 11 11 , or, 2 = 8. Substituting the values we have for 2 and 3 into the first equation produces 1 + 52 (8) − 2(−5) = 24, or, 1 = −6. Hence, the original linear system has a unique solution.

The complete solution set is {(−6 8 −5)}.

(b) First augmented matrix:

¯ ⎤ 4 3 −7 5 ¯¯ 31 ⎢ −2 −3 5 −1 ¯¯ −5 ⎥ ⎢ ⎥ ⎣ 2 −6 −2 3 ¯¯ 52 ⎦ 6 −21 −3 12 ¯ 16 ⎡

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

68

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 2 Review

The first pivot is the (1 1) entry. We make this a “1” by performing the following Type (I) row operation: ¯ 31 ⎤ ⎡ 3 5 ¯ 1 − 74 ° 4 4 ¯ 4 ⎢ ⎥ ¯ ⎢ ¯ 5 −1 ¯ −5 ⎥ (I): h1i ← 14 h1i ⎢ −2 −3 ⎥ ⎣ 2 −6 −2 3 ¯¯ 52 ⎦ 6 −21 −3 12 ¯ 16

We now target the (2 1), (3 1), and (4 1) entries, and make them “0.” We do this using the following Type (II) row operations: ¯ ⎤ ⎡ 3 31 1 − 74 45 ¯¯ ° 4 4 ⎥ ¯ ⎢ 21 ⎥ 3 3 ¯ (II): h2i ← (2) × h1i + h2i ⎢ 0 −3 2 2 2 ¯ 2 ⎥ ⎢ (II): h3i ← (−2) × h1i + h3i ⎥ ¯ ⎢ 73 ⎥ 3 1 ¯ ⎢ 0 − 15 (II): h4i ← (−6) × h1i + h4i 2 2 2 ¯ 2 ⎦ ⎣ ¯ 15 9 ¯ 0 − 51 − 61 2 2 2 2

Now we move the pivot to the (2 2) entry. We make this entry a “1” using a Type (I) row operation. ¯ ⎤ ⎡ 3 31 5 ¯ − 74 1 4 4 ¯ 4 ⎥ ⎢ ¯ ⎢ 0 1 −1 −1 ¯¯ −7 ⎥ ° ⎥ ⎢ (I): h2i ← − 23 h2i ⎥ ⎢ ¯ 3 1 ¯ 73 ⎥ ⎢ 0 − 15 2 2 2 ¯ 2 ⎦ ⎣ ¯ 15 9 ¯ 0 − 51 − 61 2 2 2 2 Now we target the (3 2) and (4 2) entries using the following Type (II) row operations. ¯ ⎤ ⎡ 5 ¯ 31 1 34 − 74 4 ¯ 4 ⎥ ⎢ ¯ ⎢ 0 ° 1 −1 −1 ¯¯ −7 ⎥ (II): h3i ← ( 15 2 ) × h2i + h3i ⎥ ⎢ ⎥ ⎢ ¯ (II): h4i ← ( 51 2 ) × h2i + h4i ⎣ 0 0 −6 −7 ¯ −16 ⎦ ¯ 0 0 −18 −21 ¯ −209

Now we move the pivot to the (3 3) entry. We make this entry a “1” using a Type (I) row operation. ¯ ⎤ ⎡ 5 ¯ 31 1 34 − 74 4 ¯ 4 ⎥ ⎢ ¯ ⎢ 0 1 −1 −1 ¯ −7 ⎥ ⎥ ⎢ ¯ 1 (I): h3i ← − 6 h3i ⎥ ⎢ ¯ 7 ¯ 8 ⎥ ⎢ 0 0 1 ° 6 ¯ 3 ⎦ ⎣ ¯ 0 0 −18 −21 ¯ −209 We target the (4 3) entry using the following Type (II) row operation: ¯ ⎤ ⎡ 31 5 ¯ 1 34 − 74 4 ¯ 4 ⎥ ⎢ ¯ ⎢ 0 1 −1 −1 ¯ −7 ⎥ ⎥ ⎢ ¯ (II): h4i ← (18) × h3i + h4i ⎥ ⎢ ¯ 7 ¯ 8 ⎥ ⎢ 0 0 ° 1 6 ¯ 3 ⎦ ⎣ ¯ 0 0 0 0 ¯ −161

The last row of this augmented matrix corresponds to the equation 0 = −161. Since this equation can not be satisfied for any values of 1 , 2 , and 3 , the original system of linear equations has no solutions. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

69

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

(c) First augmented matrix:

Chap 2 Review

¯ ⎤ 6 −2 2 −1 −6 ¯¯ −33 ⎣ −2 1 0 2 −1 ¯¯ 13 ⎦  4 −1 2 −3 1 ¯ −24 ⎡

The first pivot is the (1 1) entry. We make this a “1” by performing the following Type (I) row operation: ¯ ⎡ ⎤ 1 − 13 31 − 16 −1 ¯¯ − 11 ° 2 ⎢ ⎥ ¯ (I): h1i ← 16 h1i ⎣ −2 1 0 2 −1 ¯¯ 13 ⎦  4 −1 2 −3 1 ¯ −24

We now target the (2 1) and (3 1) entries, and make them “0.” We do this using the following Type (II) row operations: ¯ ⎤ ⎡ 1 − 13 31 − 16 −1 ¯¯ − 11 ° 2 ⎥ ⎢ ¯ (II): h2i ← (2) × h1i + h2i 1 2 5 ⎥ ⎢ 0 ¯ −3 2 3 3 3 ¯ ⎦ ⎣ (II): h3i ← (−4) × h1i + h3i ¯ 2 1 7 ¯ −2 0 − 5 3 3 3

Now we move the pivot to the (2 2) entry. We make this entry a “1” using a Type (I) row operation. ¯ ⎤ ⎡ 1 − 13 31 − 16 −1 ¯¯ − 11 2 ⎥ ⎢ ¯ ⎢ 0 ° 1 2 5 −9 ¯¯ 6 ⎥ (I): h2i ← 3 h2i ⎦ ⎣ ¯ 2 1 7 0 −3 5 ¯ −2 3 3 Now we target the (3 2) entry using the following Type (II) row operation: ¯ ⎡ ⎤ 1 − 13 31 − 16 −1 ¯¯ − 11 2 ⎢ ⎥ ¯ (II): h3i ← (− 13 ) × h2i + h3i ⎣ 0 ° 1 2 5 −9 ¯¯ 6 ⎦ 0 0 0 −4 8 ¯ −4

We attempt to move the pivot to the (3 3) entry. However, that entry is a “0”. There are no entries below it in the matrix, and so a Type (III) operation will not resolve the issue. Thus, we move the pivot horizontally to the (3 4) position. We make this entry a “1” using a Type (I) row operation: ¯ ⎡ ⎤ 1 − 13 31 − 16 −1 ¯¯ − 11 2 ⎢ ⎥ ¯ (I): h3i ← − 14 h3i ⎣ 0 ° 1 2 5 −9 ¯¯ 6 ⎦ 0 0 0 1 −2 ¯ 1

This finishes the fourth column and the third row. Since there are no more rows in the matrix, we are finished performing row operations. The linear system corresponding to the final matrix is: ⎧ 1 1 1 11 ⎪ ⎨ 1 − 3 2 + 3 3 − 6 4 − 5 = − 2 2 + 23 + 54 − 95 = 6 ⎪ ⎩ 1 4 − 25 =

The variables 3 and 5 are independent variables, because there were no pivots in the third and fifth columns. We let 3 =  and 5 = . The last equation gives 4 = 1 + 25 = 1 + 2. Plugging this expression and the values of 3 and 5 into the second equation yields 2 +2+5(1+2)−9 = c 2016 Elsevier Ltd. All rights reserved.

Copyright °

70

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 2 Review

6, which simplifies to 2 = 1 − 2 − . Substituting the expressions we have for 2 through 5 into the first equation produces 1 − 13 (1 − 2 − ) + 13 () − 16 (1 + 2) −  = − 11 2 , which simplifies to 1 = −5 −  + . Hence, the original linear system has an infinite solution set. The full solution set is {[−5 −  +  1 − 2 −   1 + 2 ] |   ∈ R}.

(2) First, plug each of the three given points into the equation  = 3 + 2 +  + . This yields the following linear system: ⎧ 120 = −27 + 9 − 3 +  ←− using (−3 120) ⎪ ⎪ ⎨ 51 = −8 + 4 − 2 +  ←− using (−2 51)  −24 = 27 + 9 + 3 +  ←− using (3 −24) ⎪ ⎪ ⎩ −69 = 64 + 16 + 4 +  ←− using (4 −69) The augmented matrix for this system is: ¯ ⎡ ⎤ −27 9 −3 1 ¯¯ 120 ⎢ −8 4 −2 1 ¯ 51 ⎥ ⎢ ¯ ⎥ ¯ ⎣ 27 9 3 1 ¯ −24 ⎦ 64 16 4 1 ¯ −69

The first pivot is the (1 1) entry. We make this a “1” by performing the following Type (I) row operation: ⎡ ⎤ 1 ¯¯ 1 1 − 13 − 27 ° − 40 9 9 ¯ ⎢ −8 4 −2 1 ¯¯ 51 ⎥ 1 ⎢ ⎥ h1i (I): h1i ← − 27 ¯ ⎣ 27 9 3 1 ¯ −24 ⎦ 64 16 4 1 ¯ −69 We now target the (2 1), (3 1) and (4 1) entries, and make them “0.” We do this using the following Type (II) row operations: ¯ ⎤ ⎡ 1 1 ¯ 40 1 − 13 − 27 ° 9 ¯ −9 ⎢ (II): h2i ← (8) × h1i + h2i 139 ⎥ 4 19 ¯¯ ⎥ ⎢ 0 − 10 3 9 27 ¯ 9 ⎥ ⎢ (II): h3i ← (−27) × h1i + h3i  ¯ ⎢ 0 18 0 2 ¯ 96 ⎥ ⎦ ⎣ (II): h4i ← (−64) × h1i + h4i ¯ 91 ¯ 1939 0 112 − 28 3

9

27

9

The second pivot is the (2 2) entry. We make this a “1” by performing the following Type (I) row operation: ¯ ⎡ ⎤ 1 ¯ 40 1 1 − 13 − 27 ° ¯ −9 9 ⎢ 139 ⎥ 19 ¯¯ ⎢ 0 ° ⎥ 1 − 56 36 ¯ 12 ⎥ 3 ⎢ (I): h2i ← 4 h2i  ⎢ 0 18 ¯ 0 2 ¯ 96 ⎥ ⎣ ⎦ ¯ 91 ¯ 1939 − 28 0 112 3 9 27 9

We now target the (3 2) and (4 2) entries, and make them “0.” We do this using the following Type (II) row operations: ¯ ⎤ ⎡ 1 ¯ 40 1 1 − 13 − 27 ° ¯ −9 9 ⎢ 19 ¯¯ 139 ⎥ ⎥ ⎢ 0 ° 1 − 56 (II): h3i ← (−18) × h2i + h3i 36 ¯ 12 ⎥ ⎢  112 ⎢ ¯ 15 225 (II): h4i ← (− 3 ) × h2i + h4i 0 15 − 2 ¯ − 2 ⎥ ⎦ ⎣ 0 ¯ 0 0 28 − 49 ¯ −217 3

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

71

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 2 Review

The third pivot is the (3 3) entry. operation:

We make this a “1” by performing the following Type (I) row ¯ ⎡ ⎤ 1 ¯ 40 1 1 − 13 − 27 ° ¯ −9 9 ⎢ 139 ⎥ 19 ¯¯ ⎢ 0 ° ⎥ 1 − 56 36 ¯ 12 ⎥ 1 ⎢ h3i (I): h3i ← 15  ⎢ 0 1 ¯ 15 ⎥ 1 0 ° −2 ¯ − 2 ⎦ ⎣ ¯ ¯ −217 0 0 28 − 49 3

We now target the (4 3) entry, and make it “0.” We do this using the following Type (II) row operation: ¯ ⎤ ⎡ 1 ¯ 40 1 1 − 13 − 27 ° ¯ −9 9 ⎢ 19 ¯¯ 139 ⎥ ⎥ ⎢ 0 ° 1 − 56 36 ¯ 12 ⎥ ⎢ (II): h4i ← (−28) × h3i + h4i  ⎢ 0 1 ¯ 15 ⎥ 1 0 ° −2 ¯ − 2 ⎦ ⎣ ¯ 0 0 0 − 73 ¯ −7

The last pivot is the (4 4) entry. We make this a “1” by performing the following Type (I) row operation: ¯ ⎡ ⎤ 1 ¯ 40 1 1 − 13 − 27 ° ¯ −9 9 ⎢ 19 ¯¯ 139 ⎥ ⎢ 0 ° ⎥ 1 − 56 36 ¯ 12 ⎥ 3 ⎢ (I): h4i ← − 7 h4i  ⎢ 0 1 ¯ 15 ⎥ 1 0 ° −2 ¯ − 2 ⎦ ⎣ ¯ 1 ¯ 0 0 0 ° 3 This matrix is in row echelon form. We solve the system using back substitution. The last equation produces  = 3. Plugging this value into the third equation yields − 32 = − 15 2 , or,  = −6. Substituting 139 = , which results in  = 5.

Finally, we use all of these into the second equation gives us  + 5 + 19 12 12 , or,  = −2.

Thus, the desired cubic polynomial values in the first equation to get  − 53 − 23 − 19 = − 40 9 is  = −23 + 52 − 6 + 3.

(4) First we find the system of linear equations corresponding to the given chemical equation by considering each element separately. This produces ⎧ =  ← Nitrogen equation ⎨  3 = 2 ← Hydrogen equation  ⎩ 2 = 2 +  ← Oxygen equation Moving all variables to the left side, and then creating an augmented matrix yields ¯ ⎡ ⎤ 1 0 −1 0 ¯¯ 0 ⎣ 3 0 0 −2 ¯¯ 0 ⎦  0 2 −2 −1 ¯ 0

The (1 1) entry is the first pivot. Since it already equals 1, we do not need a Type (I) operation.

Next, we target the (2 1) entry. (The (3 1) entry is already 0.) ¯ ⎡ ⎤ 1 0 −1 ° 0 ¯¯ 0 ⎣ 0 0 (II): h2i ← (−3) × h1i + h2i 3 −2 ¯¯ 0 ⎦  0 2 −2 −1 ¯ 0

The pivot moves to the (2 2) entry, which is zero. So, we switch the 2nd and 3rd rows to move a nonzero number into the pivot position. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

72

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

(III): h2i ↔ h3i Now we convert the (2 2) entry to 1.

(I): h2i ← ( 12 ) h2i

Chap 2 Review

¯ ⎤ 1 0 −1 0 ¯¯ 0 ⎣ 0 2 −2 −1 ¯ 0 ⎦  ¯ 0 0 3 −2 ¯ 0 ⎡

¯ ⎤ 0 ¯¯ 0 ⎢ ¯ ⎥ 1 −1 − 12 ¯ 0 ⎦  ⎣ 0 ° ¯ 0 0 3 −2 ¯ 0 ⎡

1

0 −1

The pivot moves to the (3 3) entry. We convert the (3 3) entry to 1. ¯ ⎡ ⎤ 1 0 −1 0 ¯¯ 0 ⎢ ¯ ⎥ (I): h3i ← ( 13 ) h3i ⎣ 0 1 −1 − 12 ¯ 0 ⎦  ¯ 1 − 23 ¯ 0 0 0 ° Next, we target the (1 3) and (2 3) entries:

¯ ⎤ 0 − 23 ¯¯ 0 ¯ ⎢ ⎥ ⎣ 0 1 0 − 76 ¯ 0 ⎦  ¯ 1 − 23 ¯ 0 0 0 ° ⎡

(II): h1i ← (1) × h3i + h1i

(II): h2i ← (1) × h3i + h2i

1 0

The matrix is now in reduced row echelon form. It corresponds to the following linear system: ⎧ − 23  = 0 ⎪ ⎨   − 76  = 0  ⎪ ⎩  − 23  = 0

or  = 23 ,  = 76 , and  = 23 . Setting  = 6 eliminates all fractions, and provides the smallest solution containing all positive integers. Hence, the desired solution is  = 4,  = 7,  = 4,  = 6. ⎡ ⎤ ⎡ ⎤ 2 −5 3 1 0 −1 (8) (a) A = ⎣ −1 −3 4 ⎦ row reduces to ⎣ 0 1 −1 ⎦.

7 −12 5 0 0 0 Since the reduced row echelon form matrix has 2 nonzero rows, rank(A) = 2. ⎡ ⎤ 0 0 −2 −1 ⎢ −2 0 −1 0 ⎥ ⎥ B=⎢ ⎣ −1 −2 −1 −5 ⎦ row reduces to I4 .

0 1 1 3 Since the reduced row echelon form matrix has 4 nonzero rows, rank(B) = 4. ⎡ ⎤ ⎡ ⎤ 3 −1 −5 −6 1 0 −1 0 4 8 −2 ⎦ reduces to ⎣ 0 1 2 0 ⎦.

C=⎣ 0 −2 −3 −4 0 0 0 0 1

Since the reduced row echelon form matrix has 3 nonzero rows, rank(C) = 3. (b) We apply Theorem 2.7.

The homogeneous system AX = 0 has 3 variables (because A has 3 columns). Since rank(A) = c 2016 Elsevier Ltd. All rights reserved.

Copyright °

73

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 2 Review

2  3, Theorem 2.7 implies that AX = 0 has nontrivial solutions. Hence, AX = 0 has infinitely many solutions.

The homogeneous system BX = 0 has 4 variables (because B has 4 columns). Since rank(B) = 4, Theorem 2.7 tells us that BX = 0 has only the trivial solution. Hence, BX = 0 has exactly one solution.

The homogeneous system CX = 0 has 4 variables (because C has 4 columns). Since rank(C) = 3  4, Theorem 2.7 indicates that CX = 0 has nontrivial solutions. Hence, CX = 0 has infinitely many solutions.

(10) (a) To express the vector [−34 29 −21] as a linear combination of the vectors x1 = [2 −3 −1] x2 = [5 −2 1] and x3 = [9 −8 3], we need to solve the linear system whose augmented matrix has the vectors x1  x2  and x3 as columns to the left of the augmentation bar, and has the vector [−34 29 −21] as a column to the right of the augmentation bar. A solution of this linear system gives the coefficients for the desired linear combination. Hence, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 2 5 9 ¯¯ −34 1 0 0 ¯¯ 5 ⎣ −3 −2 −8 ¯ 29 ⎦  obtaining ⎣ 0 1 0 ¯ 2 ⎦  ¯ ¯ −1 1 3 ¯ −21 0 0 1 ¯ −6 Hence [−34 29 −21] = 5[2 −3 −1] + 2[5 −2 1] − 6[9 −8 3]. We see that [−34 29 −21] is a linear combination of x1  x2  and x3 . (b) A vector is in the row space of a matrix if and only if the vector can be expressed as a linear combination of the rows of the matrix. Since the rows of the given matrix A are x1  x2  and x3 , and since, in part (a), we showed how to express [−34 29 −21] as a linear combination of x1  x2  and x3 , it follows that [−34 29 −21] is in the row space of A.

(12) To find the inverse of an  ×  matrix B, row reduce [B|I ] to obtain [I |B−1 ]. If row reduction does not produce I to the left of the augmentation bar, then B does not have an inverse. ¯ ¯ ⎡ ⎤ ⎡ ⎤ 3 4 3 5 ¯¯ 1 0 0 0 1 0 0 2 ¯¯ 5 −1 0 −5 ⎢ 4 5 5 8 ¯ 0 1 0 0 ⎥ ⎢ 0 1 0 −1 ¯ −2 0 0 3 ⎥ ¯ ⎥ ⎢ ¯ ⎥. (b) We row reduce ⎢ ¯ −2 ⎣ 7 9 8 13 ¯ 0 0 1 0 ⎦ to ⎣ 0 0 1 1 1 0 1 ⎦ ¯ ¯ 2 3 2 3 ¯ 0 0 0 1 0 0 0 0 ¯ −1 −1 1 0

Therefore, since I4 is not obtained to the left of the augmentation bar, the original matrix does not have an inverse. (Note: If you use a calculator to perform the row reduction above, it might continue to row reduce beyond the augmentation bar.) ⎤ ⎡ ⎤ ⎡ ⎤⎡ 17 4 −6 1 1 2 −1 ⎦ ⎣ 2 ⎦ = ⎣ −14 ⎦. We find the (15) The system corresponds to the matrix equation ⎣ −1 3 23 3 −5 1 inverse of the coefficient matrix by row reducing ¯ ⎡ ⎤ 4 −6 1 ¯¯ 1 0 0 ⎣ −1 2 −1 ¯¯ 0 1 0 ⎦ to obtain 3 −5 1 ¯ 0 0 1

¯ ⎤ 1 0 0 ¯¯ 3 −1 −4 ⎣ 0 1 0 ¯ 2 −1 −3 ⎦  ¯ 0 0 1 ¯ 1 −2 −2 ⎡

The inverse of the coefficient matrix is the 3 × 3 matrix to the right of the augmentation bar in the row reduced matrix. Hence, the unique solution to the system is ⎡ ⎤⎡ ⎤ ⎡ ⎤ 3 −1 −4 17 −27 ⎣ 2 −1 −3 ⎦ ⎣ −14 ⎦ = ⎣ −21 ⎦  1 −2 −2 23 −1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

74

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 2 Review

That is, the solution set for the linear system is {(−27 −21 −1)}.

(17) (a) False. Theorem 2.5 shows that this method produces the precise solution set for the system. (b) False. Theorem 2.1 states ∙ instead ¸ that (AB) = (A)B. For a counterexample ∙ to (A)B ¸ = −1 0 0 1 A(B), consider A = , B = I2 , and  = h1i ↔ h2i. Then (A)B = and 0 1 −1 0 ∙ ¸ 0 −1 A(B) = .

1 0 ¯ ⎡ ⎤ 1 2 3 ¯¯ 0 (c) False. Consider the system associated with the augmented matrix ⎣ 0 0 0 ¯¯ 1 ⎦.

0 0 0 ¯ 0

(d) True. The augmented matrices for the two systems are row equivalent. (Multiply each row of the augmented matrix for the second system by  to get the augmented matrix for the first system.)

Hence, the systems have the same solution set by Theorem 2.5. ⎡ ⎤ 1 0 0 0 ⎢ 0 0 0 1 ⎥ ⎥ (e) False. Consider ⎢ ⎣ 0 0 1 1 ⎦.

0 0 0 1

(f) True. This follows from conditions (2) and (3) for reduced row echelon form: Each successive row has its first nonzero entry in a later column, and all values below those entries are zero.

(g) True. The row operations for solving any system are those that put the coefficient matrix A into reduced row echelon form. The constants on the right side of the equation have no effect in determining the row operations used. (h) True. If X = 0 is a solution to the system AX = B, then A0 = B, implying 0 = B. (i) False. We show that if both statements are true, we obtain a contradiction. Suppose X1 is a nontrivial solution to AX = 0, and X2 is the unique solution to AX = B. Then A(X1 + X2 ) = AX1 + AX2 = 0 + B = B. Hence, (X1 + X2 ) is a solution to AX = B that is different from X2 , because X1 6= 0. This contradicts the uniqueness of X2 as a solution to AX = B.

(j) True. These are listed in Table 2.1 near the beginning of Section 2.3. ⎡ ⎤ ⎡ ⎤ 1 0 −1 1 0 −2 (k) False. Consider A = ⎣ 0 1 −1 ⎦ and B = ⎣ 0 1 −1 ⎦  The solution set for AX = 0 is 0 0 0 0 0 0 {[  ] |  ∈ R}, while £the solution set for BX = 0 is {[2  ] |  ∈ R}. For a smaller-sized ¤ £ ¤ example, consider A = 1 0 and B = 0 1 . Both matrices have rank 1. However, the solution set for AX = 0 is {[0 ] |  ∈ R}, while the solution set for BX = 0 is {[ 0] |  ∈ R}.

(Note, however, if rank(A) = rank(B) = , then both systems will have only the trivial solution, and so their solution sets will be identical.) (l) False. The rank equals the number of nonzero rows in the reduced row echelon form for the matrix. If the rank is greater than zero, the number of vectors in the row space is infinite, since every scalar multiple of any of the nonzero rows is in the row space. (m) True. By Theorem 2.15, an  ×  matrix A is nonsingular if and only if rank(A) = . But if A is nonsingular, then A−1 is also nonsingular by part (1) of Theorem 2.12. Hence rank(A−1 ) =  as well.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

75

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 2 Review

(n) True. A−1 = B(AB)−1 because A(B(AB)−1 ) = (AB)(AB)−1 = I . (Alternately, from Exercise 18(d) in Section 2.3, we know that rank(AB) ≤ rank (A), and since rank(AB) =  here, we must have rank(A) = , and hence A is nonsingular.) ⎡ ⎤ ∙ ¸ 1 0 ∙ ¸ 1 0 0 ⎣ 1 0 0 1 ⎦= (o) False. Consider .

0 1 0 0 1 0 0

(p) True. First note that since the two given systems are homogeneous, both have, at least, the trivial solution. If the determinants of both A and B are zero, then both A and B are singular (Theorem 2.14), and by part (2) of Theorem 2.16, both have an infinite number of solutions. If the determinants of A and B are both nonzero, then both A and B are nonsingular (Theorem 2.14), and by part (1) of Theorem 2.16, both have a unique solution. (q) True. This is the essence of the Inverse Method, which is justified in Section 2.4. (r) True. This follows from part (2) of Theorem 2.12 and part (2) of Theorem 2.13. (s) True. (A−1 B C)−1 = C−1 (B )−1 (A−1 )−1 = C−1 (B−1 ) A, where we used a generalized form of part (3) of Theorem 2.12 for the first equality, and parts (1) and (4) of Theorem 2.12 for the second equality.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

76

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.1

Chapter 3 Section 3.1

¯ ¯ ¯ −2 5 ¯ ¯ ¯ = (−2)(1) − (5)(3) = −17 (1) (a) ¯ 3 1 ¯ ¯ ¯ ¯ 6 −12 ¯ ¯ ¯ = (6)(8) − (−12)(−4) = 0 (c) ¯ −4 8 ¯

(e) To use basketweaving, we form a new array by taking the given matrix, and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving pattern. & & & . . .

2 −4 0 .

0 × ×

1 3

.

5 ×

7

×

−3

.

2 × ×

−4 0

&

0 × ×

1 3

&

&

Hence, the determinant equals (2)(1)(−3) + (0)(7)(0) + (5)(−4)(3) − (5)(1)(0) − (2)(7)(3) − (0)(−4)(−3) = −108 (g) To use basketweaving, we form a new array by taking the given matrix, and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving pattern. & & & . . .

5 3 −1 .

0 × ×

−2 8

.

0 × × .

0 4

5 × × &

3 −1

0 × × &

−2 8 &

Hence, the determinant equals (5)(−2)(4) + (0)(0)(−1) + (0)(3)(8) − (0)(−2)(−1) − (5)(0)(8) − (0)(3)(4) = −40 (i) To use basketweaving, we form a new array by taking the given matrix, and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

77

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.1

pattern. &

& 3 −1 3

.

&

.

1 × ×

4 1

.

−2

×

5

×

−2

.

.

3

× ×

−1 3

&

.

1

× ×

4 1

&

&

Hence, the determinant equals (3)(4)(−2) + (1)(5)(3) + (−2)(−1)(1) − (−2)(4)(3) − (3)(5)(1) − (1)(−1)(−2) = 0 (j) The determinant of a 1 × 1 matrix is defined to be the (1 1) entry of the matrix. Therefore, the determinant equals −3.

(2) Recall that the submatrix A is found by deleting the th row and the th column from A. The ( ) minor is the determinant of this submatrix. ¯ ¯ ¯ 4 3 ¯ ¯ = (4)(4) − (3)(−2) = 22. (a) |A21 | = ¯¯ −2 4 ¯ ¯ ¯ ¯ −3 0 5 ¯¯ ¯ (c) |C42 | = ¯¯ 2 −1 4 ¯¯. We will use basketweaving to compute this determinant. To do this, we ¯ 6 4 0 ¯

form a new array by taking C42 , and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving pattern. &

& −3 2 6

.

& 0

× ×

−1 4

.

.

5

× × .

4 0

× × &

. −3 2 6

.

0

× × &

−1 4 &

Hence, |C42 | = (−3)(−1)(0) + (0)(4)(6) + (5)(2)(4) − (5)(−1)(6) − (−3)(4)(4) − (0)(2)(0) = 118.

(3) The cofactor A is defined to be (−1)+ |A |, where |A | is the ( ) minor — that is, the determinant of the ( ) submatrix, obtained by deleting the th row and the th column from A. ¯ ¯ ³ ´ ¯ 4 −3 ¯ ¯ = (1) (4)(−7) − (−3)(9) = −1. (a) A22 = (−1)2+2 |A22 | = (−1)4 ¯¯ ¯ 9 −7 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

78

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.1

¯ ¯ ¯ −5 2 13 ¯¯ ¯ 2 22 ¯¯ = (−1)(−222) = 222, where we have computed (c) C43 = (−1)4+3 |C43 | = (−1)7 ¯¯ −8 ¯ −6 −3 −16 ¯ the 3 × 3 determinant |C43 | to be −222 using basketweaving as follows: To compute |C43 |, we form a new array by taking C43 , and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving pattern. &

& −5 −8 −6

.

&

.

2 × ×

2 −3

.

13 × ×

22 −16

.

× × &

. −5 −8 −6

.

2

× × &

2 −3 &

Hence, |C43 | = (−5)(2)(−16) + (2)(22)(−6) + (13)(−8)(−3) −(13)(2)(−6) − (−5)(22)(−3) −(2)(−8)(−16) = −222. ¯ ¯ ³ ´ ¯ ¯ 1+2 3¯ −4 −3 ¯ (d) D12 = (−1) |D12 | = (−1) ¯ = (−1) ( − 4)( + 2) − ( − 3)( − 1) −1 +2 ¯ ³¡ ¢ ¡ 2 ¢´ 2 = (−1)  − 2 − 8 −  − 4 + 3 = −2 + 11.

(4) In this problem, we are asked to use only the formal definition (cofactor expansion) to compute determinants. Therefore, we will not use basketweaving for 3 × 3 determinants, or the simple 11 22 − 12 21 formula for 2×2 determinants. Of course, the cofactor expansion method still produces the same results obtained in Exercise 1. ∙ ¸ −2 5 (a) Let A = . Then |A| = 21 A21 + 22 A22 = 21 (−1)2+1 (12 ) + 22 (−1)2+2 (11 ) 3 1 = (3)(−1)(5) + (1)(1)(−2) = −17. ∙ ¸ 6 −12 (c) Let A = . Then |A| = 21 A21 + 22 A22 = 21 (−1)2+1 (12 ) + 22 (−1)2+2 (11 ) −4 8

= (−4)(−1)(−12) + (8)(1)(6) = 0. ⎡ ⎤ 2 0 5 7 ⎦. Then |A| = 31 A31 + 32 A32 + 33 A33 (e) Let A = ⎣ −4 1 0 3 −3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ 0 5 ¯ ¯ ¯ ¯ ¯ ¯ + (3)(−1)3+2 ¯ 2 5 ¯ + (−3)(−1)3+3 ¯ 2 0 ¯  = (0)(−1)3+1 ¯¯ ¯ ¯ ¯ ¯ 1 7 −4 7 −4 1 ¯

Now, for a 2 × 2 matrix B, |B| = 21 B21 + 22 B22 = 21 (−1)2+1 (12 ) + 22 (−1)2+2 (11 ). ¯ ¯ ¯ ¯ ¯ 2 5 ¯ ¯ 2 0 ¯ ¯ ¯ ¯ ¯ = (−4)(−1)(0) + (1)(1)(2) = 2.

Hence, ¯ = (−4)(−1)(5) + (7)(1)(2) = 34 and ¯ −4 7 ¯ −4 1 ¯ ¯ ¯ ¯ 0 5 ¯ ¯ because its coefficient in the determinant formula is zero.

It is not necessary to compute ¯¯ 1 7 ¯ Thus, |A| = 0 + (3)(−1)(34) + (−3)(1)(2) = −108

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

79

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.1

⎡

⎤ 5 0 0 (g) Let A = ⎣ 3 −2 0 ⎦. Then |A| = 31 A31 + 32 A32 + 33 A33 −1 8 4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ 0 0 ¯ ¯ ¯ ¯ 0 ¯¯ 3+1 ¯ 3+2 ¯ 5 0 ¯ 3+3 ¯ 5 ¯ = (−1)(−1) ¯ −2 0 ¯ + (8)(−1) ¯ 3 0 ¯ + (4)(−1) ¯ 3 −2 ¯ 

Now, for a 2 × 2 matrix B, |B| = 21 B21 + 22 B22 = 21 (−1)2+1 (12 ) + 22 (−1)2+2 (11 ). ¯ ¯ ¯ ¯ ¯ 0 0 ¯ ¯ 5 0 ¯ ¯ ¯ ¯ ¯ = (3)(−1)(0) + (0)(1)(5) = 0 and Hence, ¯ = (−2)(−1)(0) + (0)(1)(0) = 0, ¯ −2 0 ¯ 3 0 ¯ ¯ ¯ ¯ 5 0 ¯¯ ¯ ¯ 3 −2 ¯ = (3)(−1)(0) + (−2)(1)(5) = −10. Thus,

|A| = (−1)(1)(0) + (8)(−1)(0) + (4)(1)(−10) = −40 ⎡ ⎤ 3 1 −2 5 ⎦. Then |A| = 31 A31 + 32 A32 + 33 A33 (i) Let A = ⎣ −1 4 3 1 −2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 3 −2 ¯ ¯ 3 1 ¯ ¯ 3+1 ¯ 1 −2 ¯ 3+2 ¯ 3+3 ¯ ¯ ¯ = (3)(−1) + (1)(−1) + (−2)(−1) ¯ 4 ¯ −1 ¯ −1 4 ¯  5 ¯ 5 ¯

Now, for a 2 × 2 matrix B, |B| = 21 B21 + 22 B22 = 21 (−1)2+1 (12 ) + 22 (−1)2+2 (11 ). ¯ ¯ ¯ ¯ ¯ 1 −2 ¯ ¯ 3 −2 ¯ ¯ ¯ ¯ ¯ = (−1)(−1)(−2) +(5)(1)(3) = 13, Hence, ¯ = (4)(−1)(−2) +(5)(1)(1) = 13, ¯ 4 5 ¯ −1 5 ¯ ¯ ¯ ¯ 3 1 ¯ ¯ = (−1)(−1)(1) + (4)(1)(3) = 13. Thus, and ¯¯ −1 4 ¯ |A| = (3)(1)(13) + (1)(−1)(13) + (−2)(1)(13) = 0

(j) By definition, the determinant of a 1 × 1 matrix is defined to be the (1,1) entry of the matrix.

Thus, the determinant of [−3] equals −3. ⎡ ⎤ 5 2 1 0 ⎢ −1 3 5 2 ⎥ ⎥ (5) (a) Let A = ⎢ ⎣ 4 1 0 2 ⎦ . Then |A| = 41 A41 + 42 A42 + 43 A43 + 44 A44 0 2 3 0 = (0)(−1)4+1 |A41 | + (2)(−1)4+2 |A42 | + (3)(−1)4+3 |A43 | + (0)(−1)4+4 |A44 | = 0 + 2|A42 | − 3|A43 | + 0 ⎡ ⎤ 5 1 0 Now, let B = A42 = ⎣ −1 5 2 ⎦ . Then |A42 | = |B| = 31 B31 + 32 B32 + 33 B33 4 0 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 0 ¯ ¯ ¯ ¯ ¯ ¯ + (0)(−1)3+2 ¯ 5 0 ¯ + (2)(−1)3+3 ¯ 5 1 ¯ ¯ 5 2 ¯ ¯ −1 2 ¯ ¯ −1 5 ¯

3+1 ¯

= (4)(−1)

= (4)(1) ((1)(2) − (0)(5)) + 0 + 2(1) ((5)(5) − (1)(−1)) = 60 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

80

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.1

⎡

⎤ 5 2 0 Also, let C = A43 = ⎣ −1 3 2 ⎦. Then |A43 | = |C| = 31 C31 + 32 C32 + 33 C33 4 1 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 0 ¯ ¯ 5 2 ¯ ¯ 3+1 ¯ 2 0 ¯ 3+2 ¯ 3+3 ¯ ¯ ¯ = (4)(−1) ¯ 3 2 ¯ + (1)(−1) ¯ −1 2 ¯ + (2)(−1) ¯ −1 3 ¯

= (4)(1) ((2)(2) − (3)(0)) + (1)(−1) ((5)(2) − (0)(−1)) + 2(1) ((5)(3) − (2)(−1)) = 40

Hence, |A| = 2|A42 | − 3|A43 | = 2(60) − 3(40) = 0. (d) Let A be the given 5 × 5 matrix. Then |A| = 51 A51 + 52 A52 + 53 A53 + 54 A54 + 55 A55 = (0)(−1)5+1 |A51 | + (3)(−1)5+2 |A52 | + (0)(−1)5+3 |A53 | + (0)(−1)5+4 |A54 | + (2)(−1)5+5 |A55 |

= (0) + (3)(−1)|A52 | + (0) + (0) + (2)(1)|A55 |. ⎡ ⎤ 0 1 3 −2 ⎢ 2 3 −1 0 ⎥ ⎥. Then, |A52 | = |B| = 41 B41 + 42 B42 + 43 B43 + 44 B44 Let B = A52 = ⎢ ⎣ 3 2 −5 1 ⎦ 1 −4 0 0 = (1)(−1)4+1 |B41 | + (−4)(−1)4+2 |B42 | + (0)(−1)4+3 |B43 | + (0)(−1)4+4 |B44 | = (1)(−1)|B41 | + (−4)(1)|B42 | + (0) + (0) ⎡ ⎤ 1 3 −2 0 ⎦. We compute |B41 | using basketweaving.

Now, B41 = ⎣ 3 −1 2 −5 1 &

&

1 3 2 .

&

3 × ×

−1 −5

.

×

.

−2 0

×

1

.

.

1 × ×

3 2

&

.

3 × ×

−1 −5

&

&

Then |B41 | = (1)(−1)(1) + (3)(0)(2) + (−2)(3)(−5) − (−2)(−1)(2) − (1)(0)(−5) − (3)(3)(1) = 16. ⎡ ⎤ 0 3 −2 0 ⎦. We compute |B42 | using basketweaving.

Next, B42 = ⎣ 2 −1 3 −5 1 &

&

0 2 3 .

&

3 × ×

−1 −5

.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

×

.

−2 0

×

1

.

0 × × &

81

.

2 3

.

3 × × &

−1 −5 &

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.1

Then |B42 | = (0)(−1)(1) + (3)(0)(3) + (−2)(2)(−5) − (−2)(−1)(3) − (0)(0)(−5) − (3)(2)(1) = 8.

Therefore, |A52 | = (1)(−1)|B41 | + (−4)(1)|B42 | = (1)(−1)(16) + (−4)(1)(8) = −48 ⎡ ⎤ 0 4 1 3 ⎢ 2 2 3 −1 ⎥ ⎥. Then, |A55 | = |C| = 41 C41 + 42 C42 + 43 C43 + 44 C44 Now, let C = A55 = ⎢ ⎣ 3 1 2 −5 ⎦ 1 0 −4 0 = (1)(−1)4+1 |C41 | + (0)(−1)4+2 |C42 | + (−4)(−1)4+3 |C43 | + (0)(−1)4+4 |C44 | = (1)(−1)4+1 |C41 | + (0) + (−4)(−1)4+3 |C43 | + (0) ⎡ ⎤ 4 1 3 Now, C41 = ⎣ 2 3 −1 ⎦. We compute |C41 | using basketweaving.

1 2 −5 &

&

4 2 1 .

&

.

1 × ×

3 2

.

3 × ×

−1 −5

.

.

4 × ×

2 1

&

.

1 × ×

3 2

&

&

Then |C41 | = (4)(3)(−5) + (1)(−1)(1) + (3)(2)(2) − (3)(3)(1) − (4)(−1)(2) − (1)(2)(−5) = −40. ⎡ ⎤ 0 4 3 Similarly, C43 = ⎣ 2 2 −1 ⎦. We compute |C43 | using basketweaving.

3 1 −5 &

&

0 2 3 .

&

.

4 × ×

2 1

.

3 × ×

−1 −5

.

.

0 × × &

2 3

.

4 × × &

2 1 &

Then |C43 | = (0)(2)(−5) + (4)(−1)(3) + (3)(2)(1) − (3)(2)(3) − (0)(−1)(1) − (4)(2)(−5) = 16.

Therefore, |A55 | = (1)(−1)4+1 |C41 | + (−4)(−1)4+3 |C43 | = (1)(−1)(−40) + (−4)(−1)(16) = 104 Finally, |A| = (0) + (3)(−1)|A52 | + (2)(1)|A55 | = (3)(−1)(−48) + (2)(1)(104) = 352 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

82

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.1

¯ ¯ ¸ ∙ ¸ ¯ 1 1 ¯ 1 0 ¯ ¯ = (1)(1) − (1)(1) = 0 and |B| = (7) Let A = , and let B = . Then |A| = ¯ 0 1 1 1 ¯ ¯ ¯ ¯ ¯ ¯ 1 0 ¯ ¯ ¯ ¯ ¯ = (1)(1) − (0)(0) = 1. Hence, |A| + |B| = 1. But, |A + B| = ¯ 2 1 ¯ = (2)(2) − (1)(1) = 3. ¯ 0 1 ¯ ¯ 1 2 ¯ ∙

1 1 1 1

Hence, |A + B| 6= |A| + |B|.

(9) According to part (1) of Theorem 3.1, the area of the parallelogram is the absolute value of the determinant of the matrix whose rows are the two given vectors. Be careful! If the determinant is negative, we must take the absolute value to find the area. ¯ ¯ ¯ 3 2 ¯ ¯. This determinant (a) The area of the parallelogram is the absolute value of the determinant ¯¯ 4 5 ¯ equals (3)(5) − (2)(4) = 7. Hence, the area is 7. ¯ ¯ ¯ 5 −1 ¯ ¯. This deter(c) The area of the parallelogram is the absolute value of the determinant ¯¯ −3 3 ¯ minant equals (5)(3) − (−1)(−3) = 12. Hence, the area is 12.

(10) Let x = [1  2 ] and y = [1  2 ]. Consider the hint in the text. Then, the area of the parallelogram is kxk khk, where h = y − projx y. Now, ¶ µ 1 x·y projx y = x= [(1 1 + 2 2 )1  (1 1 + 2 2 )2 ] 2 kxk kxk2 Hence, h = = = = = =

¶ µ 1 x·y 2 x (kxk y) − y − projx y = kxk2 kxk2 1 1 [(2 + 22 )1  (21 + 22 )2 ] − [(1 1 + 2 2 )1  (1 1 + 2 2 )2 ] kxk2 1 kxk2 1 [2 1 + 22 1 − 21 1 − 1 2 2  21 2 + 22 2 − 1 2 1 − 22 2 ] kxk2 1 1 [2 1 − 1 2 2  21 2 − 1 2 1 ] kxk2 2 1 [2 (2 1 − 1 2 ) 1 (1 2 − 2 1 )] kxk2 1 2 − 2 1 [−2  1 ] kxk2

Thus, kxk ky − projx yk = kxk

|1 2 − 2 1 | kxk2

¯ q ¯  22 + 21 = |1 2 − 2 1 | = absolute value of ¯¯ 1 1

¯ 2 ¯¯  2 ¯

(11) According to part (2) of Theorem 3.1, the volume of the parallelepiped is the absolute value of the determinant of the matrix whose rows are the three given vectors. Be careful! If the determinant is negative, we must take the absolute value to find the volume.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

83

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.1

¯ ¯ ¯ −2 3 1 ¯ ¯ ¯ (a) The volume of the parallelepiped is the absolute value of the determinant ¯¯ 4 2 0 ¯¯. We use ¯ −1 3 2 ¯ basketweaving to compute this. &

&

−2

3 ×

4

×

−1 .

&

2 3

.

.

1 × ×

0 2

.

× ×

.

−2 4 −1

&

.

3 × ×

2 3

&

&

The determinant equals (−2)(2)(2) + (3)(0)(−1) + (1)(4)(3) − (1)(2)(−1) − (−2)(0)(3) −(3)(4)(2) = −18. The volume is the absolute value of this result. Hence, the volume equals 18. ¯ ¯ ¯ −3 4 0 ¯¯ ¯ (c) The volume of the parallelepiped is the absolute value of the determinant ¯¯ 6 −2 1 ¯¯. We ¯ 0 −3 3 ¯ use basketweaving to compute this. &

&

−3 6 0 .

&

4 × ×

−2 −3

.

.

0 × × .

1 3

× × &

.

−3 6 0

.

4 × × &

−2 −3 &

The determinant equals (−3)(−2)(3) + (4)(1)(0) + (0)(6)(−3) − (0)(−2)(0) − (−3)(1)(−3) − (4)(6)(3) = −63. The volume is the absolute value of this result. Hence, the volume equals 63.

(12) First, notice from the definition of x × y in Exercise 8, that p kx × yk = (2 3 − 3 2 )2 + (1 3 − 3 1 )2 + (1 2 − 2 1 )2 

We will verify that kx × yk is equal to the area of the parallelogram determined by x and y Now, from the solution to Exercise 10 above, the area of this parallelogram is equal to kxk ky − projx yk. One can verify kx × yk = kxk ky − projx yk by a tedious, brute force, argument. (Algebraically expand and simplify kxk2 ky − projx yk2 to get (2 3 − 3 2 )2 + (1 3 − 3 1 )2 + (1 2 − 2 1 )2 ) An alternate approach is the following: Note that kx × yk2

= (2 3 − 3 2 )2 + (1 3 − 3 1 )2 + (1 2 − 2 1 )2 = 22 32 − 22 3 2 3 + 23 22 + 21 32 − 21 3 1 3 + 23 12 + 21 22 − 21 2 1 2 + 22 12 

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

84

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.1

Using some algebraic manipulation, this can be expressed as kx × yk2 = (21 + 22 + 23 )(12 + 22 + 32 ) − (1 1 + 2 2 + 3 3 )2  Therefore, kx × yk2

2

2

= kxk kyk − (x · y)2 ´ kxk2 ³ 2 2 2 = 2 kxk kyk − (x · y) kxk à !

2 2 (x · y)2 kxk kyk 2 = kxk − kxk2 kxk2 Ã Ã ! ! (x · y)2 kxk2 (y · y) (x · y)2 2 −2 = kxk + 2 2 2 kxk kxk kxk ⎞ ⎛ Ã ! Ã !2 x·y x·y 2 = kxk ⎝(y · y) − 2 (x · x)⎠ (x · y) + kxk2 kxk2 Ã ! Ã ! (x · y) x (x · y) x 2 = kxk y − · y− kxk2 kxk2 ° ! °2 Ã ° ° x·y ° 2° = kxk °y − x° 2 ° ° kxk = kxk2 ky − projx yk2 

Hence, kx × yk = kxk ky − projx yk, the area of the parallelogram determined by x and y.

Now, we determine the volume of the parallelepiped. As in the hint, let h = proj(x×y) z, the perpendicular from z to the parallelogram determined by x and y. Since the area of this parallelogram equals kx × yk, the volume of the parallelepiped equals ¯ ¯ ¯ z · (x × y) ¯ ¯ kx × yk = |z · (x × y)| = |(x × y) · z|  khk kx × yk = kproj(x×y) zk kx × yk = ¯¯ kx × yk ¯ But from the definition in Exercise 8,

|(x × y) · z| = | (2 3 − 3 2 )1 + (3 1 − 1 3 )2 + (1 2 − 2 1 )3 | = | 1 2 3 + 2 3 1 + 3 1 2 − 3 2 1 − 1 3 2 − 2 1 3 | ¯ ¯ ¯ 1 2 3 ¯ ¯ ¯ = absolute value of ¯¯ 1 2 3 ¯¯  ¯ 1 2 3 ¯

¯ ¯ ¯  2 ¯¯ (15) (a) ¯¯ = 0 =⇒ (+3)−(2)(5) = 0 =⇒ 2 +3−10 = 0 =⇒ (+5)(−2) = 0 =⇒  = −5 5 +3 ¯ or  = 2.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

85

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.2

(c) First, we compute the given determinant using basketweaving. &

& −3 0 0

.

& 5

× × .

×

−1

×

0

. −19 6 −2

.

×

. −3

× &

0 0

.

5

× ×

−1 0

&

&

The determinant equals ( − 3)( − 1)( − 2) + (5)(6)(0) + (−19)(0)(0) − (−19)( − 1)(0) − ( − 3)(6)(0) − (5)(0)( − 2) = ( − 3)( − 1)( − 2). Setting this determinant equal to zero yields  = 3,  = 1, or  = 2.

(16) (b) The given matrix is a 3 × 3 Vandermonde matrix with  ´=³2,  = 3,´and ³ ³  = −2. ´ Thus, using part (a), the determinant is ( − )( − )( − ) = 2 − 3

3 − (−2)

(−2) − 2 = 20.

(18) (a) False. The basketweaving technique only works to find determinants of 3 × 3 matrices. For larger matrices, the only method we have learned at this point is the cofactor expansion along the last row, although alternate methods are given in Sections 3.2 and 3.3. ¯ ¯ ¯ 1 2 ¯ ¯ = 1 2 − 2 1 . Part (1) of Theorem 3.1 assures us that the absolute ¯ (b) True. Note that ¯ 1 2 ¯ value of this determinant gives the area of the desired parallelogram.

(c) False in general. (True for  = 2.) An  ×  matrix has 2 cofactors — one corresponding to each entry in the matrix.

(d) False in general. (True in the special case in which B23 = 0.). By definition, B23 = (−1)2+3 |B23 | = −|B23 |, not |B23 |.

(e) True. The given formula is the cofactor expansion of A along the last row. By definition, this equals the determinant of A.

Section 3.2 (1) (a) The row operation used is (II): h1i ← −3 h2i + h1i. By Theorem 3.3, performing a Type (II) operation does not change the determinant of a matrix. Thus, since |I3 | = 1, the determinant of the given matrix also equals 1. (c) The row operation used is (I): h3i ← −4 h3i. By Theorem 3.3, performing a Type (I) operation multiplies the determinant by the constant used in the row operation. Thus, since |I3 | = 1, the determinant of the given matrix equals (−4)(1) = −4.

(f) The row operation used is (III): h1i ↔ h2i. By Theorem 3.3, performing a Type (III) row operation changes the sign of the determinant. Thus, since |I3 | = 1, the determinant of the given matrix equals −1.

(2) In each part, we use row operations to put the given matrix into upper triangular form. Notice in the solutions, below, that we stop performing row operations as soon as we obtain an upper triangular c 2016 Elsevier Ltd. All rights reserved.

Copyright °

86

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.2

matrix. We give a chart indicating the row operations used, keeping track of the variable  , as described in Example 5 in Section 3.2 of the textbook. Recall that  = 1 at the beginning of the process. We then use the final upper triangular matrix obtained and the value of  to compute the desired determinant. (a)

Effect



1 h1i ← 10 h1i

1 Multiply  by 10

h2i ← − 14 h2i

Multiply  by − 14

1 10 1 10 1 − 40 1 − 40

Row Operations (I):

(II): h3i ← 5 h1i + h3i

No change

(II): h3i ← (−1) h2i + h3i

No change

(I):

⎡

1

⎢ The upper triangular matrix obtained from these operations is B = ⎣ 0 0

(c)

2 5

1 0

21 10 − 34 − 34

⎤

⎥ ⎦. By Theorem

3.2, |B| = (1)(1)(− 34 ) = − 34 . Hence, as in Example 5 in the textbook, the determinant of the original matrix is 1 × |B| = (−40)(− 34 ) = 30.

Row Operations



Effect

(II): h2i ← 2 h1i + h2i

No change

1

(II): h3i ← 3 h1i + h3i

No change

1

(II): h4i ← (−2) h1i + h4i

No change

1

(I): h2i ← (−1) h2i

(II): h3i ← h2i + h3i

Multiply  by −1 No change

−1

(II): h4i ← (−1) h2i + h4i

No change

−1

(III): h3i ↔ h4i

Multiply  by −1

−1 1

⎡

⎤ 1 −1 5 1 ⎢ 0 1 −3 −3 ⎥ ⎥. By The upper triangular matrix obtained from these operations is B = ⎢ ⎣ 0 0 2 2 ⎦ 0 0 0 −2

(e)

Theorem 3.2, |B| = (1)(1)(2)(−2) = −4. Hence, as in Example 5 in the textbook, the determinant of the original matrix is 1 × |B| = 1(−4) = −4.

Effect



h1i ← 15 h1i h2i ← (− 15 2 ) h1i + h2i h3i ← 52 h1i + h3i

Multiply  by 15

h2i ← (− 14 ) h2i

Multiply  by − 14

1 5 1 5 1 5 1 5 1 − 20 1 − 20 1 − 20 − 15 − 15

Row Operations

(I): (II): (II):

No change No change

(II): h4i ← (−10) h1i + h4i

No change

(II): h3i ← (−3) h2i + h3i

No change

(I):

(II): h4i ← 9 h2i + h4i

No change

h4i ← 34 h3i + h4i

No change

(I): h3i ← 4 h3i (II):

Multiply  by 4

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

87

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

Section 3.2

1

⎢ ⎢ 0 The upper triangular matrix obtained from these operations is B = ⎢ ⎣ 0 0

3 5

− 85

⎤

⎥ ⎥ ⎥. By 1 −27 ⎦ 0 −7

1 − 11 4

0 0

4 5 13 4

Theorem 3.2, |B| = (1)(1)(1)(−7) = −7. Hence, as in Example 5 in the textbook, the determinant of the original matrix is 1 × |B| = (−5)(−7) = 35.

(3) In this problem, first compute the determinant by any convenient method. ¯ ¯ ¯ 5 6 ¯¯ ¯ (a) ¯ = (5)(−4) − (6)(−3) = −2. Since the determinant is nonzero, Theorem 3.5 implies −3 −4 ¯ that the given matrix is nonsingular.

(c) Using basketweaving on the given matrix produces the value (−12)(−1)(−8) + (7)(2)(3) + (−27)(4)(2) − (−27)(−1)(3) − (−12)(2)(2) − (7)(4)(−8) = −79 for the determinant. Since the determinant is nonzero, Theorem 3.5 implies that the given matrix is nonsingular. ⎡ ⎤ −6 3 −22 4 −31 ⎦. Using basketweaving (4) (a) The coefficient matrix for the given system is A = ⎣ −7 11 −6 46

yields |A| = (−6)(4)(46) + (3)(−31)(11) + (−22)(−7)(−6) −(−22)(4)(11) − (−6)(−31)(−6) − (3)(−7)(46) = −1. Because |A| 6= 0, Corollary 3.6 shows that rank(A) = 3. Hence, by part (2) of Theorem 2.7, the system has only the trivial solution.

(6) Perform the following Type (III) row operations on A:

⎡

61 62 63 ⎧ ⎢ 0 52 53 ⎢ ⎨ (III) : h1i ↔ h6i ⎢ 0 0 43 (III) : h2i ↔ h5i . This produces the matrix B = ⎢ ⎢ 0 0 0 ⎩ ⎢ (III) : h3i ↔ h4i ⎣ 0 0 0 0 0 0

64 54 44 34 0 0

65 55 45 35 25 0

⎤ 66 56 ⎥ ⎥ 46 ⎥ ⎥. 36 ⎥ ⎥ 26 ⎦ 16

Since B is upper triangular, Theorem 3.2 gives |B| = 61 52 43 34 25 16 . But, applying a Type (III) row operation to a matrix changes the sign of its determinant. Since we performed three Type (III) row operations, |B| = −(−(−|A|)), or |A| = −|B| = −61 52 43 34 25 16 .

(16) By Corollary 3.6 and part (1) of Theorem 2.7, the homogeneous system AX = 0 has nontrivial solutions. Let B be any  ×  matrix such that every column of B is a nontrivial solution for AX = 0.

Then the th column of AB = A(th column of B) = 0 for every . Hence, AB = O .

(17) (a) False in general, although Theorem 3.2 shows that it is true for upper triangular matrices. For a ∙ ¸ 1 1 counterexample to the general statement, consider the matrix A = whose determinant 1 1 is (1)(1) − (1)(1) = 0. However, the product of the main diagonal entries of A is (1)(1) = 1.

(b) True. Part (3) of Theorem 3.3 shows that performing a Type (III) row operation changes the sign of the determinant. Therefore, performing two Type (III) row operations in succession changes the sign twice. Hence, there is no overall effect on the determinant.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

88

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.3

(c) False in general. If A is a 4 × 4 matrix, Corollary 3.4 shows that |3A| = 34 |A| = 81|A|. This equals 3|A| only in the exceptional case in which |A| = 0. The matrix A = I4 provides a specific counterexample to the original statement. (d) False. If A is a matrix having (row ) = (row ), then |A| = 0 6= 1. To prove this, note that performing the Type (III) row operation hi ↔ hi on A results in A. Thus, by part (3) of Theorem 3.3, |A| = −|A|. Hence, |A| = 0. For a specific counterexample to the original statement, consider O , for any  ≥ 2. (e) False. This statement contradicts Theorem 3.5.

(f) True. This statement is the contrapositive of Corollary 3.6.

Section 3.3 (1) (a) |A| = 31 A31 + 32 A32 + 33 A33 + 34 A34

= 31 (−1)3+1 |A31 | + 32 (−1)3+2 |A32 | + 33 (−1)3+3 |A33 | + 34 (−1)3+4 |A34 | = 31 |A31 | − 32 |A32 | + 33 |A33 | − 34 |A34 |.

(c) |A| = 14 A14 + 24 A24 + 34 A34 + 44 A44

= 14 (−1)1+4 |A14 | + 24 (−1)2+4 |A24 | + 34 (−1)3+4 |A34 | + 44 (−1)4+4 |A44 | = −14 |A14 | + 24 |A24 | − 34 |A34 | + 44 |A44 |.

(2) (a) Let A be the given matrix. Then |A| = 21 A21 + 22 A22 + 23 A23 2+1 2+2 = 21 (−1) |A21 |¯+ 22 (−1) |A¯ 22 | + 23¯(−1)2+3 |A ¯ ¯ ¯ 23 | ¯ −1 ¯ ¯ ¯ ¯ 2 −1 ¯ 4 2 4 ¯ + (3) ¯ ¯ ¯ ¯ = −(0) ¯¯ ¯ 5 −3 ¯ − (−2) ¯ 5 −2 ¯ −2 −3 ¯ ³ ´ ³ ´ = 0 + (3) (2)(−3) − (4)(5) − (−2) (2)(−2) − (−1)(5) = −76.

(c) Let C be the given matrix. Then |C| = 11 C11 + 21 C21 + 31 C31 1+1 3+1 = 11 (−1) |C11¯ | + 21¯(−1)2+1 |C ¯ ¯ 21 | +¯31 (−1) ¯|C31 | ¯ −1 −2 ¯ ¯ ¯ ¯ 3 ¯¯ ¯ − (5) ¯ −2 3 ¯ + (3) ¯ −2 = (4) ¯¯ ¯ 3 2 ¯ ¯ −1 −2 ¯ 3 2 ¯ ³ ´ ³ ´ ³ ´ = (4) (−1)(2) − (−2)(3) − (5) (−2)(2) − (3)(3) + (3) (−2)(−2) − (3)(−1) = 102. ⎡

⎤ 3 −1 −1 (3) (a) Let A be the coefficient matrix ⎣ 2 −1 −2 ⎦. The matrices A1  A2 , and A3 are formed by −9 1 0 removing, respectively, the 1st, 2nd, and 3rd columns from A, and replacing them with [−8 3 39] (as a column vector). Hence, ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −8 −1 −1 3 −8 −1 3 −1 −8 3 −2 ⎦  and A3 = ⎣ 2 −1 3 ⎦ A1 = ⎣ 3 −1 −2 ⎦  A2 = ⎣ 2 39 1 0 −9 39 0 −9 1 39

Basketweaving, or any other convenient method, can be used to compute the determinant of each of these matrices, yielding |A| = −5, |A1 | = 20, |A2 | = −15, and |A3 | = 35. Therefore, Cramer’s Rule states that the unique solution to the system is given by 1 =

20 (−15) 35 |A1 | |A2 | |A3 | = = −4 2 = = = 3 and 3 = = = −7 |A| (−5) |A| (−5) |A| (−5)

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

89

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.3

The full solution set is {(−4 3 −7)}. ⎡ ⎤ −5 2 −2 1 ⎢ 2 −1 2 −2 ⎥ ⎥. The matrices A1  A2 , A3 and A4 are (d) Let A be the coefficient matrix ⎢ ⎣ 5 −2 3 −1 ⎦ −6 2 −2 1

formed by removing, respectively, the 1st, 2nd, 3rd and 4th columns from A, and replacing them with [−10 −9 7 −14] (as a column vector). Hence, ⎡ ⎡ ⎤ ⎤ −10 2 −2 1 −5 −10 −2 1 ⎢ −9 −1 ⎢ 2 −2 ⎥ 2 −2 ⎥ ⎥  A2 = ⎢ 2 −9 ⎥ A1 = ⎢ ⎣ ⎣ 7 −2 3 −1 ⎦ 5 7 3 −1 ⎦ −14 2 −2 1 −6 −14 −2 1 ⎡ ⎡ ⎤ ⎤ −5 2 −10 1 −5 2 −2 −10 ⎢ 2 −1 −9 −2 ⎥ ⎢ 2 −9 ⎥ ⎥ and A4 = ⎢ 2 −1 ⎥ A3 = ⎢ ⎣ 5 −2 ⎣ ⎦ 7 −1 5 −2 3 7 ⎦ −6 2 −14 1 −6 2 −2 −14

Row reduction or cofactor expansion can be used to compute the determinant of each of these matrices, yielding |A| = 3, |A1 | = 12, |A2 | = −3, |A3 | = −9 and |A4 | = 18. Therefore, Cramer’s Rule states that the unique solution to the system is given by 1 = 3 =

12 (−3) |A1 | |A2 | = = 4 2 = = = −1 |A| 3 |A| 3

(−9) 18 |A3 | |A4 | = = −3 and 4 = = = 6 |A| 3 |A| 3

The full solution set is {(4 −1 −3 6)}.

(4) (a) Suppose  is the Type (I) operation hi ←  hi. Then (I ) is the diagonal matrix having 1’s in every entry on the main diagonal, except for the ( ) entry, which has the value . But, since (I ) is a diagonal matrix, it is symmetric. (b) Suppose  is the Type (III) operation hi ↔ hi. Then the only nonzero entries off the main diagonal for (I ) are the ( ) entry and the ( ) entry, which are both 1’s. (The ( ) entry of (I ) has the 1 that was moved from the ( ) entry of I , and the ( ) entry of (I ) has the 1 that was moved from the ( ) entry of I .) Hence, all entries across the main diagonal from each other in (I ) are equal, and so (I ) is symmetric. (c) Suppose  is the Type (II) operation hi ←  hi + hi and  is the Type (II) operation hi ←  hi + hi. Then (I ) is the  ×  matrix having 1’s on the main diagonal and 0’s everywhere else, except for the ( ) entry, which has the value . Similarly, (I ) has 1’s all along the main diagonal and 0’s everywhere else, except for the ( ) entry, which has the value  . However, this exactly describes the matrix ((I )) . ∙ ¸ 0 − (8) (b) Try  = 2. All 2 × 2 skew-symmetric matrices are of the form . Letting  = 1 yields  0 ∙ ¸ 0 −1 A= . Note that |A| = 1 6= 0.

1 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

90

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.3

⎡

⎤ 0 1 0 (9) (b) Use trial and error, rearranging the rows of I3 . For example, A = ⎣ 1 0 0 ⎦ works, since 0 0 1

AA = AA (because A is symmetric) = I3 . (Note: all of the other four matrices obtained by rearranging the rows of I3 are also orthogonal matrices!)

(13) (b) Begin by choosing any nonsingular 2 × 2 matrix P and compute B = P−1 AP. For example, using ∙ ¸ ∙ ¸ 1 −1 2 1 (see Theorem 2.14), yielding P= , we get P−1 = −1 2 1 1 B = P−1 AP =

∙

1 −1 −1 2

¸∙

1 2 3 4

¸∙

2 1 1 1

¸

=

∙

−6 −4 16 11

¸



Next, choose another matrix P that produces a different answer for P−1 AP. For example, letting ∙ ¸ 2 −5 P= produces −1 3 ∙ ¸∙ ¸∙ ¸ ∙ ¸ 3 5 1 2 2 −5 10 −12 −1 =  B = P AP = 1 2 3 4 −1 3 4 −5 (14) (a) Let  be the Type (III) row operation hi ↔ h − 1i, let A be an  ×  matrix and let B = (A).

Then, by part (3) of Theorem 3.3, |B| = (−1)|A|. Next, notice that the submatrix A(−1) = B because the (−1)st row of A becomes the th row of B, implying the same row is being eliminated from both matrices, and since all other rows maintain their original relative positions (notice the same column is being eliminated in both cases). Hence, (−1)1 A(−1)1 + (−1)2 A(−1)2 + · · · + (−1) A(−1) = 1 (−1)(−1)+1 |A(−1)1 | + 2 (−1)(−1)+2 |A(−1)2 | + · · · +  (−1)(−1)+ |A(−1) | (because (−1) =  for 1 ≤  ≤ ) = 1 (−1) |B1 | + 2 (−1)+1 |B2 | + · · · +  (−1)+−1 |B | = (−1)(1 (−1)+1 |B1 | + 2 (−1)+2 |B2 | + · · · +  (−1)+ |B |) = (−1)|B| (by applying part (1) of Theorem 3.11 along the th row of B, since we have assumed this result is valid for  = ) = (−1)|(A)| = (−1)(−1)|A| = |A|, finishing the proof. (b) The definition of the determinant is the Base Step for an induction proof on the row number, counting down from  to 1. Part (a) is the Inductive Step.

(15) (a) Case 1: Assume 1 ≤    and 1 ≤   . Then the ( ) entry of (A ) = ( ) entry of A = ( ) entry of A = ( ) entry of A = ( ) entry of (A ) .

Case 2: Assume  ≤    and 1 ≤   . Then the ( ) entry of (A ) = ( ) entry of A = ( + 1 ) entry of A = (  + 1) entry of A = ( ) entry of (A ) .

Case 3: Assume 1 ≤    and  ≤   . Then the ( ) entry of (A ) = ( ) entry of A = (  + 1) entry of A = ( + 1 ) entry of A = ( ) entry of (A ) .

Case 4: Assume  ≤    and  ≤   . Then the ( ) entry of (A ) = ( ) entry of A = ( + 1  + 1) entry of A = ( + 1  + 1) entry of A = ( ) entry of (A ) .

Hence, the corresponding entries of (A ) and (A ) are all equal, proving that the matrices themselves are equal. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

91

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

1+

2+

Section 3.3

+

(b) 1 A1 + 2 A2 |A¯2 | + · · ·¯ +  (−1) |A | ¯ +···+ ¯  A = 1 (−1) ¯ ¯|A1 | + 2 (−1) 1+ ¯ ¯ 2+ ¯ ¯ + ¯ ¯ = 1 (−1) ¯(A1 ) ¯ + 2 (−1) ¯(A2 ) ¯ + · · · +  (−1) ¯(A ) ¯ (by Theorem 3.10) ¯¡ ¢ ¯ ¯¡ ¢ ¯ ¯¡ ¢ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 1 (−1)+1 ¯ A 1 ¯ + 2 (−1)+2 ¯ A 2 ¯ + · · · +  (−1)+ ¯ A  ¯ (by part (a)) ¯ ¯ = ¯A ¯ (using part (1) of Theorem 3.11 along the th row of A ) = |A|, (by Theorem 3.10).

(18) Assume in what follows that A is an  ×  matrix. Let A be the th matrix (as defined in Theorem 3.12) for the matrix A. (a) Let C = ([A|B]) We must show that for each , 1 ≤  ≤  the matrix C (as defined in Theorem 3.12) for C is identical to (A ) First we consider the columns of C other than the th column. If 1 ≤  ≤  with  6= , then the th column of C is the same as the th column of C = ([A|B]). But since the th column of [A|B] is identical to the th column of A , it follows that the th column of ([A|B]) is identical to the th column of (A ). Therefore, for 1 ≤  ≤  with  6= , the th column of C is the same as the th column of (A ).

Finally, we consider the th column of C . Now, the th column of C is identical to the last column of ([A|B]) = [(A)|(B)], which is (B) But since the th column of A equals B, it follows that the th column of (A ) = (B). Thus, the th column of C is identical to the th column of (A ).

Therefore, since C and (A ) agree in every column, we have C = (A ) (b) Consider each type of operation in turn. First, if  is the Type (I) operation hi ←  hi for some  6= 0, then by part (1) of Theorem 3.3, |A | |A | |(A )| = =  |(A)| |A| |A| If  is any Type (II) operation, then by part (2) of Theorem 3.3, |A | |(A )| =  |(A)| |A| If  is any Type (III) operation, then (−1)|A | |A | |(A )| = =  |(A)| (−1)|A| |A| (c) In the special case when  = 1, we have A = [1]. Let B = []. Hence, A1 (as defined in Theorem 3.12) = B = [] Then the formula in Theorem 3.12 gives 1 =

 |A1 | = =  |A| 1

which is the correct solution to the equation AX = B in this case. For the remainder of the proof, we assume   1.

If A = I , then the solution to the system is X = B. Therefore we must show that, for each , 1 ≤  ≤ , the formula given in Theorem 3.12 yields  =  (the th entry of B). First, note that |A| = |I | = 1. Now, the th matrix A (as defined in Theorem 3.12) for A is identical to I except in its th column, which equals B. Therefore, the th row of A has  as its th entry, and zeroes elsewhere. Thus, a cofactor expansion along the th row of A yields |A | =  (−1)+ |I−1 | =   c 2016 Elsevier Ltd. All rights reserved.

Copyright °

92

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.3

Hence, for each , the formula in Theorem 3.12 produces  =

 |A | = =   |A| 1

completing the proof. (d) Because A is nonsingular, [A|B] row reduces to [I |X], where X is the unique solution to the system. Let [C|D] represent any intermediate augmented matrix during this row reduction process.

Now, by part (a) and repeated use of part (b), the ratio |A | |C | is identical to the ratio |C| |A| obtained from the original augmented matrix, for each , 1 ≤  ≤ . But part (c) proves that for the final augmented matrix, [I |X], this common ratio gives the correct solution for  , for each , 1 ≤  ≤ . Since all of the systems corresponding to these intermediate matrices have the same unique solution, the formula in Theorem 3.12 gives the correct solution for the original system, [A|B], as well. Thus, Cramer’s Rule is validated.

1 . But Theorem 3.10 asserts that |A| = |A |. Com(20) (a) True. Corollary 3.8 shows that |A−1 | = |A|

bining these two equations produces the desired result. (b) True. According to Theorem 3.11, both cofactor expansions produce |A| as an answer. (The size of the square matrix is not important.) (c) False. Type (III) column operations applied to square matrices change the sign of the determinant, just as Type (III) row operations do. Hence, |B| = −|A|. The equation |B| = |A| can only be true in the special case in which |A| = 0. For a particular counterexample to |B| = |A|, consider ∙ ¸ 0 1 . Note that A = I2 , and use the column operation (III): hcol  1i ↔ hcol  2i. Then B = 1 0 |B| = −1, but |A| = 1.

(d) True. This fact was stated in the paragraph after Example 2 in this section. To prove it, suppose  A ¯ has ¯ two identical columns. Then A ¯ has¯ two identical rows. By Exercise 12 of Section 3.2, ¯A ¯ = 0. But, by Theorem 3.10, |A| = ¯A ¯. Hence, |A| = 0.

(e) True. This fact was stated in the paragraph after Example 2 in this section. ¯ To¯ prove it, suppose A is lower triangular. Then A is upper triangular. By Theorem 3.2, ¯A ¯ = ¯ 11 22 ¯ · · ·  , the product of the entries along its main diagonal. But, by Theorem 3.10, |A| = ¯A ¯. Hence, |A| = 11 22 · · ·  as well.

(f) True. This follows directly from Cramer’s Rule. The coefficient matrix A for the given system is upper triangular, and so |A| is easily found to be (4)(−3)(1) = −12 (see Theorem 3.2). Notice ⎡ ⎤ 4 −6 −1 5 4 ⎦. Cramer’s Rule then states that that A2 = ⎣ 0 0 3 1 2 =

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

1 |A2 | = − |A2 | |A| 12

93

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

Section 3.4 (1) In each part, let A represent the given matrix. ¯ ¯ ¯ −3 −1 ¯¯ ¯ (a) A () = |I2 − A| = ¯ = ( − 3)( − 4) − (2)(−1) = 2 − 7 + 14.

2 −4 ¯ ¯ ¯ ¯  − 2 −1 1 ¯¯ ¯  − 6 0 ¯¯. At this point, we could use basketweaving, but instead, (c) A () = |I3 −A| = ¯¯ 6 ¯ −3 0  ¯ we will use a cofactor expansion on the last column: ¯ ¯ ¯ ¯ ¯ ¯  − 6 ¯¯ −1 ¯¯ 1+3 ¯ 6 3+3 ¯  − 2 + (0) + (−1) A () = (1)(−1) ¯ −3 ¯ 6 0 ¯ −6 ¯

= (1)(0 + 3( − 6)) +  (( − 2)( − 6) + 6) = 3 − 82 + 21 − 18

¯ ¯  1 0 ¯ ¯ 5 −2 1 (e) A () = |I4 − A| = ¯¯ −1 −1 ¯ 0 ¯ −4 1 −3 along the third row: ¯ ¯  0 ¯ 1 A () = (0) + (−1)(−1)3+2 ¯¯ 5 ¯ −4 −3

¯ −1 ¯¯ −2 ¯¯ . Following the hint, we do a cofactor expansion 0 ¯¯  ¯ ¯ ¯ ¯ ¯  −1 ¯¯ 1 −1 ¯¯ ¯ −2 ¯¯ + ( − 1)(−1)3+3 ¯¯ 5  − 2 −2 ¯¯ + (0) ¯ −4  ¯ 1  ¯

Using ³ basketweaving produces A () = ´ (1) ()(1)() + (0)(−2)(−4) + (−1)(5)(−3) −(−1)(1)(−4) − ()(−2)(−3) − (0)(5)() + ³ ´ ( − 1) ()( − 2)() + (1)(−2)(−4) + (−1)(5)(1) −(−1)( − 2)(−4) − ()(−2)(1) − (1)(5)() ,

which, after algebraic simplification, equals 4 − 33 − 42 + 12.

(2) In each part, let A represent the given matrix. Then, to find the eigenspace  , we form the matrix [(I − A)|0] and row reduce to solve the associated linear system. ¯ ¯ ∙ ¸ ∙ ¸ 1 −1 ¯¯ 0 1 −1 ¯¯ 0 (a) Since  = 2, [(I2 −A)|0] = [(2I2 −A)|0] = .

This row reduces to , 2 −2 ¯ 0 0 0 ¯ 0 ½ 1 − 2 = 0  The independent variable is 2 . Let which corresponds to the system 0 = 0

2 = . The first equation then yields 1 =  as well. Hence, 2 , the solution set for this system, is {[1 1] |  ∈ R}. A fundamental eigenvector for  = 2 is [1 1]. ¯ ⎡ ⎤ 4 −2 0 ¯¯ 0 (c) Since  = −1, [(I3 − A)|0] = [((−1)I3 − A)|0] = ⎣ 8 −4 0 ¯¯ 0 ⎦ . This reduces to −4 2 0 ¯ 0 ¯ ⎧ ⎡ ⎤ 1 − 12 0 ¯ 0 ⎨ 1 − 12 2 = 0 ¯ ¯ ⎣ 0 ⎦ 0 = 0  The 2nd and 0 0 ¯ 0 , which corresponds to the system ⎩ 0 = 0 0 0 0 ¯ 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

94

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

3rd columns are nonpivot columns, and so 2 and 3 are independent variables. Let 2 =  and 3 = . The first equation then yields 1 = 12 . We can eliminate fractions by letting  = 2, giving 1 =  and 2 = 2. Hence, −1 , the solution set for this system, is {[ 2 ] |   ∈ R} = {[1 2 0] + [0 0 1] |   ∈ R}. Fundamental eigenvectors for  = −1 are [1 2 0] and [0 0 1].

(3) In each part, let A represent the given matrix. We will follow the first three steps of the Diagonalization Method, as described in Section 3.4 of the textbook. ¯ ¯ ¯ −1 −3 ¯¯ (a) A () = |I2 − A| = ¯¯ = ( − 1)2 . The eigenvalues of A are the roots of A ().

0 −1 ¯

Clearly,  = 1 is the only root of ( − 1)2 . Since the linear factor ( − 1) appears to the 2nd power in A (), this eigenvalue has algebraic multiplicity 2. Next, we solve for the eigenspace 1 by solving the homogeneous system whose augmented matrix is [(1I2 − A)|0]. Now, [(1I2 − A)|0] = ¯ ¯ ∙ ¸ ∙ ¸ ½ 0 −3 ¯¯ 0 0 1 ¯¯ 0 2 = 0 .  which reduces to .

The associated linear system is 0 = 0 0 0 ¯ 0 0 0 ¯ 0

Since column 1 is not a pivot column, 1 is an independent variable. Let 1 = . Then the solution set is 1 = {[ 0] |  ∈ R} = {[1 0] |  ∈ R}. A fundamental eigenvector for  = 1 is [1 0]. ¯ ¯ ¯ −1 0 −1 ¯¯ ¯ −2 3 ¯¯ = ( − 1)( − 2)( + 5), where we have used the (c) A () = |I3 − A| = ¯¯ 0 ¯ 0 0 +5 ¯ fact that |I3 − A| is upper triangular to calculate the determinant quickly, using Theorem 3.2.

The eigenvalues of A are the roots of A (). These are clearly 1 = 1, 2 = 2, and 3 = −5.

Since each of the linear factors ( − 1), ( − 2), and ( + 5) appears in A () raised to the 1st power, each of these eigenvalues has algebraic multiplicity 1. Next, we solve for each eigenspace.

Eigenvalue 1 = 1: We solve the homogeneous system whose augmented matrix is [(1I3 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ 0 0 −1 ¯¯ 0 0 1 0 ¯¯ 0 3 ¯¯ 0 ⎦  which reduces to ⎣ 0 0 1 ¯¯ 0 ⎦. The associated Now, [(1I3 − A)|0] = ⎣ 0 −1 0 0 6 ¯ 0 0 0 0 ¯ 0 ⎧ = 0 ⎨ 2 3 = 0 . Since column 1 is not a pivot column, 1 is an independent linear system is ⎩ 0 = 0 variable. Let 1 = . Then the solution set is 1 = {[ 0 0] |  ∈ R} = {[1 0 0] |  ∈ R}. A fundamental eigenvector for 1 = 1 is [1 0 0].

Eigenvalue 2 = 2: We solve the homogeneous system whose augmented matrix is [(2I3 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 0 −1 ¯¯ 0 1 0 0 ¯¯ 0 3 ¯¯ 0 ⎦  which reduces to ⎣ 0 0 1 ¯¯ 0 ⎦. The associated Now, [(2I3 − A)|0] = ⎣ 0 0 0 0 7 ¯ 0 0 0 0 ¯ 0 ⎧ = 0 ⎨ 1 3 = 0 . Since column 2 is not a pivot column, 2 is an independent linear system is ⎩ 0 = 0 variable. Let 2 = . Then the solution set is 2 = {[0  0] |  ∈ R} = {[0 1 0] |  ∈ R} A fundamental eigenvector for 2 = 2 is [0 1 0].

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

95

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

Eigenvalue 3 = −5: We solve the homogeneous system whose augmented matrix is ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 ¯ 1 0 6 ¯ 0 −6 0 −1 ¯¯ 0 ¯ ⎢ ⎥ 3 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 − 37 ¯ 0 ⎦. The associ[((−5)I3 − A)|0] = ⎣ 0 −7 ¯ 0 0 0 ¯ 0 0 0 0 ¯ 0 ⎧ + 16 3 = 0 ⎪ ⎨ 1 2 − 37 3 = 0 . Since column 3 is not a pivot column, 3 is ated linear system is ⎪ ⎩ 0 = 0

an independent variable. Let 3 = . Then 1 = − 16 3 = − 16  and 2 = 37 3 = 37 . Hence, the ©£ ¤ ª © £ ¤ ª solution set is −5 = − 16  37   |  ∈ R =  − 16  37  1 |  ∈ R . We can eliminate fractions

by multiplying by 42 to obtain −5 = { [−7 18 42] |  ∈ R}  A fundamental eigenvector for 3 = −5 is [−7 18 42]. ¯ ¯ ¯ −4 0 2 ¯¯ ¯ 6 ¯¯ . Cofactor expansion along the second column (e) A () = |I3 − A| = ¯¯ −6  − 2 ¯ −4 0 +2 ¯ ¯ ¯ ³ ´ ¯ −4 2 ¯¯ + (0) = ( − 2) ( − 4)( + 2) − (−4)(2) = yields A () = (0) + ( − 2)(−1)2+2 ¯¯ ¯ −4  + 2 ( − 2)(2 − 2) = ( − 2)2 . The eigenvalues of A are the roots of A (). These are clearly 1 = 0, and 2 = 2. Since the linear factor ( − 0) =  appears in A () raised to the 1st power, 1 = 0 has algebraic multiplicity 1. Similarly, since the linear factor ( − 2) appears to the 2nd power in A (), 2 = 2 has algebraic multiplicity 2. Next, we solve for each eigenspace.

Eigenvalue 1 = 0: We solve the homogeneous system whose augmented matrix is [(0I3 − A)|0] = ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 0 − 12 ¯¯ 0 −4 0 2 ¯¯ 0 ¯ ⎢ ⎥ [−A|0]. Now, [−A|0] = ⎣ −6 −2 6 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 − 32 ¯ 0 ⎦. The ¯ −4 0 2 ¯ 0 0 0 0 ¯ 0 ⎧ 1 − 2 3 = 0 ⎪ ⎨ 1 2 − 32 3 = 0 . Since column 3 is not a pivot column, associated linear system is ⎪ ⎩ 0 = 0 3 is an independent variable. Let 3 = 2, where the coefficient “2” is used to eliminate fractions.

Then 1 = 12 3 = , and 2 = 32 3 = 3. Hence, the solution set is 0 = {[ 3 2] |  ∈ R} =

{[1 3 2] |  ∈ R}. A fundamental eigenvector for 1 = 0 is [1 3 2].

Eigenvalue 2 = 2: We solve the homogeneous system whose augmented matrix is [(2I3 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ −2 0 2 ¯¯ 0 1 0 −1 ¯¯ 0 0 ¯¯ 0 ⎦. The associated Now, [(2I3 − A)|0] = ⎣ −6 0 6 ¯¯ 0 ⎦  which reduces to ⎣ 0 0 −4 0 4 ¯ 0 0 0 0 ¯ 0 ⎧ ⎨ 1 − 3 = 0 0 = 0 . Since columns 2 and 3 are not pivot columns, 2 and 3 linear system is ⎩ 0 = 0 are independent variables. Let 2 =  and 3 = . Then 1 = 3 = , and the solution set is 2 = {[  ] |   ∈ R} = {[0 1 0] + [1 0 1] |   ∈ R}. A pair of fundamental eigenvectors for

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

96

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

2 = 2 is [0 1 0] and [1 0 1]. ¯ ¯ ¯ −3 1 −4 1 ¯¯ ¯ ¯ 0 −3 3 −3 ¯¯ (h) A () = |I4 − A| = ¯¯ ¯ . We use a cofactor expansion along the first 6 −2  + 8 −2 ¯ ¯ ¯ 6 4 2 +4 ¯ column. This gives

¯ ¯ ¯ ¯ ¯ −3 ¯ 1 3 −3 ¯¯ −4 1 ¯¯ ¯ ¯ −3 ¯¯ A () = ( − 3)(−1)1+1 ¯¯ −2  + 8 −2 ¯¯ + (0) + (6)(−1)3+1 ¯¯  − 3 3 ¯ 4 ¯ 4 2 +4 ¯ 2 +4 ¯ ¯ ¯ ¯ 1 −4 1 ¯¯ ¯ 3 −3 ¯¯  + (6)(−1)4+1 ¯¯  − 3 ¯ −2  + 8 −2 ¯

Using basketweaving to find each of the 3 × 3 determinants, we get

A () = ( − 3)( − 3)( + 8)( + 4) + (3)(−2)(4) + (−3)(−2)(2) − (−3)( + 8)(4) − ( − 3)(−2)(2) − (3)(−2)( + 4) + 6(1)(3)( + 4) + (−4)(−3)(4) + (1)( − 3)(2) − (1)(3)(4) − (1)(−3)(2) − (−4)( − 3)( + 4) − 6(1)(3)(−2) + (−4)(−3)(−2) + (1)( − 3)( + 8) − (1)(3)(−2) − (1)(−3)( + 8) − (−4)( − 3)(−2) = ( − 3)(3 + 92 + 18) + 6(42 + 9) − 6(2 ) = 4 + 63 + 92 = 2 (2 + 6 + 9) = 2 ( + 3)2  The eigenvalues of A are the roots of A (). Hence, there are two eigenvalues, 1 = 0 and 2 = −3. Since each of the linear factors ( − 0) =  and ( + 3) appears in A () raised to the 2nd power, each of these eigenvalues has algebraic multiplicity 2. Next, we solve for each eigenspace.

Eigenvalue 1 = 0: We solve the homogeneous system whose augmented matrix is [(0I4 − A)|0] = ¯ ¯ ⎡ ⎤ ⎡ ⎤ −3 1 −4 1 ¯¯ 0 1 0 1 0 ¯¯ 0 ⎢ 0 −3 ⎢ ¯ ⎥ 3 −3 ¯¯ 0 ⎥ ⎥  reducing to ⎢ 0 1 −1 1 ¯ 0 ⎥. The [−A|0]. Now, [−A|0] = ⎢ ¯ ¯ ⎣ 6 −2 ⎦ ⎣ 8 −2 ¯ 0 0 0 0 0 ¯ 0 ⎦ ¯ 6 4 2 4 0 0 0 0 0 ¯ 0 ⎧ + 3 = 0 1 ⎪ ⎪ ⎨ 2 − 3 + 4 = 0 . Since columns 3 and 4 are not associated linear system is 0 = 0 ⎪ ⎪ ⎩ 0 = 0 pivot columns, 3 and 4 are independent variables. Let 3 =  and 4 = . Thus, 1 = −3 = − and 2 = 3 − 4 =  − . Hence, the solution set is 0 = {[−  −   ] |   ∈ R} = {[−1 1 1 0]+[0 −1 0 1] |   ∈ R}. A pair of fundamental eigenvectors for 1 = 0 is [−1 1 1 0] and [0 −1 0 1].

Eigenvalue 2 = −3: We solve the homogeneous system whose augmented matrix is ¯ ⎡ ⎤ ⎤ ⎡ 1 ¯¯ 1 0 0 −6 1 −4 1 ¯¯ 0 2 ¯ 0 ⎢ 0 −6 ⎢ 3 −3 ¯¯ 0 ⎥ 0 ¯¯ 0 ⎥ ⎥. The asso⎥  reducing to ⎢ 0 1 0 [((−3)I4 − A)|0] = ⎢ ¯ ⎣ 6 −2 ⎦ ⎣ 5 −2 ¯ 0 0 0 1 −1 ¯¯ 0 ⎦ 6 4 2 1 ¯ 0 0 0 0 0 ¯ 0 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

97

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

ciated linear system is

⎧ 1 ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

+ 2 3

−

1 2 4

Section 3.4

= 0

= 0 . Since column 4 is not a pivot column, 4 = 0 0 = 0

4 is an independent variable. Let 4 = 2, where we have included the factor “2” to eliminate fractions. Thus, 1 = − 12 4 = −, 2 = 0, and 3 = 4 = 2. Hence, the solution set is

−3 = {[− 0 2 2] |  ∈ R} = {[−1 0 2 2] |  ∈ R}. A fundamental eigenvector for 2 = −3 is [−1 0 2 2]. ¯ ¯ ¯  − 19 48 ¯¯ (4) (a) Step 1: A () = |I2 − A| = ¯¯ = ( − 19)( + 21) − (48)(−8) = 2 + 2 − 15 = −8  + 21 ¯ ( − 3)( + 5).

Step 2: The eigenvalues of A are the roots of A (). Hence, there are two eigenvalues, 1 = 3 and 2 = −5.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 3: We solve the homogeneous system whose augmented matrix is [(3I2 − A)|0]. ¯ ¯ ∙ ¸ ∙ ¸ −16 48 ¯¯ 0 1 −3 ¯¯ 0 Now, [(3I2 − A)|0] =  which reduces to . The associated linear −8 24 ¯ 0 0 0 ¯ 0 ½ 1 − 32 = 0 . Since column 2 is not a pivot column, 2 is an independent system is 0 = 0 variable. Let 2 = 1. Then 1 = 32 = 3. This yields the fundamental eigenvector [3 1].

Eigenvalue 2 = −5: We solve the homogeneous system whose augmented matrix is ¯ ¯ ∙ ¸ ∙ ¸ −24 48 ¯¯ 0 1 −2 ¯¯ 0 [((−5)I2 − A)|0] =  which reduces to . The associated linear −8 16 ¯ 0 0 0 ¯ 0 ½ 1 − 22 = 0 . Since column 2 is not a pivot column, 2 is an independent system is 0 = 0 variable. Let 2 = 1. Then 1 = 22 = 2. This yields the fundamental eigenvector [2 1].

Step 4: Since  = 2, and since we have found 2 fundamental eigenvectors, A is diagonalizable.

Step 5: We form the¸2 × 2 matrix whose columns are the two fundamental eigenvectors we have ∙ 3 2 found: P = .

1 1

Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ∙ ¸ ∙ ¸ 1 −2 3 0 −1 . It is easy to verify for D= . Also, using Theorem 2.14, we get P = −1 3 0 −5

these matrices that D = P−1 AP. ¯ ¯ ¯  − 13 34 ¯¯ (c) Step 1: A () = |I2 − A| = ¯¯ = ( − 13)( + 13) − (34)(−5) = 2 + 1. −5  + 13 ¯

Step 2: The eigenvalues of A are the roots of A (). But A () = 2 + 1 has no real roots.

Therefore, A has no eigenvalues.

Step 3: Because A has no eigenvalues, there are no associated eigenvectors, fundamental or otherwise.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

98

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

Step 4: Since there are no eigenvectors for A, A is not diagonalizable. ¯ ¯ ¯  + 13 3 −18 ¯¯ ¯  + 4 −26 ¯¯. Using basketweaving gives (d) Step 1: A () = |I3 − A| = ¯¯ 20 ¯ 14 3  − 19 ¯

A () = ( + 13)( + 4)( − 19) + (3)(−26)(14) + (−18)(20)(3) − (−18)( + 4)(14) − ( + 13)(−26)(3) − (3)(20)( − 19) 3 =  − 22 −  + 2 = (2 − 1)( − 2) = ( − 1)( + 1)( − 2)

Step 2: The eigenvalues of A are the roots of A (). Hence, there are three eigenvalues, 1 = 1, 2 = −1, and 3 = 2.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 1: We solve the homogeneous system whose augmented matrix is [(1I3 − A)|0]. ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 0 − 65 ¯¯ 0 14 3 −18 ¯¯ 0 ¯ ⎢ ⎥ Now, [(1I3 − A)|0] = ⎣ 20 5 −26 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 − 25 ¯ 0 ⎦. The asso¯ 14 3 −18 ¯ 0 0 0 0 ¯ 0 ⎧ 6 − 5 3 = 0 ⎪ ⎨ 1 2 − 25 3 = 0 . Since column 3 is not a pivot column, 3 ciated linear system is ⎪ ⎩ 0 = 0 is an independent variable. We choose 3 = 5 to eliminate fractions. Then 1 = 65 3 = 6 and 2 = 25 3 = 2. This yields the fundamental eigenvector [6 2 5].

Eigenvalue 2 = −1: We solve the homogeneous system whose augmented matrix is ¯ ¯ ⎡ ⎤ ⎡ ⎤ 12 3 −18 ¯¯ 0 1 0 −1 ¯¯ 0 [((−1)I3 − A)|0] = ⎣ 20 3 −26 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 −2 ¯¯ 0 ⎦. The associated 14 3 −20 ¯ 0 0 0 0 ¯ 0 ⎧ − 3 = 0 ⎨ 1 2 − 23 = 0 . Since column 3 is not a pivot column, 3 is an linear system is ⎩ 0 = 0 independent variable. Let 3 = 1. Then 1 = 3 = 1 and 2 = 23 = 2. This yields the fundamental eigenvector [1 2 1].

Eigenvalue 3 = 2: we solve the homogeneous system whose augmented matrix is [(2I3 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ 15 3 −18 ¯¯ 0 1 0 −1 ¯¯ 0 Now, [(2I3 − A)|0] = ⎣ 20 6 −26 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 −1 ¯¯ 0 ⎦. The asso14 3 −17 ¯ 0 0 0 0 ¯ 0 ⎧ − 3 = 0 ⎨ 1 2 − 3 = 0 . Since column 3 is not a pivot column, 3 is ciated linear system is ⎩ 0 = 0 an independent variable. Let 3 = 1. Then 1 = 3 = 1 and 2 = 3 = 1. This yields the fundamental eigenvector [1 1 1].

Step 4: Since  = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable.

Step 5: We form the 3 × 3 matrix whose columns are the 3 fundamental eigenvectors we have

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

99

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

⎡

⎤ 6 1 1 found: P = ⎣ 2 2 1 ⎦.

5 1 1

Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ⎡ ⎤ ⎡ ⎤ 1 0 −1 1 0 0 1 −4 ⎦. It is easy D = ⎣ 0 −1 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ 3 −8 −1 10 0 0 2

to verify for these matrices that D = P−1 AP. ¯ ¯ ¯ −5 8 12 ¯¯ ¯  − 3 −4 ¯¯. Using basketweaving gives us (f) Step 1: A () = |I3 − A| = ¯¯ 2 ¯ −4 6 +9 ¯

A () = ( − 5)( − 3)( + 9) + (8)(−4)(−4) + (12)(2)(6) − (12)( − 3)(−4) − ( − 5)(−4)(6) − (8)(2)( + 9) 3 =  + 2 −  − 1 = (2 − 1)( + 1) = ( − 1)( + 1)2 

Step 2: The eigenvalues of A are the roots of A (). Hence, there are two eigenvalues, 1 = 1, and 2 = −1.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 1: We solve the homogeneous system whose augmented matrix is [(1I3 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ −4 8 12 ¯¯ 0 1 0 −1 ¯¯ 0 1 ¯¯ 0 ⎦. The assoNow, [(1I3 − A)|0] = ⎣ 2 −2 −4 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 ¯ −4 6 10 0 0 0 0 ¯ 0 ⎧ − 3 = 0 ⎨ 1 2 + 3 = 0 . Since column 3 is not a pivot column, 3 is ciated linear system is ⎩ 0 = 0 an independent variable. Let 3 = 1. Then 1 = 3 = 1 and 2 = −3 = −1. This yields the fundamental eigenvector [1 −1 1].

Eigenvalue 2 = −1: We solve the homogeneous system whose augmented matrix is ¯ ¯ ⎡ ⎤ ⎡ ⎤ −6 8 12 ¯¯ 0 1 0 −2 ¯¯ 0 0 ¯¯ 0 ⎦. The associated [((−1)I3 − A)|0] = ⎣ 2 −4 −4 ¯¯ 0 ⎦  reducing to ⎣ 0 1 −4 6 8 ¯ 0 0 0 0 ¯ 0 ⎧ − 23 = 0 ⎨ 1 = 0 . Since column 3 is not a pivot column, 3 is an 2 linear system is ⎩ 0 = 0 independent variable. Let 3 = 1. Then 1 = 23 = 2 and 2 = 0. This yields the fundamental eigenvector [2 0 1].

Step 4: Since  = 3, and the process has produced only 2 fundamental eigenvectors, A is not diagonalizable. ¯ ¯ ¯ −2 0 0 ¯¯ ¯  − 4 −1 ¯¯ . Using a cofactor expansion along the first (g) Step 1: A () = |I3 − A| = ¯¯ 3 ¯ −3 2 −1 ¯ c 2016 Elsevier Ltd. All rights reserved.

Copyright °

100

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

row gives ¯ ¯ ¯  − 4 −1 ¯ ¯ = ( − 2) (( − 4)( − 1) − (−1)(2)) A () = ( − 2)(−1)1+1 ¯¯ 2 −1 ¯ = ( − 2)(2 − 5 + 6) = ( − 2)2 ( − 3)

Step 2: The eigenvalues of A are the roots of A (). Hence, there are two eigenvalues, 1 = 2, and 2 = 3.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 2: We solve the homogeneous system whose augmented matrix is [(2I3 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 − 23 − 13 ¯ 0 0 0 0 ¯¯ 0 ¯ Now, [(2I3 − A)|0] = ⎣ 3 −2 −1 ¯¯ 0 ⎦  which reduces to ⎣ 0 0 0 ¯¯ 0 ⎦. The as−3 2 1 ¯ 0 0 0 0 ¯ 0 ⎧ 2 1 ⎨ 1 − 3 2 − 3 3 = 0 sociated linear system is 0 = 0 . Since columns 2 and 3 are not pivot ⎩ 0 = 0 columns, 2 and 3 are independent variables. First, we choose 2 = 3 (to eliminate fractions)

and 3 = 0. Then 1 = 23 2 + 13 3 = 2, producing the fundamental eigenvector [2 3 0]. Second, we choose 2 = 0 and 3 = 3 (to eliminate fractions). This gives 1 = 23 2 + 13 3 = 1, yielding the fundamental eigenvector [1 0 3].

Eigenvalue 2 = 3: We solve the homogeneous system whose augmented matrix is [(3I3 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 0 0 ¯¯ 0 1 0 0 ¯¯ 0 Now, [(3I3 − A)|0] = ⎣ 3 −1 −1 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 1 ¯¯ 0 ⎦. The associ−3 2 2 ¯ 0 0 0 0 ¯ 0 ⎧ = 0 ⎨ 1 2 + 3 = 0 . Since column 3 is not a pivot column, 3 is an ated linear system is ⎩ 0 = 0 independent variable. Let 3 = 1. Then 1 = 0 and 2 = −3 = −1. This yields the fundamental eigenvector [0 −1 1].

Step 4: Since  = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable.

Step 5: We ⎡form the 3 ×⎤3 matrix whose columns are the 3 fundamental eigenvectors we have 2 1 0 found: P = ⎣ 3 0 −1 ⎦.

0 3 1

Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ⎤ ⎡ ⎡ ⎤ 1 − 13 − 13 2 0 0 ⎢ 2 2 ⎥ D = ⎣ 0 2 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ −1 3 3 ⎦. It is easy to 0 0 3 3 −2 −1 verify for these matrices that D = P−1 AP.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

101

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

¯ ¯ ¯  − 3 −1 6 2 ¯¯ ¯ ¯ −4  6 4 ¯¯ (i) Step 1: A () = |I4 − A| = ¯¯ . Using a cofactor expansion along the 0 +3 2 ¯¯ ¯ −2 ¯ 0 ¯ −1 2 −1 last row gives

¯ ¯ ¯ ¯ ¯ −3 ¯  − 3 −1 2 ¯ 6 2 ¯¯ ¯ ¯ ¯ 6 4 ¯¯ + (2)(−1)4+3 ¯¯ −4  4 ¯¯ A () = (0) + (−1)(−1)4+2 ¯¯ −4 ¯ −2  + 3 2 ¯ ¯ −2 0 2 ¯ ¯ ¯ ¯  − 3 −1 6 ¯¯ ¯ 4+4 ¯  6 ¯¯  + ( − 1)(−1) ¯ −4 ¯ −2 0 +3 ¯

Using basketweaving on each 3 × 3 determinant produces

A () = −( − 3)(6)(2) + (6)(4)(−2) + (2)(−4)( + 3) − (2)(6)(−2) − ( − 3)(4)( + 3) − (6)(−4)(2) − 2( − 3)()(2) + (−1)(4)(−2) + (2)(−4)(0) − (2)((−2) − ( − 3)(4)(0) − (−1)(−4)(2) + ( − 1)( − 3)()( + 3) + (−1)(6)(−2) + (6)(−4)(0) − (6)()(−2) − ( − 3)(6)(0) − (−1)(−4)( + 3) = −(−42 + 4) − 2(22 − 2) + ( − 1)(3 − ) = 4( − 1) − 4( − 1) + ( − 1)(2 − 1) = ( − 1)2 ( + 1) Step 2: The eigenvalues of A are the roots of A (). Hence, there are three eigenvalues, 1 = 1, 2 = −1 and 3 = 0.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 1: We solve the homogeneous system whose augmented matrix is [(1I4 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ −2 −1 6 2 ¯¯ 0 1 0 −2 −1 ¯¯ 0 ⎢ −4 ⎢ 1 6 4 ¯¯ 0 ⎥ 0 ¯¯ 0 ⎥ ⎥  reducing to ⎢ 0 1 −2 ⎥. The asNow, [(1I4 − A)|0] = ⎢ ⎣ −2 ⎣ 0 0 0 4 2 ¯¯ 0 ⎦ 0 0 ¯¯ 0 ⎦ 0 −1 2 0 ¯ 0 0 0 0 0 ¯ 0 ⎧ − 23 − 4 = 0 1 ⎪ ⎪ ⎨ = 0 2 − 23 . Since columns 3 and 4 are not sociated linear system is 0 = 0 ⎪ ⎪ ⎩ 0 = 0

pivot columns, 3 and 4 are independent variables. First, we let 3 = 1 and 4 = 0. Then 1 = 23 + 4 = 2 and 2 = 23 = 2. This yields the fundamental eigenvector [2 2 1 0]. Second, we let 3 = 0 and 4 = 1. Then 1 = 23 + 4 = 1 and 2 = 23 = 0, producing the fundamental eigenvector [1 0 0 1].

Eigenvalue 2 = −1: We solve the homogeneous system whose augmented matrix is ¯ ¯ ⎡ ⎤ ⎡ ⎤ −4 −1 6 2 ¯¯ 0 1 0 −1 0 ¯¯ 0 ⎢ −4 −1 6 ⎢ ¯ ⎥ 4 ¯¯ 0 ⎥ ⎥  reducing to ⎢ 0 1 −2 0 ¯ 0 ⎥. The associ[((−1)I4 − A)|0] = ⎢ ¯ ¯ ⎣ −2 ⎦ ⎣ 0 2 2 ¯ 0 0 0 0 1 ¯ 0 ⎦ ¯ 0 −1 2 −2 0 0 0 0 0 ¯ 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

102

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

ated linear system is

⎧ 1 ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

2

− 3 − 23

= = 4 = 0 =

Section 3.4

0 0 . Since column 3 is not a pivot column, 3 0 0

is an independent variable. Let 3 = 1. Then 1 = 3 = 1, 2 = 23 = 2, and 4 = 0. This yields the fundamental eigenvector [1 2 1 0].

Eigenvalue 3 = 0: We solve the homogeneous system whose augmented matrix is [(0I4 − A)|0] = ¯ ¯ ⎡ ⎤ ⎡ ⎤ −3 −1 6 2 ¯¯ 0 1 0 0 −1 ¯¯ 0 ⎢ −4 ⎢ 0 6 4 ¯¯ 0 ⎥ 1 ¯¯ 0 ⎥ ⎥  reducing to ⎢ 0 1 0 ⎥. The [−A|0]. Now, [−A|0] = ⎢ ¯ ⎣ −2 ⎦ ⎣ 0 3 2 ¯ 0 0 0 1 0 ¯¯ 0 ⎦ 0 −1 2 −1 ¯ 0 0 0 0 0 ¯ 0 ⎧ − 4 = 0 1 ⎪ ⎪ ⎨ + 4 = 0 2 . Since column 4 is not a pivot colassociated linear system is = 0  ⎪ 3 ⎪ ⎩ 0 = 0 umn, 4 is an independent variable. Let 4 = 1. Then 1 = 4 = 1, 2 = −4 = −1, and 3 = 0.

This yields the fundamental eigenvector [1 −1 0 1].

Step 4: Since  = 4, and we have found 4 fundamental eigenvectors, A is diagonalizable.

Step 5: We ⎡ form the 4 × 4 matrix whose columns are the four fundamental eigenvectors we have ⎤ 2 1 1 1 ⎢ 2 0 2 −1 ⎥ ⎥. found: P = ⎢ ⎣ 1 0 1 0 ⎦ 0 1 0 1

Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ⎡ ⎡ ⎤ ⎤ 1 0 −1 −1 1 0 0 0 ⎢ 0 ⎢ 0 1 1 −2 1 ⎥ 0 0 ⎥ −1 ⎢ ⎥. It ⎥ D=⎢ ⎣ 0 0 −1 0 ⎦. Also, by row reducing [P|I4 ], we get P = ⎣ −1 0 2 1 ⎦ 0 −1 2 0 0 0 0 0

is easy to verify for the given matrices that D = P−1 AP.

(5) In each part, we use the Diagonalization Method to find a matrix P and a diagonal matrix D such that A = PDP−1 . Then, A = PD P−1 . For a diagonal matrix D, D is easily computed as the diagonal matrix whose ( ) entry is  . ¯ ¯ ¯ −4 6 ¯¯ (a) First, we diagonalize A. Step 1: A () = |I2 − A| = ¯¯ −3  + 5 ¯ = ( − 4)( + 5) − (6)(−3) = 2 +  − 2 = ( − 1)( + 2).

Step 2: The eigenvalues of A are the roots of A (). Hence, there are two eigenvalues, 1 = 1 and 2 = −2.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 1: We solve the homogeneous system whose augmented matrix is [(1I2 − A)|0]. ¯ ¯ ∙ ¸ ∙ ¸ −3 6 ¯¯ 0 1 −2 ¯¯ 0 Now, [(1I2 − A)|0] =  which reduces to . The associated linear −3 6 ¯ 0 0 0 ¯ 0 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

103

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

system is

½

1

Section 3.4

− 22 = 0 . Since column 2 is not a pivot column, 2 is an independent 0 = 0

variable. Let 2 = 1. Then 1 = 22 = 2. This yields the fundamental eigenvector [2 1].

For 2 = −2, we solve the homogeneous system whose augmented matrix is [((−2)I2 − A)|0] = ¯ ¯ ∙ ¸ ∙ ¸ −6 6 ¯¯ 0 1 −1 ¯¯ 0  which reduces to . The associated linear system is −3 3 ¯ 0 0 0 ¯ 0 ½ 1 − 2 = 0 . Since column 2 is not a pivot column, 2 is an independent variable. Let 0 = 0 2 = 1. Then 1 = 2 = 1. This yields the fundamental eigenvector [1 1].

Step 4: Since  = 2, and we have found 2 fundamental eigenvectors, A is diagonalizable.

Step 5: We form the¸2 × 2 matrix whose columns are the two fundamental eigenvectors we have ∙ 2 1 found: P = .

1 1 Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ∙ ¸ ∙ ¸ 1 −1 1 0 . Thus, D= . Also, using Theorem 2.14, we get P−1 = −1 2 0 −2 A

15

15

−1

= PD P

=

∙

2 1 1 1

¸"

115

0

0

(−2)15

#∙

1 −1 −1 2

¸

=

∙

32770 −65538 32769 −65537

¸



¯ ¯ ¯  − 11 6 12 ¯¯ ¯ 16 ¯¯. Using bas(c) First, we diagonalize A. Step 1: A () = |I3 − A| = ¯¯ −13  + 6 ¯ −5 3 +5 ¯ ketweaving gives A () = ( − 11)( + 6)( + 5) + (6)(16)(−5) + (12)(−13)(3) − (12)( + 6)(−5) − ( − 11)(16)(3) − (6)(−13)( + 5) 3 =  −  = (2 − 1) = ( − 1)( + 1)

Step 2: The eigenvalues of A are the roots of A (). Hence, there are three eigenvalues, 1 = 0, 2 = 1 and 3 = −1.

Normally, at this point, we solve for fundamental eigenvectors. However, as we will see, we do not need to compute the actual matrix P in this case. Note that we get (at least) one fundamental eigenvector from each eigenvalue, so we obtain 3 fundamental eigenvectors.

Hence, A ⎡ ⎤ 0 0 0 0 ⎦ is the diagonal is diagonalizable, and A = PDP−1 for some matrix P, where D = ⎣ 0 1 0 0 −1 ⎤ ⎡ 0 0 0 ⎥ ⎢ 0 matrix whose diagonal entries are the eigenvalues of A. Now D49 = ⎣ 0 149 ⎦ = D. 0

49

Thus, A

49

−1

= PD P

−1

= PDP

= A.

0

(−1)49

¯ ¯ ¯ −7 −9 12 ¯¯ ¯ 22 ¯¯. Using bas(e) First, we diagonalize A. Step 1: A () = |I3 − A| = ¯¯ −10  − 16 ¯ −8 −12  + 16 ¯

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

104

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

ketweaving gives A () = ( − 7)( − 16)( + 16) + (−9)(22)(−8) + (12)(−10)(−12) − (12)( − 16)(−8) − ( − 7)(22)(−12) − (−9)(−10)( + 16) = 3 − 72 + 14 − 8 = ( − 1)( − 2)( − 4) Step 2: The eigenvalues of A are the roots of A (). Hence, there are three eigenvalues, 1 = 1, 2 = 2 and 3 = 4.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 1: We solve the homogeneous system whose augmented matrix is [(1I3 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 32 0 ¯ 0 −6 −9 12 ¯¯ 0 ¯ Now, [(1I3 − A)|0] = ⎣ −10 −15 22 ¯¯ 0 ⎦  which reduces to ⎣ 0 0 1 ¯¯ 0 ⎦. The asso−8 −12 17 ¯ 0 0 0 0 ¯ 0 ⎧ 3 = 0 ⎨ 1 + 2 2 ciated linear system is 3 = 0 . Since column 2 is not a pivot column, 2 is ⎩ 0 = 0 an independent variable. We choose 2 = 2 to eliminate fractions. Then 1 = − 32 2 = −3 and 3 = 0, producing the fundamental eigenvector [−3 2 0].

Eigenvalue 2 = 2: We solve the homogeneous system whose augmented matrix is [(2I3 − A)|0]. ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 0 − 32 ¯¯ 0 −5 −9 12 ¯¯ 0 ¯ ⎢ ⎥ Now, [(2I3 − A)|0] = ⎣ −10 −14 22 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 − 12 ¯ 0 ⎦. The as¯ −8 −12 18 ¯ 0 0 0 0 ¯ 0 ⎧ 3 − 2 3 = 0 ⎪ ⎨ 1 2 − 12 3 = 0 . Since column 3 is not a pivot column, 3 sociated linear system is ⎪ ⎩ 0 = 0 is an independent variable. We choose 3 = 2 to eliminate fractions. Then 1 = 32 3 = 3 and 2 = 12 3 = 1, yielding the fundamental eigenvector [3 1 2].

Eigenvalue 3 = 4: We solve the homogeneous system whose augmented matrix is [(4I3 − A)|0]. ¯ ¯ ⎡ ⎤ ⎡ ⎤ −3 −9 12 ¯¯ 0 1 0 −1 ¯¯ 0 Now, [(4I3 − A)|0] = ⎣ −10 −12 22 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 −1 ¯¯ 0 ⎦. The as−8 −12 20 ¯ 0 0 0 0 ¯ 0 ⎧ − 3 = 0 ⎨ 1 2 − 3 = 0 . Since column 3 is not a pivot column, 3 sociated linear system is ⎩ 0 = 0 is an independent variable. Let 3 = 1. Then 1 = 3 = 1 and 2 = 3 = 1. This yields the fundamental eigenvector [1 1 1].

Step 4: Since  = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable.

Step 5: We ⎡form the 3 ×⎤3 matrix whose columns are the 3 fundamental eigenvectors we have −3 3 1 found: P = ⎣ 2 1 1 ⎦.

0 2 1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

105

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ⎡ ⎤ ⎡ ⎤ −1 −1 2 1 0 0 5 ⎦. Thus, D = ⎣ 0 2 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ −2 −3 4 6 −9 0 0 4 ⎡ ⎤ ⎤ ⎡ ⎤ 110 0 0 ⎡ −1 −1 2 −3 3 1 ⎢ ⎥ 5 ⎦ 0 ⎦ ⎣ −2 −3 A10 = PD10 P−1 = ⎣ 2 1 1 ⎦ ⎣ 0 210 4 6 −9 0 2 1 10 0 0 4 ⎡ ⎤ 4188163 6282243 −9421830 = ⎣ 4192254 6288382 −9432060 ⎦  4190208 6285312 −9426944

−1 (7) (b) If A has all eigenvalues nonnegative, then A has a square root. √ Suppose A2 = PDP for some diagonal matrix D. Let C be the diagonal matrix with  =  . Thus, C = D. Let B = PCP−1 . Then, B2 = (PCP−1 )2 = (PCP−1 )(PCP−1 ) = PC(P−1 P)CP−1 = PCI CP−1 = PC2 P−1 = PDP−1 = A. ⎡ ⎤ 15 −14 −14 16 17 ⎦. We will use the Diagonalization Method to find a matrix P and a diag(8) Let B = ⎣ −13 20 −22 −23 √ onal matrix D such that B = PDP−1 . If C is the diagonal matrix with  = 3  , then C3 = D. So, −1 if A = PCP−1 , then A3 = (PCP−1 )3 = (PCP−1 )(PCP−1 )(PCP−1 ) = PC(P P)C(P−1 P)CP−1 = PCI CI CP−1 = PC3 P−1 = PDP−1 = B. Hence, to solve this problem, we first need to find P and D. To do this, we diagonalize B. ¯ ¯ ¯  − 15 14 14 ¯¯ ¯  − 16 −17 ¯¯.

Step 1: B () = |I3 − B| = ¯¯ 13 ¯ −20 22  + 23 ¯

Using basketweaving, we obtain

B () = ( − 15)( − 16)( + 23) + (14)(−17)(−20) + (14)(13)(22) − (14)( − 16)(−20) − ( − 15)(−17)(22) − (14)(13)( + 23) = 3 − 82 −  + 8 = (2 − 1)( − 8) = ( − 1)( + 1)( − 8) Step 2: The eigenvalues of B are the roots of B (). Hence, there are three eigenvalues, 1 = 1, 2 = −1 and 3 = 8.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 1: We solve the homogeneous system whose augmented matrix is [(1I3 − B)|0]. Now, ¯ ¯ ⎡ ⎤ ⎡ ⎤ −14 14 14 ¯¯ 0 1 0 1 ¯¯ 0 [(1I3 −B)|0] = ⎣ 13 −15 −17 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 2 ¯¯ 0 ⎦. The associated linear −20 22 24 ¯ 0 0 0 0 ¯ 0 ⎧ + 3 = 0 ⎨ 1 2 + 23 = 0 . Since column 3 is not a pivot column, 3 is an independent system is ⎩ 0 = 0 variable. Let 3 = 1. Then 1 = −3 = −1 and 2 = −23 = −2. This yields the fundamental eigenvector [−1 −2 1].

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

106

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

Eigenvalue 2 = −1: We solve the homogeneous system whose augmented matrix is [((−1)I3 − B)|0] = ¯ ¯ ⎡ ⎤ ⎡ ⎤ −16 14 14 ¯¯ 0 1 0 0 ¯¯ 0 ⎣ 13 −17 −17 ¯ 0 ⎦  which reduces to ⎣ 0 1 1 ¯ 0 ⎦. The associated linear system is ¯ ¯ −20 22 22 ¯ 0 0 0 0 ¯ 0 ⎧ = 0 ⎨ 1 2 + 3 = 0 . Since column 3 is not a pivot column, 3 is an independent variable. ⎩ 0 = 0

Let 3 = 1. Then 1 = 0 and 2 = −3 = −1, producing the fundamental eigenvector [0 −1 1].

Eigenvalue 3 = 8: We solve the homogeneous system whose augmented matrix is [(8I3 − B)|0]. Now, ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 0 −1 ¯¯ 0 −7 14 14 ¯¯ 0 ⎢ ⎥ 1 ¯ [(8I3 − B)|0] = ⎣ 13 −8 −17 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 2 ¯¯ 0 ⎦. The associated lin−20 22 31 ¯ 0 0 0 0 ¯ 0 ⎧ − 3 = 0 ⎪ ⎨ 1 1 2 + 2 3 = 0 . Since column 3 is not a pivot column, 3 is an independent ear system is ⎪ ⎩ 0 = 0 variable. We choose 3 = 2 to eliminate fractions. Then 1 = 3 = 2 and 2 = − 12 3 = −1, yielding the fundamental eigenvector [2 −1 2].

Step 4: Since  = 3, and we have found 3 fundamental eigenvectors, B is diagonalizable.

Step ⎡5: We form the 3⎤× 3 matrix whose columns are the 3 fundamental eigenvectors we have found: −1 0 2 P = ⎣ −2 −1 −1 ⎦.

1 1 2 Step 6: The matrix whose diagonal entries are the eigenvalues of B (in the correct order) is ⎡ ⎤ ⎡ ⎤ 1 −2 −2 1 0 0 4 5 ⎦.

D = ⎣ 0 −1 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ −3 1 −1 −1 0 0 8

Next, we compute the diagonal matrix C whose main diagonal entries are the cube roots of the eigen⎡ ⎤ 1 0 0 values of B. That is, C = ⎣ 0 −1 0 ⎦. Then 0 0 2 ⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤ −1 0 2 1 0 0 1 −2 −2 3 −2 −2 4 5 ⎦ = ⎣ −7 10 11 ⎦  A = PCP−1 = ⎣ −2 −1 −1 ⎦ ⎣ 0 −1 0 ⎦ ⎣ −3 1 1 2 0 0 2 1 −1 −1 8 −10 −11

Direct calculation of A3 verifies that A3 = B. ∙ ¸ 0 −1 (10) (b) Consider the matrix A = . Now A has no eigenvalues, because A () = |I2 − A| = 1 0 ¯ ¯ ¯  1 ¯ 2 4 ¯ ¯ ¯ −1  ¯ =  +1, which has no real roots. However, A = I2 has 1 as an eigenvalue. Thus, with √  = 4, 4 = 1 is an eigenvalue for A4 , but neither solution for  = ± 4 1 = ±1 is an eigenvalue for A. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

107

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 3.4

(24) (a) True. This is the definition of what it means for 5 to be an eigenvalue of A. (b) False. The eigenvalues of A are the solutions of A () = |I − A| = 0. The statement given in the text is missing the determinant symbol, and also has a zero matrix on the right side of the equation. (c) True. This is part of Theorem 3.13. (d) True. This fact is stated in Step 6 of the Diagonalization Method. (e) False. If every ∙column ¸ of P is a multiple of the same eigenvector, then |P| = 0, and P is singular.

1 1 Consider A = , which has  = 1 as an eigenvalue (why?). The Diagonalization Method 0 2 produces [1 0] as a fundamental eigenvector for . Note that [2 0] is ∙also an eigenvector for . If ¸ 1 2 we use the eigenvectors [1 0] and [2 0] as columns for P, we get P = , which is singular.

0 0

(f) True. The given characteristic polynomial indicates that the algebraic multiplicity of the eigenvalue  = −1 is 1. We stated in Section 3.4 (without proof) that the number of fundamental eigenvectors produced for a given eigenvalue in Step 3 of the Diagonalization Method is always less than or equal to the algebraic multiplicity of the eigenvalue. (This is proven in Chapter 5.)

Hence, no more than one fundamental eigenvector for  can be produced in the Diagonalization Method. (Note that since the Diagonalization Method always produces at least one fundamental eigenvector for each eigenvalue, in this case the method must produce exactly one fundamental eigenvector for the eigenvalue .)

(g) True. Since Step 3 of the Diagonalization Method always produces at least one fundamental eigenvector for each eigenvalue of A, the method must yield (at least) three fundamental eigenvectors.

Step 4 then assures us that A is diagonalizable. (h) False. In the discussion in Section 3.4 about using eigenvalues and eigenvectors to compute large powers of a matrix, we noted that, in general, A = PDP−1 implies A = PD P−1 (where D is a diagonal matrix). Neither P nor P−1 should be raised to the th power. For a specific counterexample to the equation A = P D (P−1 ) given in the problem, consider ∙ ¸ 19 −48 A= , from Exercise 4(a). In the solution to that exercise (see above), we found that 8 −21 ∙ ¸ ∙ ¸ 3 2 3 0 −1 −1 D = P AP, and so A = PDP , for the matrices P = and D = . Now, a 1 1 0 −5 ∙ ¸ ∙ ¸ −23 96 −503 1408 , but P2 D2 (P−1 )2 = instead. short computation shows that A2 = −16 57 −192 537 Note, however, that PD2 P−1 = A2 .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

108

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 3 Review

Chapter 3 Review Exercises

¯ ¯ ¯ 4 −5 2 ¯¯ ¯ 1 −2 ¯¯. We calculate this determinant (1) (b) The (3,4) cofactor = A34 = (−1)3+4 |A34 | = − ¯¯ −6 ¯ −7 0 −1 ¯ using basketweaving: &

&

4 −6 −7 .

× ×

&

−5 1 0

.

.

2 × ×

−2 −1

.

.

4 × ×

−6 −7

&

× × &

.

−5 1 0 &

¯ ¯ ¯ 4 −5 −3 ¯ ¯ ¯ ¯ −6 1 −4 ¯¯ = (4 × 1 × (−1)) + ((−5) × (−2) × (−7)) + (2 × (−6) × 0) ¯ ¯ 3 −8 2 ¯

− (2 × 1 × (−7)) − (4 × (−2) × 0) − ((−5) × (−6) × (−1)) = −4 − 70 + 0 + 14 + 0 + 30 = −30

Hence, A34 = −(−30) = 30.

(d) Using cofactor expansion along the second column, we have: |A| = 12 A12 + 22 A22 + 32 A32 + 42 A42 ¯ ¯ ¯ ¯ ¯ −6 −2 −4 ¯ ¯ 4 2 −3 ¯¯ ¯ ¯ ¯ 5 2 ¯¯ + 1(−1)2+2 ¯¯ 3 5 2 ¯¯ = (−5)(−1)1+2 ¯¯ 3 ¯ −7 −1 ¯ −7 −1 9 ¯ 9 ¯ ¯ ¯ ¯ 4 2 −3 ¯¯ ¯ + (−8)(−1)3+2 ¯¯ −6 −2 −4 ¯¯ + 0 ¯ −7 −1 9 ¯

Using cofactor expansion along the first row for each 3 × 3 matrix yields: ¯ ¯ ¯ ¯ ¯ ¯¶ µ ¯ 5 2 ¯ ¯ ¯ ¯ 5 ¯¯ ¯ − (−2) ¯ 3 2 ¯ + (−4) ¯ 3 |A| = 5 (−6) ¯¯ ¯ −7 9 ¯ ¯ −7 −1 ¯ −1 9 ¯ ¯ ¯ ¯ ¯ ¯¶ µ ¯ ¯ 5 2 ¯ ¯ ¯ ¯ 5 ¯¯ ¯ − 2 ¯ 3 2 ¯ + (−3) ¯ 3 + 1 4 ¯¯ ¯ −7 9 ¯ ¯ −7 −1 ¯ −1 9 ¯ ¯ ¯ ¯ ¯ ¯¶ µ ¯ ¯ −2 −4 ¯ ¯ ¯ ¯ ¯ ¯ − 2 ¯ −6 −4 ¯ + (−3) ¯ −6 −2 ¯ + 8 4 ¯¯ ¯ ¯ ¯ ¯ −1 9 −7 9 −7 −1 ¯ = 5 ((−6) × 47 + 2 × 41 − 4 × 32) + 1 (4 × 47 − 2 × 41 − 3 × 32) + 8 (4 × (−22) − 2 × (−82) − 3 × (−8)) = 5 × (−328) + 10 + 8 × 100 = −1640 + 10 + 800 = −830 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

109

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 3 Review

(3) We use row operations to put A into upper triangular form. Notice in the solution below that we stop performing row operations as soon as we obtain an upper triangular matrix. We give charts indicating the row operations used, keeping track of the variable  , as described in Example 5 in Section 3.2 of the textbook. Recall that  = 1 at the beginning of the process. We then use the final upper triangular matrix obtained and the value of  to compute the desired determinant. ⎡ ⎤ 2 0 −3 3 ⎢ 4 −2 −1 3 ⎥ ⎥  Pivoting at the (1 1) entry: Initial Matrix: A = ⎢ ⎣ 1 −1 0 −2 ⎦ 2 1 −2 1 Effect



Multiply  by 12

1 2 1 2 1 2 1 2

Row Operations h1i ← 12 h1i

(I):

(II): h2i ← (−4) h1i + h2i

No change

(II): h4i ← (−2) h1i + h4i

No change

(II): h3i ← (−1) h1i + h3i

No change

⎡

1

0 − 32

⎢ ⎢ 0 −2 gives the result ⎢ ⎢ 0 −1 ⎣ 0

1

3 2

⎤

⎥ −3 ⎥ ⎥ 3 7 ⎥ − 2 2 ⎦ 1 −2 5

We move on to the second column: Row Operations

Effect



(I): h2i ← − 12 h2i

Multiply  by − 12

− 14

(II): h4i ← (−1) h2i + h4i

No change

− 14

(II): h3i ← (1) h2i + h3i

No change

− 14

⎡

⎢ ⎢ 0 1 − 52 gives the result ⎢ ⎢ 0 0 −1 ⎣ 0 0

We move on to the third column: Row Operations (I): h3i ← (−1) h3i (II):

h4i ← (− 72 ) h3i + h4i

Effect



Multiply  by −1

1 4 1 4

No change

1 0 − 32

⎡

7 2

1 0 − 32

⎢ ⎢ 0 1 − 52 gives the result ⎢ ⎢ 0 0 1 ⎣ 0 0

3 2 3 2

⎤

3 2 3 2

⎤

⎥ ⎥ ⎥ −2 ⎥ ⎦ − 72 ⎥ ⎥ ⎥ = B, 2 ⎥ ⎦

0 − 21 2

the desired matrix in upper triangular form. The final value of  is 14 . By Theorem 3.2, |B| =

21 1 21 (1)(1)(1)(− 21 2 ) = − 2 . Hence, as in Example 5 in Section 3.2, |A| =  × |B| = 4 × (− 2 ) = −42.

(5) (a) By part (1) of Theorem 3.3, |B| = (−4) × |A| = (−4) × (−15) = 60.

(b) By part (2) of Theorem 3.3, performing a Type (II) row operation does not change the determinant of a matrix. Hence, |B| = |A| = −15.

(c) By part (3) of Theorem 3.3, performing a Type (III) row operation changes the sign of the determinant of a matrix. Hence, |B| = −|A| = −(−15) = 15.

(7) |−3A B−1 | = (−3)3 |A B−1 | (by Corollary 3.4) = (−27)|A ||B−1 | (by Theorem 3.7) = (−27)|A||B−1 | 1 (by Corollary 3.8) = (−27)(−7)(1(12)) = (−27)(−7)(2) = 378. (by Theorem 3.10) = (−27)|A| |B| ∙ ¸ ∙ ¸ 4 3 (8) (a) Let B = A =A . Then AX = B has at least two solutions, and so by Table 3.1 at −7 1 the end of Section 3.2, |A| = 0.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

110

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 3 Review

(b) By Table 3.1, AX = 0 has infinitely many solutions. One nontrivial solution is ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ 1 4 3 1 4 3 0 = − because A =A −A = . −8 −7 1 −8 −7 1 0 ⎡ ⎤ ⎡ ⎤ 2 −3 2 11 4 3 ⎦ and B = ⎣ −9 ⎦. Following (10) We need to solve the system AX = B, where A = ⎣ 3 −1 2 1 3 Cramer’s Rule, we first construct ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 11 −3 2 2 11 2 2 −3 11 4 3 ⎦ , A2 = ⎣ 3 −9 3 ⎦ , and A3 = ⎣ 3 4 −9 ⎦ .

A1 = ⎣ −9 3 2 1 −1 3 1 −1 2 3 Computing the required determinants results in:

|A| = 34, |A1 | = −136, |A2 | = −102, and |A3 | = 170.

Hence, 1 =

−136 −102 170 |A1 | |A2 | |A3 | = = −4, 2 = = = −3, and 3 = = = 5. |A| 34 |A| 34 |A| 34

Therefore, the solution set for the system is {(−4 −3 5)}.

(11) (a) Using Theorem 3.7 three times shows that for any square matrix A, |A4 | = |A|4 . Hence, p |A| = 4 |A4 |. Now, using basketweaving, or any other method, we see that ¯ ¯ ¯ 5 −4 −2 ¯ ¯ ¯ ¯ −8 −3 3 ¯¯ = −289 ¯ ¯ −2 4 7 ¯ √ Thus, if such a matrix A exists, then |A4 | = −289, implying |A| = 4 −289. However, negative numbers do not have real fourth roots. Therefore, such a matrix A, having real entries, can not exist. ⎡ ⎤ 3 −2 5 1 4 ⎦. A short computation shows that |B| = 0. But, if B = A−1 , (b) Suppose that B = ⎣ −1 1 0 13 then, by part (1) of Theorem 2.12, B = A−1 is nonsingular, and so |B| 6= 0 by Theorem 3.5, which gives a contradiction.

(12) Since B is similar to A, there is a nonsingular matrix P such that B = P−1 AP. (b) |B | = |B| (by Theorem 3.10) = |P−1 AP| = |P−1 ||A||P| (by using Theorem 3.7 twice) = 1  |P| |A||P| (by Corollary 3.8) = |A| = |A | (by Theorem 3.10).

(e) B + I = P−1 AP + I = P−1 AP + P−1 I P = P−1 (A + I )P. Hence, B + I is similar to A + I . ¯ ¯ ¯ −5 −16 16 ¯¯ ¯  + 67 −64 ¯¯. Using basketweaving gives (14) (b) Step 1: A () = |I3 − A| = ¯¯ 32 ¯ 32 64  − 61 ¯ A () = ( − 5)( + 67)( − 61) + (−16)(−64)(32) + (16)(32)(64) − (16)( + 67)(32) − ( − 5)(−64)(64) − (−16)(32)( − 61) 3 =  + 2 − 21 − 45 = ( − 5)(2 + 6 + 9) = ( − 5)( + 3)2 

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

111

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 3 Review

Step 2: The eigenvalues of A are the roots of A (). Hence, there are two eigenvalues, 1 = 5, and 2 = −3.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 5: We solve the homogeneous system whose augmented matrix is [(5I3 − A)|0]. ¯ ⎡ ⎤ ⎡ ⎤ 1 ¯¯ 1 0 0 −16 16 ¯¯ 0 4 ¯ 0 72 −64 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 −1 ¯¯ 0 ⎦. The Now, [(5I3 − A)|0] = ⎣ 32 32 64 −56 ¯ 0 0 0 0 ¯ 0 ⎧ + 14 3 = 0 ⎨ 1 associated linear system is 3 = 0 . Since column 3 is not a pivot column, 2 − ⎩ 0 = 0 3 is an independent variable. Let 3 = 4 (to eliminate fractions). Then 1 = − 14 3 = −1

and 2 = 3 = 4. This yields the fundamental eigenvector [−1 4 4]. The eigenspace 5 = {[−1 4 4] |  ∈ R}.

Eigenvalue 2 = −3: We solve the homogeneous system whose augmented matrix is ¯ ¯ ⎡ ⎤ ⎡ ⎤ −8 −16 16 ¯¯ 0 1 2 −2 ¯¯ 0 64 −64 ¯¯ 0 ⎦  reducing to ⎣ 0 0 0 ¯¯ 0 ⎦. The associated [((−3)I3 − A)|0] = ⎣ 32 ¯ 32 64 −64 0 0 0 0 ¯ 0 ⎧ ⎨ 1 + 22 − 23 = 0 0 = 0 . Since columns 2 and 3 are not pivot columns, linear system is ⎩ 0 = 0 2 and 3 are independent variables. Letting 2 = 1 and 3 = 0 produces 1 = −2, thus yielding the fundamental eigenvector [−2 1 0]. Letting 2 = 0 and 3 = 1 produces 1 = 2, thus yielding the fundamental eigenvector [2 0 1]. The eigenspace −3 = {[−2 1 0] + [2 0 1] |   ∈ R}.

Step 4: Since  = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable.

Step 5: We ⎡form the 3 × 3⎤matrix whose columns are the 3 fundamental eigenvectors we have −1 −2 2 1 0 ⎦. found: P = ⎣ 4 4 0 1

Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ⎡ ⎤ ⎡ ⎤ 5 0 0 −1 −2 2 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ 4 9 −8 ⎦. It is easy D = ⎣ 0 −3 0 0 −3 4 8 −7 to verify for these matrices that D = P−1 AP.

(15) (b) From the hint in the text, A () = 4 + 63 + 92 , which factors as 2 ( + 3)2 . Thus, we have two eigenvalues: 1 = 0 and 2 = −3. Each of these eigenvalues has algebraic multiplicity 2.

We continue from Step 3 of the Diagonalization Method, looking for fundamental eigenvectors: Eigenvalue 1 = 0: We solve the homogeneous system whose augmented matrix is [(0I4 − A)|0] ¯ ¯ ⎡ ⎤ ⎤ ⎡ 1 12 0 − 94 ¯ 0 468 234 754 −299 ¯¯ 0 ¯ ⎢ −324 −162 −525 ⎢ 204 ¯¯ 0 ⎥ 1 ¯¯ 0 ⎥ ⎥. The ⎥  reducing to ⎢ 0 0 1 = [−A|0] = ⎢ ¯ ⎣ −144 −72 −231 ⎦ ⎣ 93 ¯ 0 0 0 0 0 ¯¯ 0 ⎦ 108 54 174 −69 ¯ 0 0 0 0 0 ¯ 0 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

112

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

associated linear system is

⎧ 1 ⎪ ⎪ ⎨

+

1 2 2

⎪ ⎪ ⎩

3

− +

9 4 4

Chap 3 Review

= 0

4 = 0 . Since columns 2 and 4 are 0 = 0 0 = 0

not pivot columns, 2 and 4 are independent variables. First, we let 2 = 1 and 4 = 0.

Then 1 = − 12 2 + 94 4 = − 12 and 3 = −4 = 0. Multiplying these results by 2 to eliminate fractions yields the fundamental eigenvector [−1 2 0 0]. Second, we let 2 = 0 and 4 = 1. Then 1 = − 12 2 + 94 4 = 94 and 3 = −4 = −1. Multiplying these results by 4 to eliminate fractions produces the fundamental eigenvector [9 0 −4 4].

Eigenvalue 2 = −3: We solve the homogeneous system whose augmented matrix is ¯ ⎤ ⎡ ¯ ¯ ⎡ ⎤ 1 0 0 − 13 3 ¯ 0 465 234 754 −299 ¯¯ 0 ⎥ ⎢ 5 ¯¯ ⎢ −324 −165 −525 ⎥ ⎢ 0 1 0 204 ¯¯ 0 ⎥ 2 ¯ 0 ⎥ ⎢ ⎥ ⎢ [(−3I4 − A)|0] = ⎣  reducing to ⎢ ⎥.

3 ¯ −144 −72 −234 93 ¯¯ 0 ⎦ ⎣ 0 0 1 2 ¯ 0 ⎦ ¯ 108 54 174 −72 ¯ 0 0 0 0 0 ¯ 0 ⎧ 1 − 13 ⎪ ⎪ 3 4 = 0 ⎪ ⎪ 5 ⎨ + 2 2 4 = 0 . Since column 4 is not a pivot The associated linear system is 3 ⎪ 3 ⎪ 2 4 = 0 ⎪ ⎪ ⎩ 0 = 0

13 column, 4 is an independent variable. Setting 4 = 1 in the system produces 1 = 13 3 4 = 3 , 5 5 3 3 2 = − 2 4 = − 2 , and 3 = − 2 4 = − 2 . Multiplying these results by 6 to eliminate fractions produces the fundamental eigenvector [26 −15 −9 6].

Step 4: Since  = 4, and the process has produced only 3 fundamental eigenvectors, A is not diagonalizable.

Notice that we actually did more work than necessary to solve this problem. In fact, if we had started by working on the eigenvalue 2 = −3, we would have quickly discovered that there would only be 1 fundamental eigenvector for this eigenvalue, even though its algebraic multiplicity is 2. In general, we can conclude that a matrix is not diagonalizable if any eigenvalue has fewer fundamental eigenvectors than its algebraic multiplicity. ¯ ¯ ¯  + 21 −22 −16 ¯¯ ¯  − 29 −20 ¯¯. Using basketweaving, (16) First, we diagonalize A. Step 1: A () = |I3 −A| = ¯¯ 28 ¯ −8 8 +5 ¯ A () = ( + 21)( − 29)( + 5) + (−22)(−20)(−8) + (−16)(28)(8) − (−16)( − 29)(−8) − ( + 21)(−20)(8) − (−22)(28)( + 5) 3 =  − 32 −  + 3 = ( − 3)(2 − 1) = ( − 3)( − 1)( + 1)

Step 2: The eigenvalues of A are the roots of A (). Hence, there are three eigenvalues, 1 = 1, 2 = −1 and 3 = 3.

Step 3: Now we solve for fundamental eigenvectors.

Eigenvalue 1 = 1: We solve the homogeneous system whose augmented matrix is [(1I3 − A)|0]. Now, ¯ ¯ ⎡ ⎤ ⎡ ⎤ 22 −22 −16 ¯¯ 0 1 −1 0 ¯¯ 0 0 1 ¯¯ 0 ⎦. The associated lin[(1I3 − A)|0] = ⎣ 28 −28 −20 ¯¯ 0 ⎦  which reduces to ⎣ 0 ¯ −8 8 6 0 0 0 0 ¯ 0 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

113

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

ear system is

⎧ ⎨ 1 ⎩

− 2

Chap 3 Review

= 0 3 = 0 . Since column 2 is not a pivot column, 2 is an independent 0 = 0

variable. We choose 2 = 1. Then 1 = 2 = 1 and 3 = 0, producing the fundamental eigenvector [1 1 0].

For 2 = −1, we solve the homogeneous system whose augmented matrix is [(−1I3 − A)|0] = ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 0 52 ¯ 0 20 −22 −16 ¯¯ 0 ¯ ⎣ 28 −30 −20 ¯ 0 ⎦  which reduces to ⎣ 0 1 3 ¯ 0 ⎦. The associated linear system is ¯ ¯ −8 8 4 ¯ 0 0 0 0 ¯ 0 ⎧ + 52 3 = 0 ⎨ 1 2 + 33 = 0 . Since column 3 is not a pivot column, 3 is an independent variable. ⎩ 0 = 0 We choose 3 = 2 to eliminate fractions. Then 1 = − 52 3 = −5 and 2 = −33 = −6, producing the fundamental eigenvector [−5 −6 2].

Eigenvalue 3 = 3: We solve the homogeneous system whose augmented matrix is [(3I3 − A)|0]. Now, ¯ ¯ ⎡ ⎤ ⎡ ⎤ 24 −22 −16 ¯¯ 0 1 0 3 ¯¯ 0 [(3I3 − A)|0] = ⎣ 28 −26 −20 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 4 ¯¯ 0 ⎦. The associated linear −8 8 8 ¯ 0 0 0 0 ¯ 0 ⎧ + 33 = 0 ⎨ 1 2 + 43 = 0 . Since column 3 is not a pivot column, 3 is an independent system is ⎩ 0 = 0 variable. We choose 3 = 1. Then 1 = −33 = −3 and 2 = −43 = −4, producing the fundamental eigenvector [−3 −4 1].

Step 4: Since  = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable.

Step 5: We form the 3 × 3 matrix whose columns are the 3 fundamental eigenvectors we have found: ⎡ ⎤ 1 −5 −3 P = ⎣ 1 −6 −4 ⎦.

0 2 1

Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ⎡ ⎤ ⎡ ⎤ 2 −1 2 1 0 0 1 1 ⎦. Thus, D = ⎣ 0 −1 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ −1 2 −2 −1 0 0 3 ⎤ ⎤ ⎡ ⎤ ⎡ 113 0 0 ⎡ 2 −1 2 1 −5 −3 ⎢ ⎥ 1 1 ⎦ 0 ⎦ ⎣ −1 A13 = PD13 P−1 = ⎣ 1 −6 −4 ⎦ ⎣ 0 (−1)13 2 −2 −1 0 2 1 0 0 313 ⎡ ⎤ −9565941 9565942 4782976 6377300 ⎦  = ⎣ −12754588 12754589 3188648 −3188648 −1594325

(17) Because D = P−1 AP, with D being a diagonal matrix, A is diagonalizable with its eigenvalues appearing on the main diagonal of D. The columns of P are fundamental eigenvectors corresponding c 2016 Elsevier Ltd. All rights reserved.

Copyright °

114

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 3 Review

to these eigenvalues. (a) The eigenvalues of A are the main diagonal entries of D: 1 = 2, 2 = −1, and 3 = 3.

(b) Because 1 = 2 is the (1 1) entry of D, the 1st column of P is a fundamental eigenvector for A corresponding to 1 . Hence, 2 = {[1 −2 1 1] |  ∈ R}.

Next, 2 = −1 appears as both the (2 2) and (3 3) entries of D. Therefore, both the 2nd and 3rd columns of P are fundamental eigenvectors for A corresponding to 2 . Hence, −1 = {[1 0 0 1] + [3 7 −3 2] |   ∈ R}.

Finally, because 3 = 3 is the (4 4) entry of D, the 4th column of P is a fundamental eigenvector for A corresponding to 3 . Hence, 3 = {[2 8 −4 3] |  ∈ R}.

(c) The matrices D and A are similar to each other. Therefore, by part (g) of Exercise 13 in Section 3.3, |A| = |D|. But, since D is diagonal, it is also upper triangular. Hence, by Theorem 3.2, |D| = 2 × (−1) × (−1) × 3 = 6, the product of its main diagonal entries. Therefore, |A| = 6. ∙ ¸ ∙ ¸ 1 0 2 3 (18) (a) False. For example, and both have determinant 1, but are not equal as 0 1 1 2 matrices. ¯ ¯ ∙ ¸ ¯ 0 1 ¯ 0 1 ¯ ¯ = −1. (b) False. For example, the matrix has all zeroes on its main diagonal but ¯ 1 0 1 0 ¯ (c) False. If A is an  ×  matrix with  ≥ 3, A is a matrix, while (−1)+ A is a real number. For a specific counterexample, suppose A = I3 . Then A11 = I2 , a 2 × 2 matrix, while (−1)1+1 A11 = (−1)1+1 (−1)1+1 22 = 1.

(d) False. By part (1) of Theorem 3.1, to get the area of the parallelogram, we need to take the absolute value of this determinant in case the determinant is negative. For example, consider the square whose corners are at (0 0), (1 0), (0 1), and (1 1),¯ whose ¯area is 1. The vectors [1 0] ¯ 0 1 ¯ ¯ = −1, which does not equal and [0 1] determine this square. However, the determinant ¯¯ 1 0 ¯ the area of the square. (Note that if this problem had said “in some order” rather than “in either order,” the answer would have been true. This is because switching the two rows (that is, switching the order of the two vectors) reverses the sign of the determinant.) (e) True. Part (2) of Theorem 3.1 tells us that the volume of the parallelepiped is the absolute value of the determinant of the matrix whose rows are the given vectors. However, Theorem 3.10 assures us that the determinant of its transpose, that is, the matrix whose columns determine the parallelepiped, has the same determinant. (f) True. See the comments in the text just after Example 2 in Section 3.3. (A formal proof is as follows: If A is lower triangular, then A is upper triangular. By Theorem 3.2, |A | is the product of its main diagonal entries. But these are the same as the main diagonal entries of A, one of which is zero. Hence, |A | equals zero. Thus, |A| = |A | (by Theorem 3.10) = 0. Theorem 3.5 then asserts that A is singular.) (g) True. Swapping the columns of a matrix can be performed by first taking the transpose (which does not change the determinant, by Theorem 3.10), then swapping the corresponding rows, and then taking the transpose again. Such a sequence of column operations effectively changes the sign of the original determinant once for each row operation involved. If the number of sign changes is even, the original and final determinants will be equal. If the number of sign changes is odd, the original and final determinants will have opposite signs.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

115

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 3 Review

(h) True. Since Ae1 = 0, e1 is a nontrivial solution to the homogeneous system AX = 0. By Table 3.1 (at the end of Section 3.2), rank(A)  . Hence, by Corollary 3.6, |A| = 0.

(i) False. As the text says at the beginning of Section 3.2, using row reduction to compute determinants for large matrices is, in general, more efficient than cofactor expansion. (It can be shown that for an  ×  matrix, row reduction takes less than a constant times 3 steps, while cofactor expansion takes more than a constant times ! steps. Because of the comparative rates of increase between 3 and !, for large values of , row reduction takes much less effort.)

(j) True. Suppose |A| = |B| 6= 0. By Theorem 3.5, both A and B are nonsingular. Hence, by the Inverse Method for finding the inverse of a matrix, both A and B are row equivalent to I .

Therefore, since they are both row equivalent to the same matrix, they are row equivalent to each other, by an argument similar to that in Exercise 3(c) in Section 2.3. ∙ ¸ ∙ ¸ 0 0 1 1 (k) False. For example, A = and B = both have determinant 0, but they are not 0 0 1 1 row equivalent because they have different ranks. (It is also easy to find counterexamples where ⎡ ⎤ ⎡ ⎤ 1 0 1 1 0 0 the matrices have the same rank. Consider A = ⎣ 0 1 0 ⎦ = I3 and B = ⎣ 0 1 1 ⎦, where 0 0 1 0 0 1 A and B have the same determinant (= 1) and the same rank (= 3), but A cannot be obtained from B using a single Type (II) row operation.) (l) False. In fact, if the coefficient matrix has a nonzero determinant, Theorem 3.5 and part (1) of Theorem 2.16 imply that the system has only one solution. Since the system is homogeneous, that solution is the trivial solution. (m) True. By Theorem 3.7, |AB| = 0 implies that |A||B| = 0, which implies that |A| = 0 or |B| = 0. ∙ ¸ 1 0 (n) False. For a particular counterexample, consider A = , which has determinant 2. If 0 2 1  is the row operation (I): h2i ← 2 h2i, then  (A) = I2 , so A is row equivalent to I2 . But |I2 | = 1 6= |A|. ¯ ¯ ∙ ¸ ∙ ¸ ¯ 2 0 ¯ 1 0 1 0 ¯ ¯ = 2, (o) False. For example, consider A = and B = . Then |A + B| = ¯ 0 0 0 1 0 1 ¯ but |B| = 1. ¯ ¯ ¯ 0 1 ¯ ¯ ¯ = −1. (p) False. For example, ¯ 1 0 ¯

(q) False. If  is an eigenvalue for an  ×  matrix A, and v is any nonzero eigenvector for A corresponding to , then every distinct scalar multiple of v is also an eigenvector for A corresponding to . Hence,  contains an infinite number of elements. However, the Diagonalization Method produces at most a finite number of fundamental eigenvectors: one for each independent variable encountered when solving the homogeneous system (I − A)X = 0. (r) True. This follows from the definition of the algebraic multiplicity of an eigenvalue and the fact that an th degree polynomial can not have more than  roots, counting multiplicities. (See the comments right before Example 2 in Section 3.4.) (s) True. See part (b) of Exercise 23 in Section 3.4.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

116

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 3 Review

(t) False. In general, if A = P−1 BP, then B = PAP−1 , not P−1 AP. For a particular counterexam∙ ¸ ∙ ¸ ∙ ¸∙ ¸∙ ¸ 1 0 1 −1 0 1 1 0 1 −1 ple, consider B = and P = . Then A = = 0 0 1 0 −1 1 0 0 1 0 ∙ ¸∙ ¸∙ ¸ ∙ ¸ ∙ ¸ 0 1 0 0 1 −1 0 1 0 0 = 6= B. , and P−1 AP = −1 1 −1 1 1 0 0 1 −1 1

(u) False. The only  ×  matrix similar to I is I itself. This is because, for any nonsingular  ×  matrix P, P−1 I P = P−1 (I P) = P−1 P = I . For a specific counterexample, consider any nonsingular matrix other than I . (v) True. The discussion that precedes Theorem 3.13 in the text shows that if  is a root of pA () then the system (I − A)X = 0 has a nontrivial solution. Hence there is at least one nonzero vector X such that AX = X.

(w) True. First, D−1 = R (see Exercise 5(c) in Section 2.4). Next, since D = P−1 AP, A = PDP−1 .

Hence, by parts (3) and (1) of Theorem 2.12, A−1 = (P−1 )−1 D−1 P−1 = PRP−1 . (x) True. Any nontrivial solution to (I − A)X = 0 would be an eigenvector for A, making  an eigenvalue for A. ∙ ¸ ∙ ¸ ∙ ¸ 1 0 0 1 1 1 (y) False. It is easy to show that and are both diagonalizable, but is 0 0 0 1 0 1 not. ∙ ¸ ∙ ¸ 1 1 2 1 (z) False. It is easy to show that and are both diagonalizable (see Exercise 16 in 0 2 0 1 ∙ ¸ 2 2 Section 3.4), but is not.

0 2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

117

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.2

Chapter 4 Section 4.1

∙

¸ ∙ ¸ ∙ ¸ 1 2 2 3 1 1 (5) The set is not closed under addition. For example, + = . Both 2 4 3 5 1 1 ∙ ¸ ∙ ¸ 1 2 2 3 and are singular (their determinants are zero), but is nonsingular (its determinant 2 4 3 5 1 1 1 1

¸

∙

is 1 6= 0).

(8) Properties (2), (3), and (6) are not satisfied. Property (4) makes no sense without Property (3). The following is a counterexample for Property (2): 3 ⊕ (4 ⊕ 5) = 3 ⊕ 18 = 42 but (3 ⊕ 4) ⊕ 5 = 14 ⊕ 5 = 38 While this single counterexample for just one of the properties is enough to prove that the set is not a vector space, we will also illustrate why properties (3) and (6) do not hold.

For Property (3), suppose  ∈ R is the zero vector. Then, for all  ∈ R,  ⊕  =  ⇒ 2( + ) =  ⇒  = − 2 . But then different values of  would produce different values of . For if  = 2,  = −1, and if  = 4,  = −2. However, the value of  must be unique.

For a counterexample to Property (6), note that (1 + 2)(3) = (3)(3) = 9 but (1)(3) ⊕ (2)(3) = 3 ⊕ 6 = 18 (19) (a) False. The two Closure Properties and the 8 remaining Properties must be true for all elements in R in order for it to be a vector space under some “addition” and “scalar multiplication.”

For example, Example 10 in Section 4.1 illustrates operations on R2 for which R2 is not a vector space. Also, Exercise 8, whose answer appears above, provides a counterexample in R. (b) False. This set is not closed under addition. Both 7 +  and −7 are in this set, but their sum is , which does not have degree 7. (Also note that the zero polynomial is not in the set.) (c) True. This is the vector space P7 described in Example 4 of Section 4.1. (d) True. This statement is logically equivalent to part (4) of Theorem 4.1. Note that part (4) of Theorem 4.1 is stated in the form “If  then  or ,” with  = “x = 0,”  = “ = 0,” and  = “x = 0” (where we use  and x instead of  and v). Now, in Section 1.3 we found that “If  then  or ” is logically equivalent to “If  is true and  is false, then .” Here,  is given to be false (since  6= 0), and so  implies . That is, when  6= 0, “x = 0” implies “x = 0.” (e) False. Scalar multiplication by the zero scalar always results in the zero vector, not a scalar. This is part (2) of Theorem 4.1. (f) True. This is the statement of part (3) of Theorem 4.1. (g) True. Closure is easy to check, as is Property (1). Example 6 asserts that Properties (2), and (5) through (8) are true. Properties (3) and (4) also hold since the zero function is in the given set, and the additive inverses of functions in the set are also in the set.

Section 4.2 (1) In each part, we will refer to the given set as V. If V is a subspace of R2 , we will prove this by using Theorem 4.2 to show that it is nonempty and that it is closed under both vector addition and scalar multiplication. If V is not a subspace, we will either show that it lacks at least one of these properties, or that it does not contain the zero vector [0 0] (since every subspace of R2 contains the zero vector). c 2016 Elsevier Ltd. All rights reserved.

Copyright °

118

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.2

(a) This set V is not a subspace of R2 because the zero vector, [0 0], is not a unit vector, and so is not in V. (Note: It can also be shown that V is not closed under either operation.)

(c) This set V is a subspace of R2 . V is nonempty since [0 0] ∈ V because it is of the form [ 2] with  = 0. Notice that the vectors in V are precisely those of the form [1 2], where  is a scalar. Thus, V is closed under addition because 1 [1 2] + 2 [1 2] = (1 + 2 )[1 2], which has the required form since 1 + 2 is a scalar. Similarly, V is closed under scalar multiplication because ([1 2]) = ()[1 2] which has the required form since  is a scalar. (e) This set V is not a subspace of R2 since the zero vector, [0 0], does not equal [1 2], and so is not in V. (Note: It can also be shown that V is not closed under either operation.)

(g) This set V is not a subspace of R2 because it is not closed under addition. The counterexample [1 1] + [1 −1] = [2 0] ∈  V (since |2| 6= |0|) proves this.

(j) This set V is not a subspace of R2 because it is not closed under addition. To prove this, consider the following counterexample: [1 1] + [2 4] = [3 5] ∈  V, since 5 6= 32 . (Note: It can also be shown that V is not closed under scalar multiplication.) (l) This set V is not a subspace of R2 because it is not closed under scalar multiplication. To prove this, consider the following counterexample: 2[075 0] = [15 0] ∈  V. (Note: It can also be shown that V is not closed under addition.)

(2) In each part, we will refer to the given set as V. If V is a subspace of M22 , we will prove this by using Theorem 4.2 to show that it is nonempty and that it is closed under both vector addition and scalar multiplication. If V is not a subspace, we will either show that it lacks at least one of these properties, or that it does not contain the zero matrix O22 (since every subspace of M22 contains the zero matrix). (a) This set V is a subspace of M22 . V is nonempty since setting  =  = 0 shows that O22 ∈ V. Notice ∙ ¸ ∙ ¸ ∙ ¸  − 1 −1 0 0 that the matrices in V are precisely those of the form = +   0 0 0 1 0 Now, V is closed under matrix addition since µ ∙ ¸ ∙ ¸¶ µ ∙ ¸ ∙ ¸¶ 1 −1 0 0 1 −1 0 0 1 + 1 + 2 + 2 0 0 1 0 0 0 1 0 ∙ ¸ ∙ ¸ 1 −1 0 0 + (1 + 2 )  = (1 + 2 ) 0 0 1 0 which is of the required form since 1 + 2 and 1 + 2 are scalars. Similarly, V is closed under scalar multiplication because µ ∙ ¸ ∙ ¸¶ ∙ ¸ ∙ ¸ 1 −1 0 0 1 −1 0 0   + = () + ()  0 0 1 0 0 0 1 0 which is of the required form since  and  are scalars. (c) This set V is a subspace of M22 . V is clearly nonempty since O22 is symmetric, and hence in V. Notice that the matrices in V are precisely those of the form ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸   1 0 0 1 0 0 = + +    0 0 1 0 0 1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

119

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

V is closed under matrix addition since µ ∙ ¸ ∙ ¸ ∙ ¸¶ µ ∙ 1 0 0 1 0 0 1 1 + 1 + 1 + 2 0 0 1 0 0 1 0 ∙ ¸ ∙ ¸ ∙ 1 0 0 1 0 + (1 + 2 ) + (1 + 2 ) = (1 + 2 ) 0 0 1 0 0

0 0 0 1

¸

¸

+ 2

Section 4.2

∙

0 1 1 0

¸

+ 2

∙

0 0 0 1

¸¶



which is of the required form, since 1 + 2  1 + 2  and 1 + 2 are scalars. Similarly, V is closed under scalar multiplication because ∙ ¸ ∙ ¸ ∙ ¸ µ ∙ ¸ ∙ ¸ ∙ ¸¶ 1 0 0 1 0 0 1 0 0 1 0 0 + (1 ) + (1 )    + + = (1 ) 0 0 1 0 0 1 0 0 1 0 0 1 which is of the required form, since 1 , 1 , and 1 are scalars. (e) This set V is a subspace of M22 . Since the sum of the entries of O22 is zero, O22 ∈ V. Thus V ∙ ¸   is nonempty. Note that a typical element of V has the form , where  = −( +  + ).   Therefore, the matrices in V are precisely those of the form ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸   1 0 0 1 0 0 = + +   −( +  + ) 0 −1 0 −1 1 −1 Now, V is closed under addition because µ ∙ ¸ ∙ ¸ ∙ ¸¶ 1 0 0 1 0 0 1 + 1 + 1 0 −1 0 −1 1 −1 ¸ ∙ ¸ ∙ ¸¶ µ ∙ 1 0 0 1 0 0 + 2 + 2 + 2 0 −1 0 −1 1 −1 ∙ ¸ ∙ ¸ ∙ ¸ 1 0 0 1 0 0 + (1 + 2 ) + (1 + 2 )  = (1 + 2 ) 0 −1 0 −1 1 −1 which is of the required form, since 1 + 2  1 + 2  and 1 + 2 are scalars. Finally, V is closed under scalar multiplication since µ ∙ ¸ ∙ ¸ ∙ ¸¶ 1 0 0 1 0 0   + + 0 −1 0 −1 1 −1 ∙ ¸ ∙ ¸ ∙ ¸ 1 0 0 1 0 0 = () + () + ()  0 −1 0 −1 1 −1 which is of the required form, since , , and  are scalars. ∙ ¸ 1 3 . Because O22 B = O22 , O22 ∈ V, and so (g) This set V is a subspace of M22 . Let B = −2 −6

V is nonempty. Next we show that V is closed under addition. If A C ∈ V, then (A + C)B = AB + CB = O22 + O22 = O22 . Hence, (A + C) ∈ V. Finally, we show that V is closed under scalar multiplication. If A ∈ V, then (A)B = (AB) = O22 = O22 , and so (A) ∈ V. (h) This set V is not a subspace of M22 since it is not closed under addition. This is proven by the ∙ ¸ ∙ ¸ ∙ ¸ 1 1 0 0 1 1 following counterexample: + = ∈  V (because the product (1)(1)(1)(1) 0 0 1 1 1 1 of its entries is not zero).

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

120

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.2

(3) In each part, we will refer to the given set as V. If V is a subspace of P5 , we will prove this by using Theorem 4.2 to show that it is nonempty and that it is closed under both vector addition and scalar multiplication. If V is not a subspace, we will either show that it lacks at least one of these properties, or that it does not contain z, the zero polynomial (since every subspace of P5 contains the zero polynomial). (a) This set V is a subspace of P5 . Clearly z ∈ V because both the fifth degree term and the first degree term of z are zero. Thus V is nonempty. Next we show that V is closed under addition: (5 + 4 + 3 + 2 +  + ) + (5 + 4 + 3 + 2 +  + ) = ( + )5 + ( + )4 + ( + )3 + ( + )2 + ( + ) + ( + ) which is in V since the coefficient of 5 equals the coefficient of . Finally, we show that V is closed under scalar multiplication: (5 + 4 + 3 + 2 +  + ) = ()5 + ()4 + ()3 + ()2 + () + () which is in V, since the coefficient of 5 equals the coefficient of .

(b) This set V is a subspace of P5 . Now z(3) = 0, so z ∈ V. Hence V is nonempty. Suppose p q ∈ V. Then V is closed under addition because (p+q)(3) = p(3)+q(3) = 0+0 = 0, proving (p+q) ∈ V. Also, V is closed under scalar multiplication because (p)(3) = p(3) = (0) = 0, which shows that (p) ∈ V.  V, since the degree of z is 0. (Note: It can also (e) This set V is not a subspace of P5 . Note that z ∈ be shown that V is not closed under either operation.)

(g) This set V is a subspace of P5 . Now z0 (4) = 0, so z ∈ V. Hence V is nonempty. Suppose p q ∈ V. Then V is closed under addition because (p + q)0 (4) = p0 (4) + q0 (4) = 0 + 0 = 0, proving (p + q) ∈ V. Also, V is closed under scalar multiplication because (p)0 (4) = p0 (4) = (0) = 0, which shows that (p) ∈ V. (12) (e) No; if |A| 6= 0 and  = 0, then |A| = |O | = 0. (For a specific counterexample, consider A = I .)

(15) If  is not an eigenvalue for A, then, by the definition of an eigenvalue, there are no nonzero vectors X such that AX = X. Thus, no nonzero vector can be in . However, A0 = 0 = 0, hence 0 ∈ .

Therefore,  = {0}, the trivial subspace of R .

(20) (a) Suppose that W is a subspace of a vector space V. We give a proof by induction on .

Base Step: If  = 1, then we must show that if v1 ∈ W and 1 is a scalar, then 1 v1 ∈ W But this is certainly true since the subspace W is closed under scalar multiplication.

Inductive Step: Assume that the theorem is true for any linear combination of  vectors in W. We must prove the theorem holds for a linear combination of  + 1 vectors. Suppose v1  v2      v  v+1 are vectors in W, and 1  2        +1 are scalars. We must show that 1 v1 + 2 v2 + · · · +  v + +1 v+1 ∈ W. However, by the inductive hypothesis, we know that 1 v1 +2 v2 +· · ·+ v ∈ W. Also, +1 v+1 ∈ W, since W is closed under scalar multiplication.

But since W is also closed under addition, the sum of any two vectors in W is again in W, so (1 v1 + 2 v2 + · · · +  v ) + (+1 v+1 ) ∈ W. (22) (a) False. By Theorem 4.2, a nonempty subset W of a vector space V must be closed under addition and scalar multiplication in order to be a subspace of V. For example, in Example 5 in the textbook, the first quadrant in R2 is a nonempty subset of R2 that is not a subspace of R2 . c 2016 Elsevier Ltd. All rights reserved.

Copyright °

121

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.3

(b) True. Every vector space has itself as a subspace. The trivial subspace {0} is also a subspace of every vector space. (Note that if V = {0}, these two subspaces are equal.)

(c) False. The plane W must pass through the origin, or else 0 ∈  W, and so W is not a subspace.

For example, the plane  +  +  = 1 is not a subspace for this reason.

(d) True. Let V be the set of lower triangular 5 × 5 matrices. To show that V is a subspace of M55 , we must show that it is nonempty, and that it is closed under both addition and scalar multiplication. Since O5 is lower triangular, O5 ∈ V, and so V is nonempty. Next, let A B ∈ V. Then  =  = 0 for all   . If C = A + B, then for   ,  =  +  = 0 + 0 = 0, and so C ∈ V. Hence, V is closed under addition. Similarly, if D = A, then for   ,  = ( ) = (0) = 0, proving D ∈ V. Therefore V is also closed under scalar multiplication.

(e) True. Let V be the given subset of R4 . To show that V is a subspace of R4 , we must show that it is nonempty, and that it is closed under both matrix addition and scalar multiplication. Letting  =  = 0 shows that [0 0 0 0] ∈ V, so V is nonempty. Notice that the vectors in V are precisely those of the form [0   0] = [0 1 0 0] + [0 0 1 0]. Now, V is closed under addition because (1 [0 1 0 0] + 1 [0 0 1 0])+(2 [0 1 0 0] + 2 [0 0 1 0]) = (1 + 2 ) [0 1 0 0]+(1 + 2 ) [0 0 1 0] which is a vector of the same form, since 1 + 2 and 1 + 2 are scalars. Finally, V is closed under scalar multiplication since ([0 1 0 0] + [0 0 1 0]) = ()[0 1 0 0] + ()[0 0 1 0] which is a vector of the same form, since  and  are scalars. (f) False. While containing the zero vector is a necessary condition for a subset of a vector space to be a subspace, it is not a sufficient condition. For example, the subset W = {[0 0] [1 0]} of R2 contains the zero vector but is not a subspace of R2 . It is not closed under scalar multiplication because 2[1 0] = [2 0] ∈  W. (Note that W also fails to be closed under addition.)

(g) True. This is the statement of Theorem 4.3.

(h) True. Theorem 4.4 asserts that the eigenspace corresponding to any eigenvalue of any × matrix is a subspace of R .

Section 4.3 (1) In each part we follow the three steps of the Simplified Span Method. ∙ ¸ 1 1 0 (a) Step 1: Form the matrix A = .

2 −3 −5 ∙ ¸ 1 0 −1 Step 2: A row reduces to C = .

0 1 1

Step 3: span() = {[1 0 −1] + [0 1 1] |   ∈ R} = {[  − + ] |   ∈ R}. ⎡ ⎤ 1 −1 1 3 ⎦. (c) Step 1: Form the matrix A = ⎣ 2 −3 0 1 −1 ⎡ ⎤ 1 0 0 Step 2: A row reduces to C = ⎣ 0 1 −1 ⎦.

0 0 0 Step 3: Using only the nonzero rows of C gives span() = {[1 0 0] + [0 1 −1] |   ∈ R} = {[  −] |   ∈ R}.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

122

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

1 ⎢ 0 (e) Step 1: Form the matrix A = ⎢ ⎣ 0 1 ⎡ 1 ⎢ 0 Step 2: A row reduces to C = ⎢ ⎣ 0 0

Section 4.3

⎤ 1 1 ⎥ ⎥.

1 ⎦ 4

3 0 1 5

0 1 0 1

0 1 0 0

⎤ 0 −2 0 1 ⎥ ⎥.

1 1 ⎦ 0 0

Step 3: Using only the nonzero rows of C gives span() = {[1 0 0 −2] + [0 1 0 1] + [0 0 1 1] |    ∈ R} = {[   −2 +  + ] |    ∈ R}.

(2) (a) First we convert the polynomials in  into vectors in R4 : (3 − 1) → [1 0 0 −1], (2 − ) → [0 1 −1 0], and ( − 1) → [0 0 1 −1]. Now, we use the Simplified Span Method on these vectors.

To do this, we row reduce the matrix ⎡ ⎤ ⎡ ⎤ 1 0 0 −1 1 0 0 −1 0 ⎦ to obtain C = ⎣ 0 1 0 −1 ⎦  A = ⎣ 0 1 −1 0 0 1 −1 0 0 1 −1 We must convert the nonzero rows of C to polynomial form: [1 0 0 −1] → (3 −1), [0 1 0 −1] → (2 − 1), and [0 0 1 −1] → ( − 1). Hence, span() is the set of linear combinations of these 3 polynomials. That is, span() = {(3 − 1) + (2 − 1) + ( − 1) |    ∈ R} = {3 + 2 +  − ( +  + ) |    ∈ R}.

(c) First we convert the polynomials in  into vectors in R4 : (3 −  + 5) → [1 0 −1 5], (33 −3+10) → [3 0 −3 10], (53 −5−6) → [5 0 −5 −6] and (6−63 −13) → [−6 0 6 −13].

Now we use the Simplified Span Method on these vectors. To do this, we row reduce the matrix ⎡ ⎤ ⎡ ⎤ 1 0 −1 5 1 0 −1 0 ⎢ 3 0 −3 ⎢ 10 ⎥ 0 1 ⎥ ⎥ to obtain C = ⎢ 0 0 ⎥ A=⎢ ⎣ 5 0 −5 −6 ⎦ ⎣ 0 0 0 0 ⎦ −6 0 6 −13 0 0 0 0

We must convert the nonzero rows of C to polynomial form: [1 0 −1 0] → (3 − ), and [0 0 0 1] → (1). Hence, span() is the set of linear combinations of these 2 polynomials. That is, span() = {(3 − ) + (1) |   ∈ R} = {3 −  +  |   ∈ R}.

(3) (a) First we convert the matrices in  into vectors in R4 : ∙ ¸ ∙ ¸ ∙ ¸ −1 1 0 0 −1 0 → [−1 1 0 0] → [0 0 1 −1] and, → [−1 0 0 1] 0 0 1 −1 0 1 Now we use the Simplified Span Method on these vectors. To do this, we row reduce the matrix ⎡ ⎤ ⎡ ⎤ −1 1 0 0 1 0 0 −1 A = ⎣ 0 0 1 −1 ⎦ to obtain C = ⎣ 0 1 0 −1 ⎦  −1 0 0 1 0 0 1 −1 We must convert the nonzero rows of C to matrix form: ∙ ¸ ∙ ¸ ∙ ¸ 1 0 0 1 0 0 [1 0 0 −1] →  [0 1 0 −1] →  and, [0 0 1 −1] →  0 −1 0 −1 1 −1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

123

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.3

Hence, span() is the set of linear combinations of these 3 matrices. That is, span() = ½ ∙ ¸ ∙ ¸ ∙ ¸¯ ¾ ½∙ ¸¯ ¾ ¯ 1 0 0 1 0 0 ¯¯   ¯    ∈ R .  + +    ∈ R = 0 −1 0 −1 1 −1 ¯  − −  −  ¯

(c) First we convert the matrices in  into vectors in R4 : ∙ ¸ ∙ ¸ 1 −1 2 −1 → [1 −1 3 0] → [2 −1 8 −1] 3 0 8 −1 ∙ ¸ ∙ ¸ −1 4 3 −4 → [−1 4 4 −1] and, → [3 −4 5 6] 4 −1 5 6

Now we use the Simplified Span Method on these vectors. To do this, we row reduce the matrix ⎡ ⎤ ⎡ ⎤ 1 −1 3 0 1 0 0 0 ⎢ 2 −1 8 −1 ⎥ ⎢ ⎥ ⎥ to obtain C = ⎢ 0 1 0 0 ⎥  A=⎢ ⎣ −1 ⎣ 0 0 1 0 ⎦ 4 4 −1 ⎦ 3 −4 5 6 0 0 0 1

We must convert the nonzero rows of C to matrix form: ∙ ¸ ∙ ¸ 1 0 0 1 [1 0 0 0] →  [0 1 0 0] →  0 0 0 0 ∙ ¸ ∙ ¸ 0 0 0 0 [0 0 1 0] →  and, [0 0 0 1] →  1 0 0 1

Hence, span() is the set of linear combinations of these 4 matrices. ¾ ½ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ¯ ¯ 1 0 0 1 0 0 0 0 ¯ That is, span() =  + + + ¯     ∈ R 0 0 0 0 1 0 0 1 ½∙ ¸¯ ¾   ¯¯ =     ∈ R = M22 .   ¯

(4) (a) W = {[ +   +   +  } |    ∈ R} = {[1 1 0 0] + [1 0 1 0] + [0 1 1 1] |    ∈ R} ⎡ ⎤ 1 1 0 0 = row space of A, where A = ⎣ 1 0 1 0 ⎦.

0 1 1 1 ⎤ ⎡ 1 0 0 − 12 ⎥ ⎢ 1 ⎥ (b) A row reduces to B = ⎢ 2 ⎦. ⎣ 0 1 0

1 0 0 1 2 (c) From the Simplified Span Method, we know that W also equals the row space of B, which equals {[1 0 0 − 12 ] + [0 1 0 12 ] + [0 0 1 12 ] |    ∈ R} = {[   − 12  + 12  + 12 ] |    ∈ R}.

(11) Suppose (3 − 22 +  − 3) + (23 − 32 + 2 + 5) + (42 +  − 3) + (43 − 72 + 4 − 1) = 33 − 82 + 2 + 16. Equating the coefficients of the 3 term, the 2 term, etc., from each side of this equation produces the following linear system: ⎧  + 2 + 4 = 3 ← 3 terms ⎪ ⎪ ⎨ −2 − 3 + 4 − 7 = −8 ← 2 terms   + 2 +  + 4 = 2 ←  terms ⎪ ⎪ ⎩ −3 + 5 − 3 −  = 16 ← constant terms

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

124

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.3

Using either Gaussian elimination or the Gauss-Jordan method on this system produces the solution set {[−2 − 1 − + 2 −1 ] |  ∈ R}. Choosing  = 0 produces the particular solution [−1 2 −1 0].

This vector provides coefficients for the desired linear combination. That is, 33 −82 +2+16 = −1(3 −22 +−3)+2(23 −32 +2+5)−1(42 +−3)+0(43 −72 +4−1) Other particular solutions to the system will yield alternate linear combinations that produce 33 − 82 + 2 + 16. Finally, note that the augmented matrix that was row reduced in this solution has the coefficients of each polynomial as its columns. ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 32 − 45 ¯¯ 0 10 15 −8 ¯¯ 0 ⎢ ¯ ⎥ 0 ¯ 0 ⎦.

(15) (a) Row reduce [(1I3 − A)|0] = ⎣ 10 15 −8 ¯¯ 0 ⎦ to obtain ⎣ 0 0 ¯ ¯ 30 45 −24 0 0 0 0 ¯ 0 The solution set to the associated linear system is {[− 32  + 45   ] |   ∈ R}. Choosing  = 2 (to eliminate fractions) and  = 0 produces the fundamental eigenvector [−3 2 0]. Choosing  = 0 and  = 5 (to eliminate fractions) yields the fundamental eigenvector [4 0 5]. Hence, the desired set of two fundamental eigenvectors is {[−3 2 0] [4 0 5]}.

(20) Theorem 1.15 shows that every  ×  matrix A can be expressed as the sum of a symmetric matrix S and a skew-symmetric matrix V. Hence, every matrix in M is a linear combination of a matrix S in 1 and a matrix V in 2 . Therefore, M ⊆ span(1 ∪ 2 ). But since span(1 ∪ 2 ) ⊆ M , we have span(1 ∪ 2 ) = M .

(22) First, 1 ⊆ 2 ⊆ span(2 ), by Theorem 4.5, part (1). Then, since span(2 ) is a subspace of V containing 1 (by Theorem 4.5, part (2)), span(1 ) ⊆ span(2 ) (by Theorem 4.5, part (3)).

(25) (b) Suppose 1 = {[1 0 0] [0 1 0]} and 2 = {[0 1 0] [0 0 1]}. Then 1 ∩ 2 = {[0 1 0]}, and so span(1 ∩ 2 ) = {[0 1 0] |  ∈ R} = {[0  0] |  ∈ R} Next, Similarly,

span(1 ) = {[1 0 0] + [0 1 0] |   ∈ R} = {[  0] |   ∈ R} span(2 ) = {[0 1 0] + [0 0 1] |   ∈ R} = {[0  ] |   ∈ R}

We want to show that span(1 ∩ 2 ) = span(1 ) ∩ span(2 ). This is done by proving the two set inclusions span(1 ∩ 2 ) ⊆ span(1 ) ∩ span(2 ) and span(1 ∩ 2 ) ⊇ span(1 ) ∩ span(2 ).

The first of these is already proven for any sets 1 and 2 in part (a). To prove the second, we must show that if x ∈ span(1 ) ∩ span(2 ), then x ∈ span(1 ∩ 2 ).

So, suppose x ∈ span(1 ) ∩ span(2 ). Thus, x is in both span(1 ) and span(2 ), and so must have the correct form for both sets. But x = [1  2  3 ] ∈ span(1 ) implies that 3 = 0.

Similarly, x ∈ span(2 ) implies that 1 = 0. Hence, x must have the form [0 2  0], implying x ∈ span(1 ∩ 2 ), since it has the proper form.

(c) Suppose 1 = {[1 0 0] [0 1 0]} and 2 = {[1 0 0] [1 1 0]}. Then 1 ∩ 2 = {[1 0 0]}, and so span(1 ∩ 2 ) = {[1 0 0] |  ∈ R} = {[ 0 0] |  ∈ R}

Now x = [0 1 0] ∈ 1 , and so, x ∈ span(1 ), by part (1) of Theorem 4.5. Also, x = [0 1 0] = (−1)[1 0 0] + (1)[1 1 0] shows that x ∈ span(2 ). Hence, x ∈ span(1 ) ∩ span(2 ). But, since x has “1” as its second coordinate, it is not in span(1 ∩ 2 ). Therefore, we have span(1 ∩ 2 ) 6= span(1 ) ∩ span(2 ). c 2016 Elsevier Ltd. All rights reserved.

Copyright °

125

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.3

(26) (c) Suppose 1 = {5 } and 2 = {4 }. Then 1 ∪ 2 = {5  4 }. Now span(1 ) = {5 |  ∈ R} and span(2 ) = {4 |  ∈ R}. Consider p = 5 + 4 . Then p is clearly in neither span(1 ) nor span(2 ), since it does not fit the description of the polynomials in either set. Hence, we have p∈  span(1 ) ∪ span(2 ). However, p ∈ span(1 ∪ 2 ) because it is a linear combination of 5 and 4 . Therefore, span(1 ∪ 2 ) 6= span(1 ) ∪ span(2 ).

(28) (a) Let v1 and v2 be two vectors in span(). By the definition of span(), both v1 and v2 can be expressed as finite linear combinations of vectors from . That is, there are finite subsets 1 = {w1     w } and 2 = {x1     x } of  such that v1 = 1 w1 +· · ·+ w and v2 = 1 x1 +· · ·+ x for some real numbers 1       1        (b) The natural thing to do at this point would be to combine the expressions for v1 and v2 by adding corresponding coefficients. However, each of the subsets {w1     w } or {x1     x } may contain elements not found in the other. Therefore, we create a larger set 3 = 1 ∪ 2 containing all of the vectors in both subsets. We rename the elements of the finite subset 3 as {z1     z }. Then v1 = 1 w1 + · · · +  w can be expressed as 1 z1 + · · · +  z , where  =  if z = w , and  {w1     w } Similarly, v2 = 1 x1 +· · ·+ x can be expressed as 1 z1 +· · ·+ z ,  = 0 if z ∈  {x1     x } In this way, both v1 and v2 can be where  =  if z = x , and  = 0 if z ∈ expressed as linear combinations of the vectors in 3 . (c) From parts (a) and (b), we have v1 + v2 =

 X

 z +

=1

 X =1

 z =

 X

( +  ) z 

=1

a linear combination of the vectors z in the subset 3 of . Thus, v1 + v2 ∈ span().

(30) (a) False. The set  is permitted to be infinite. The restriction is that span() contains only the finite linear combinations from the set . That is, in each separate linear combination, only a finite number of the vectors in  are used. (See Example 3 of Section 4.3 in the textbook for an example of the span of an infinite set.) (b) True. Span() is defined to be the set of all such finite linear combinations. (c) False. Theorem 4.5 shows that span() is the smallest subspace of V containing , not the smallest set in V containing . The smallest set containing  is clearly just  itself (no matter what  is).

So, for example, if  = {[1 0]} ⊆ R2 , then span() = {[1 0] |  ∈ R}. Hence span() contains vectors such as [2 0] and [−1 0], and so is bigger than  itself. (d) False. Part (3) of Theorem 4.5 shows that the set inclusion given in this problem should be reversed. That is, span() ⊆ W. For a specific counterexample to the statement in the problem, consider V = W = R2 and  = {[1 0]}. Then W contains , but span() = {[1 0] |  ∈ R} clearly does not contain all of W, since [0 1] ∈ W, but [0 1] ∈  span(). (e) False. The row space of an  ×  matrix is a subspace of R , since it consists of the set of linear combinations of the rows of the matrix, each of which has  entries. Therefore, the row space of a 4 × 5 matrix is a subspace of R5 , not R4 . (f) True. This statement summarizes the Simplified Span Method.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

126

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.4

Section 4.4 (1) In each part, let  represent the given set. (a) Because  contains only 1 element,  is linearly independent since that element is not the zero vector. (See the discussion in the textbook just before Example 4.) (b) Because  contains 2 nonzero vectors,  is linearly dependent if and only if either vector in  is a scalar multiple of the other vector in . But since neither vector in  is a scalar multiple of the other,  is linearly independent instead. (See the discussion in the textbook directly after Example 4.) (c) Because  contains 2 nonzero vectors,  is linearly dependent if and only if either vector in  is a scalar multiple of the other vector in . Since [−2 −4 10] = (−2)[1 2 −5],  is linearly dependent. (See the discussion in the textbook directly after Example 4.) (d)  is linearly dependent because  contains the zero vector [0 0 0], as explained after Example 5 in the textbook. (e)  is a subset of R3 that contains 4 distinct vectors. Hence,  is linearly dependent by Theorem 4.7.

(2) In each part, let  represent the given set of vectors. (a) We follow the Independence Test Method. ⎡ ⎤ 1 3 −2 5 ⎦ , whose columns are the vectors in .

Step 1: Form the matrix A = ⎣ 9 4 −2 5 −7 ⎡ ⎤ 1 0 1 Step 2: A row reduces to B = ⎣ 0 1 −1 ⎦.

0 0 0

Step 3: There is no pivot in column 3, so  is linearly dependent. (b) We follow the Independence Test Method. ⎡ ⎤ 2 4 −2 0 ⎦, whose columns are the vectors in .

Step 1: Form the matrix A = ⎣ −1 −1 3 6 2 Step 2: A row reduces to B = I3 .

Step 3: Since every column in I3 is a pivot column,  is linearly independent. (e) We follow the Independence Test Method. ⎡ ⎤ 2 4 1 ⎢ 5 3 −1 ⎥ ⎥ , whose columns are the vectors in .

Step 1: Form the matrix A = ⎢ ⎣ −1 1 1 ⎦ 6 4 −1 ⎤ ⎡ 1 0 − 12 ⎢ 1 ⎥ ⎢ 0 1 2 ⎥ Step 2: A row reduces to B = ⎢ ⎥. ⎣ 0 0 0 ⎦ 0 0 0 Step 3: There is no pivot in column 3, so  is linearly dependent.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

127

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.4

(3) In each part, let  represent the given set of polynomials. (a) First, we convert the polynomials in  into vectors in R3 : 2 +  + 1 → [1 1 1] 2 − 1 → [1 0 −1] and, 2 + 1 → [1 0 1] Next, we use the Independence Test Method on this set of 3-vectors. ⎡ ⎤ 1 1 1 0 0 ⎦, whose columns are the vectors converted from the Step 1: Form the matrix A = ⎣ 1 1 −1 1 polynomials in .

Step 2: A row reduces to B = I3 .

Step 3: Since every column in I3 is a pivot column,  is linearly independent. (c) First, we convert the polynomials in  into vectors in R2 : 4 − 2 → [4 −2] 60 − 25 → [60 −25] and, − 24 + 9 → [−24 9] Next, we use the Independence Test Method on this set of 2-vectors. ∙ ¸ 4 60 −24 Step 1: Form the matrix A = , whose columns are the polynomials in  −2 −25 9 converted to vectors. " # 1 0 3 Step 2: A row reduces to B = .

0 1 − 35 Step 3: There is no pivot in column 3, so  is linearly dependent. (Note that, in this problem, once we converted the polynomials to vectors in R2 , we could easily see from Theorem 4.7 that the vectors are linearly dependent, confirming the result we obtained using the Independence Test Method.)

(4) In each part, let  represent the given set of polynomials. (a) First, we convert the polynomials in  into vectors in R3 : 2 − 1 → [1 0 −1] 2 + 1 → [1 0 1] and, 2 +  → [1 1 0] We use the Independence Test Method on this set of 3-vectors. ⎡ ⎤ 1 1 1 Step 1: Form the matrix A = ⎣ 0 0 1 ⎦, whose columns are the polynomials in  converted −1 1 0 into vectors.

Step 2: A row reduces to B = I3 .

Step 3: Since every column in I3 is a pivot column,  is linearly independent. (c) First, we convert the polynomials in  into vectors in R3 : 42 + 2 → [4 0 2] 2 +  − 1 → [1 1 −1]  → [0 1 0] and 2 − 5 − 3 → [1 −5 −3] But this set of four vectors in R3 is linearly dependent by Theorem 4.7. Hence,  is linearly dependent.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

128

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.4

(e) Linearly independent. See the remarks just before Example 13 in the textbook. No polynomial in the set can be expressed as a finite linear combination of the others since no single power of  can be expressed as a combination of other powers of .

(7) It is useful to have a simplified description of the vectors in span(). Using the Simplified Span Method, # " ∙ ¸ 1 0 − 12 1 1 0 . Hence, span() = {[  (− 12 + 12 )] |   ∈ R}. we row reduce to obtain 1 −2 0 1 0 1 2

(b) There are many possible choices. For example, we can let v = [0 1 0], or, v = [0 0 1].

How did we arrive at these choices? From part (a), we already know that  itself is linearly independent. We will want our choices for v to be outside of span() so that if we list the vectors in  ∪{v} with v listed last, each vector in  ∪{v} would be outside of the span of those that come before it. Thus,  ∪ {v} would be linearly independent by the “boxed” alternate characterization of linear independence just before Example 11 in the textbook. Using the simplified form of span() we gave at the beginning of this solution, both [0 1 0] and [0 0 1] are easily seen not to be in span() and have a simple form. Hence, they make good choices to extend . (c) Any nonzero vector u ∈ span() other than [1 1 0] or [−2 0 1] will provide the example we need.

In particular, consider u = [1 1 0] + [−2 0 1] = [−1 1 1]. Then  ∪ {u} = {[1 1 0] [−2 0 1] [−1 1 1]}. You can verify that  ∪ {u} is linearly dependent by using the Independence Test Method. (Also, since u ∈ span(),  ∪{u} is linearly dependent by the Alternate Characterization of linear dependence in Table 4.1.)

(11) In each part, there are many different correct answers, but only one possibility is given here. (a) Let  = {e1  e2  e3  e4 }. We will use the definition of linear independence to verify that  is linearly independent. Now 1 e1 + 2 e2 + 3 e3 + 4 e4 = [1  2  3  4 ], which clearly equals the zero vector [0 0 0 0] if and only if 1 = 2 = 3 = 4 = 0. Therefore,  is linearly independent. (c) Let  = {1  2  3 }. We will use the definition of linear independence to verify that  is linearly independent. Now 1 (1) + 2  + 3 2 + 4 3 clearly equals the zero polynomial if and only if 1 = 2 = 3 = 4 = 0. Therefore,  is linearly independent. ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ 0 1 0 0 0 1 0 0 0 ⎬ ⎨ 1 0 0 (e) Let  = ⎣ 0 0 0 ⎦  ⎣ 1 0 0 ⎦  ⎣ 0 0 0 ⎦  ⎣ 0 1 0 ⎦ . ⎭ ⎩ 0 0 0 0 0 0 1 0 0 0 0 0

Notice that each matrix in  is symmetric. We will use the definition of linear independence to verify that  is linearly independent. Now, ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 1 0 0 0 1 0 0 0 1 0 0 0 1 2 3 0 ⎦ 1 ⎣ 0 0 0 ⎦ + 2 ⎣ 1 0 0 ⎦ + 3 ⎣ 0 0 0 ⎦ + 4 ⎣ 0 1 0 ⎦ = ⎣ 2 4 0 0 0 0 0 0 0 0 1 0 0 0 0 0 3

which clearly equals the zero matrix O3 if and only if 1 = 2 = 3 = 4 = 0. Therefore,  is linearly independent.

(13) (b) Let  be the given set of three vectors. We prove that v = [0 0 −6 0] is redundant. To do this, we need to show that span( − {v}) = span(). We will do this by applying the Simplified Span Method to both  − {v} and .

For span( − {v}):

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

129

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ∙

1 1 0 0 1 1 1 0

¸

row reduces to

∙

1 1 0 0 0 0 1 0

Section 4.4

¸ .

For span(): ⎡ ⎤ ⎡ ⎤ 1 1 0 0 1 1 0 0 ⎣ 1 1 1 0 ⎦ row reduces to ⎣ 0 0 1 0 ⎦.

0 0 −6 0 0 0 0 0

Since the reduced row echelon form matrices are the same, except for the extra row of zeroes, the two spans are equal, and v = [0 0 −6 0] is a redundant vector.

For an alternate approach to proving that span( − {v}) = span(), first note that  − {v} ⊆ , and so span( − {v}) ⊆ span() by Corollary 4.6. Next, [1 1 0 0] and [1 1 1 0] are clearly in span( − {v}) by part (1) of Theorem 4.5. But v ∈ span( − {v}) as well, because [0 0 −6 0] = (6)[1 1 0 0] + (−6)[1 1 1 0]. Hence,  is a subset of the subspace span( − {v}). Therefore, by part (3) of Theorem 4.5, span() ⊆ span( − {v}). Thus, since we have proven subset inclusion in both directions, span( − {v}) = span().

(17) (b) The converse to the statement is: If  = {v1      v } is a linearly independent subset of R , then  = {Av1      Av } is a linearly independent subset of R with Av1      Av distinct.

We can construct specific counterexamples that contradict either (or both) of the conclusions of the converse. For a specific counterexample in which Av1      Av are distinct but  is linearly ∙ ¸ 1 1 dependent, let A = and  = {[1 0] [1 2]}. Then  = {[1 1] [3 3]}. Note that  is 1 1 linearly independent, but the vectors Av1 and Av2 are distinct and  is linearly dependent. For a specific∙counterexample in which Av1      Av are not distinct but  is linearly independent, ¸ 1 1 use A = and  = {[1 2] [2 1]}. Then  = {[3 3]}. Note that Av1 and Av2 both equal 1 1 [3 3], and that both  and  are linearly independent. For a final counterexample in which both parts of the conclusion are false, let A = O22 and  = {e1  e2 }. Then Av1 and Av2 both equal 0,  is linearly independent, but  = {0} is linearly dependent.

(21) (b)  is linearly dependent iff every vector v in span() can be expressed in more than one way as a linear combination of vectors in  (ignoring zero coefficients).

(24) Suppose that  is linearly independent, and suppose v ∈ span() can be expressed both as v = 1 u1 + · · · +  u and v = 1 v1 + · · · +  v  for distinct u1     u ∈  and distinct v1     v ∈ , where these expressions differ in at least one nonzero term. Since the u ’s might not be distinct from the v ’s, we consider the set  = {u1      u } ∪ {v1      v } and label the distinct vectors in  as {w1     w } Then we can express v = 1 u1 + · · · +  u as v = 1 w1 + · · · +  w  and also v = 1 v1 + · · · +  v as v = 1 w1 + · · · +  w  choosing the scalars     1 ≤  ≤  with  =  if w = u ,  = 0 otherwise, and  =  if w = v ,  = 0 otherwise. Since the original linear combinations for v are distinct, we know that  6=  for some . Now, v − v = 0 = (1 − 1 )w1 + · · · + ( −  )w  Since {w1     w } ⊆  a linearly independent set, each  −  = 0 for every  with 1 ≤  ≤ . But this is a contradiction since  6=   Conversely, assume every vector in span() can be uniquely expressed as a linear combination of elements of  Since 0 ∈ span(), there is exactly one linear combination of elements of  that equals 0.

Now, if {v1      v } is any finite subset of , we have 0 = 0v1 + · · · + 0v . Because this representation is unique, it means that in any linear combination of {v1      v } that equals 0, the only possible coefficients are zeroes. Thus, by definition,  is linearly independent.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

130

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.5

(25) (a) False. [−8 12 −4] = (−4)[2 −3 1]. Thus, one vector is a scalar multiple (hence a linear combination) of the other vector. Therefore, the set is linearly dependent by the remarks after Example 4 in the textbook. (b) True. This follows directly from Theorem 4.8. (c) True. See the discussion in the textbook directly before Example 4 in the textbook. (d) False. This statement directly contradicts the first “boxed” alternate characterization for linear independence following Example 10 in the textbook. (e) True. This is the definition of linear independence for a finite nonempty set. (f) True. This follows directly from Theorem 4.7. (g) False. In order to determine linear independence, we need to row reduce the matrix whose columns are the vectors in , not the matrix whose rows are the vectors in . (The matrix A whose rows are the vectors in  is used, instead, in the Simplified Span Method to find a simplified form for span().) For a specific counterexample to the given statement, consider  = {i j 0} in R2 . Then ⎡ ⎤ 1 0 A = ⎣ 0 1 ⎦, which row reduces to A itself. Note that there is a pivot in both columns of A. 0 0

However,  is linearly dependent since it contains the zero vector (see the remarks after Example 5 in the textbook). (h) True. This follows directly from Theorem 4.9. (i) True. This follows directly from Theorem 4.8, since the given statement indicates that v3 is a linear combination of v1 and v2 . (If 5v1 − 3v2 equals either v1 or v2 , then {v1  v2 } is linearly dependent, and so the statement is true in this case as well.)

Section 4.5 (4) (a) The set  is not a basis because  can not span R4 . For if  spans R4 , then by part (1) of Theorem 4.12, 4 = dim(R4 ) ≤ || = 3, a contradiction. (c) The set  is a basis. To prove this, we will show that  spans R4 , and then apply Theorem 4.12. ⎡ ⎤ 7 1 2 0 ⎢ 8 0 1 −1 ⎥ ⎥ We show that  spans R4 using the Simplified Span Method. The matrix A = ⎢ ⎣ 1 0 0 −2 ⎦ 3 0 1 −1 whose rows are the vectors in  reduces to I4 . Hence, span() = span({e1  e2  e3  e4 }) = R4 .

Now, || = 4 = dim(R4 ), and so part (1) of Theorem 4.12 implies that  is a basis for R4 . (e) The set  is not a basis since  is not linearly independent. For if  is linearly independent, then by part (2) of Theorem 4.12, 5 = || ≤ dim(R4 ) = 4, a contradiction.

(5) (a) W is nonempty, since 0 ∈ W Let X1  X2 ∈ W  ∈ R Then A(X1 + X2 ) = AX1 + AX2 = 0 + 0 = 0 Also, A(X1 ) = (AX1 ) = 0 = 0 Hence W is closed under addition and scalar multiplication, and so is a subspace by Theorem 4.2. ⎡ ⎤ 2 1 1 0 15 5 ⎢ ⎥ ⎢ 0 1 2 − 1 −1 ⎥ 5 5 ⎢ ⎥ (b) A row reduces to ⎢ ⎥. ⎢ 0 0 0 0 0 ⎥ ⎣ ⎦ 0 0

0

0

0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

131

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.5

So, the solution set for the homogeneous system AX = 0 is ¸ ¾ ½∙ 2 2 1 1 −  −  −  −  +  +     |    ∈ R .

5 5 5 5 We find basis vectors for W by setting each of , , and  equal to 1, in turn, while the others equal zero, to obtain ¸ ∙ ¸ ¾ ½∙ 2 2 1 1 −  −  1 0 0  −   0 1 0  [−1 1 0 0 1] .

5 5 5 5 We can multiply each of the first two vectors in this basis by 5 to eliminate fractions, producing the following basis for W: {[−1 −2 5 0 0] [−2 1 0 5 0] [−1 1 0 0 1]}.

(c) dim(W) = 3; rank(A) = 2; 3 + 2 = 5

(9) (b) Part (a) shows that the given set  is a basis for V. Since || = 5, the definition of dimension yields dim(V) = 5. (c) We claim that ª ©  = 3 ( − 2)( − 3) 2 ( − 2)( − 3) ( − 2)( − 3) ( − 2)( − 3)

is a basis for W. First, we show that  spans W. Let p ∈ W. Then p(2) = p(3) = 0. Therefore, ( − 2) and ( − 3) are factors of p. Thus, since p has degree at most 5, dividing p by ( − 2)( − 3) produces a polynomial of the form 3 + 2 +  + . That is, p() = ( − 2)( − 3)(3 + 2 +  + ).

Hence, p() = 3 ( − 2)( − 3) + 2 ( − 2)( − 3) + ( − 2)( − 3) + ( − 2)( − 3), which shows that p ∈ span(). Therefore, W ⊆ span(). Since  is clearly a subset of W, part (3) of Theorem 4.5 shows that span() ⊆ W. Thus span() = W. To see why  is linearly independent, notice that the polynomials in  are listed in order of increasing degree. Therefore, none of these polynomials can be expressed as a linear combination of the previously listed polynomials. Hence, by Exercise 19 of Section 4.4,  is linearly independent.

Therefore, since  spans W and is linearly independent,  is a basis for W.

(d) Part (c) shows that the set  defined there is a basis for W. Since || = 4, the definition of dimension yields dim(W) = 4.

(10) (b) We illustrate two approaches: First approach: We check that  and  both span the same subspace of R4 by using the Simplified Span Method and showing that both  and  produce the same bases for their spans.

For : ⎡ ⎡ ⎤ 2 ⎤ 1 0 − 35 5 1 4 1 −2 ⎢ 2 3 ⎥ ⎢ −1 1 ⎥ 0 1 − 1 −1 ⎢ 5 5 ⎥ ⎢ ⎥ row reduces to ⎢ ⎥. ⎣ 3 2 −1 0 ⎦ ⎣ 0 0 0 0 ⎦ 2 3 0 −1 0 0 0 0 For :

∙

2 3 0 −1 −1 1 1 −1

¸

row reduces to

"

1 0 − 35 0 1

2 5

2 5 − 35

#

.

Since both row reduced matrices have identical nonzero rows, span() and span() have identical bases (the set of nonzero rows of the row reduced matrices). Hence, they must be the same c 2016 Elsevier Ltd. All rights reserved.

Copyright °

132

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.5

subspaces. Also, since the basis produced for span() contains two vectors, dim(span()) = 2.

Finally, because  is a linearly independent subset of span() with || = dim(span()), part (2) of Theorem 4.12 implies that  is a basis for span().

Second approach: In part (a) we showed that no larger subset of  containing  is linearly independent; that is, each of the remaining vectors in  can be expressed as a linear combination of the vectors in . Therefore,  ⊆ span(). Then, part (3) of Theorem 4.5 implies that span() ⊆ span(). We also know that  ⊆ , and so Corollary 4.6 implies that span() ⊆ span().

Therefore, span() = span(). Since  was shown to be linearly independent in part (a),  is a basis for span(). Hence, dim(span()) = || = 2.

(c) No. Span() 6= R4 because dim(span()) = 2 6= 4 = dim(R4 ). If span() and R4 were the same vector space, they would have the same dimension.

(12) (a) Let V = R3 , and let  = {[1 0 0] [2 0 0] [3 0 0]}. Now dim(V) = 3 = ||. However,  does not span R3 . Every element of span() is of the form [1 0 0] + [2 0 0] + [3 0 0], which must have zeroes in the second and third coordinates. Therefore, vectors in R3 such as [0 1 0] and [0 0 1] are not in span() because they are not of the correct form. (b) Let V = R3 , and let  = {[1 0 0] [2 0 0] [3 0 0]}. Now dim(V) = 3 = | |, but  is not linearly independent. This is clear, since [3 0 0] = [1 0 0] + [2 0 0], showing that one vector in  is a linear combination of the other vectors in  .

(21) Let V be a finite dimensional vector space and let W be a subspace of V. Consider the set  of nonnegative integers defined by  = { | a set  exists with  ⊆ W | | =  and  linearly independent} (a) The empty set is linearly independent and is a subset of W. Hence 0 ∈ . Therefore,  is not the empty set.

(b) Since V and W share the same operations, every linearly independent subset  of W is also a linearly independent subset of V (by the definition of linear independence). Hence, using part (2) of Theorem 4.12 on  in V shows that  = | | ≤ dim(V). Therefore,  contains no numbers larger than dim(V), and so has a finite number of elements. (c) Because  is a finite nonempty set of numbers, it has a largest number . Therefore, because  ∈ , the definition of the set  implies that there must exist some set  such that  ⊆ W, | | = , and  is linearly independent. Let  = {v1      v } be such a set.

Now, we are given that  ⊆ W, and so span( ) ⊆ W by part (3) of Theorem 4.5.

(d) Suppose w ∈ W with w ∈  span( ). Then w ∈   , and so  ∪ {w} is a subset of W containing  + 1 vectors. But since  is the largest element of ,  + 1 is not in . Hence, the set  ∪ {w} must fail to satisfy at least one of the conditions in the description of . Since  ∪ {w} ⊆ W, we must have that  ∪ {w} is not linearly independent. Therefore,  ∪ {w} is linearly dependent.

(e) Suppose w ∈ W, but w ∈  span( ) as in part (d). Because  ∪ {w} = {v1      v  w} is linearly dependent (by part (d)), there exist scalars 1         such that 1 v1 + · · · +  v + w = 0 with not all coefficients equal to zero. But if  = 0, then 1 v1 + · · · +  v = 0 with not all  = 0, contradicting the fact that  is linearly independent. Hence,  6= 0 and thus w = (−

1  )v1 + · · · + (− )v   

(f) We assumed that w ∈  span( ). However, the conclusion of part (e) shows that w ∈ span( ).

Since both clearly can not be true, we have a contradiction. Hence, our assumption that there is c 2016 Elsevier Ltd. All rights reserved.

Copyright °

133

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.6

a w ∈ W with w ∈  span( ) is false. That is, there are no vectors in W outside of span( ). Thus, W ⊆ span( ).

(g) From the conclusions of parts (c) and (f), W = span( ). Now,  is a basis for W, since it is linearly independent by assumption (part (c)), and spans W. Hence, W is finite dimensional because  is a finite basis (containing  elements). Also, using part (b), dim(W) = | | =  ≤ dim(V) (h) Suppose dim(W) = dim(V), and let  be a basis for W. Then  is a linearly independent subset of V (see the discussion in part (b)) with || = dim(W) = dim(V). Part (2) of Theorem 4.12 then shows that  is a basis for V. Hence, W = span() = V. (i) The converse of part (h) is “If W = V, then dim(W) = dim(V).” This is obviously true, since the dimension of a finite dimensional vector space is unique.

(24) (a) True. This is the definition of a basis for a vector space. (b) False. As we saw in Example 8 in Section 4.5, dim(P ) =  + 1. Hence, dim(P4 ) = 5, implying that every basis for P4 has five elements, not four. (c) False. As we saw in Example 9 in Section 4.5, dim(M ) =  · . Hence, dim(M43 ) = 4 · 3 = 12, not 7.

(d) False. Part (1) of Theorem 4.12 shows, instead, that || ≥ . For a specific counterexample to the statement in the problem, consider  = {[1 0] [0 1] [1 1]}, a subset of R2 . Now  spans R2 because every 2-vector can be expressed as a linear combination of vectors in  as follows: [ ] = [1 0] + [0 1]. However, || = 3  2 = dim(R2 ).

(e) False. Part (2) of Theorem 4.12 shows, instead, that | | ≤ . For a specific counterexample to the equation | | =  in the given statement, consider  = {[1 0]}, a subset of R2 . Then since [1 0] 6= [0 0],  is linearly independent by the comments in the textbook directly before Example 4 in Section 4.4. But, | | = 1  2 = dim(R2 ). (f) True. This is proven using both parts of Theorem 4.12 as follows: | | ≤ dim(W) (by part (2)) ≤ || (by part (1)).

(g) False. Recall that V is a subspace of itself. Letting W = V gives an case in which dim(W) = dim(V), contradicting the strict inequality given in the problem. (h) False. Consider the infinite dimensional vector space P. P5 is a finite dimensional subspace of P. (Note: dim(P5 ) = 6.) (i) True. Since A is nonsingular, A row reduces to I4 . The Simplified Span Method then shows that the span of the rows of A equals the span of the rows of I4 , which equals R4 . But, since A has exactly 4 rows, the size of this spanning set equals dim(R4 ) = 4. Part (1) of Theorem 4.12 then proves that the rows of A form a basis for R4 .

Section 4.6 (1) (a) In accordance with the Simplified Span Method, we row reduce the matrix whose rows are the vectors in the given set . Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 2 3 −1 0 1 0 0 2 −2 ⎢ 3 ⎢ 6 8 −2 0 ⎥ 0 1 ⎥ ⎢ ⎥ to obtain ⎢ 0 1 0 ⎥ ⎣ −1 −1 −3 ⎣ 0 0 1 −1 1 1 ⎦ 0 ⎦ −2 −3 −5 1 1 0 0 0 0 0 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

134

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.6

The set of nonzero rows of the row reduced matrix is a basis  for span(). That is,  = {[1 0 0 2 −2] [0 1 0 0 1] [0 0 1 −1 0]}.

(d) In accordance with the Simplified Span Method, we row reduce the matrix whose rows are the vectors in the given set . Hence, we row reduce ⎤ ⎡ ⎡ ⎤ 1 0 0 −2 − 13 4 1 1 1 1 1 ⎢ 9 ⎥ ⎥ ⎢ 1 2 3 4 ⎢ 0 1 0 3 5 ⎥ 2 ⎥ ⎢ ⎥ to obtain ⎢  ⎢ 0 0 1 ⎣ 0 1 2 3 1 ⎥ 4 ⎦ 0 −4 ⎦ ⎣ 0 0 4 0 −1 0 0 0 0 0 The set of nonzero rows of the row reduced matrix is a basis  for span(). That is, 9 1  = {[1 0 0 −2 − 13 4 ] [0 1 0 3 2 ] [0 0 1 0 − 4 ]}.

(2) First, we convert the polynomials in  into vectors in R4 : (3 − 32 + 2) → [1 −3 0 2], (23 − 72 +  − 3) → [2 −7 1 −3], and (43 − 132 +  + 5) → [4 −13 1 5]. Next, using the Simplified Span Method, we row reduce the matrix whose rows are the converted vectors, above. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 −3 0 2 1 0 −3 0 ⎣ 2 −7 1 −3 ⎦ to obtain ⎣ 0 1 −1 0 ⎦  4 −13 1 5 0 0 0 1

A basis  for span() is found by converting the nonzero rows of the reduced matrix into polynomials in P3 . This produces  = {3 − 3 2 −  1}.

(3) First, we convert the matrices in  into vectors in R6 . This is done by changing each 3 × 2 matrix A into its corresponding 6-vector [11  12  21  22  31  32 ]. Doing this for each of the 4 matrices in  yields the vectors [1 4 0 −1 2 2], [2 5 1 −1 4 9], [1 7 −1 −2 2 −3], and [3 6 2 −1 6 12]. Next, using the Simplified Span Method, we row reduce the matrix whose rows are these vectors, above.

Hence, we row reduce ⎤ ⎡ ⎡ ⎤ 1 4 2 0 1 0 3 3 1 4 0 −1 2 2 ⎥ ⎢ ⎢ 2 5 1 −1 4 9 ⎥ 0 1 − 13 − 13 0 0 ⎥ ⎢ ⎥ to obtain ⎢ ⎥ ⎢ ⎣ 1 7 −1 −2 2 −3 ⎦ ⎣ 0 0 0 0 0 1 ⎦ 3 6 2 −1 6 12 0 0 0 0 0 0 A basis  for span() is found by converting the nonzero rows of the row reduced matrix back into 3 × 2 matrices. Hence, ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ 0 1 0 0 ⎬ ⎨ 1 0  = ⎣ 43 31 ⎦  ⎣ − 13 − 13 ⎦  ⎣ 0 0 ⎦  ⎭ ⎩ 0 1 2 0 0 0

(4) In each part, we use the Independence Test Method, as described in Section 4.6. That is, we create a matrix whose columns are the vectors in the given set . We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis for span() which is a subset of .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

135

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.6

(a) Creating the matrix whose columns are the vectors in , we get ⎡ ⎤ ⎡ ⎤ 1 2 3 0 −2 1 0 −3 0 0 ⎣ 3 1 −6 1 1 ⎦ , row reducing to ⎣ 0 1 3 0 −1 ⎦  −2 4 18 −1 −6 0 0 0 1 2

Since columns 1, 2 and 4 are pivot columns, we choose the 1st, 2nd and 4th columns of the original matrix for our basis for span(). Hence,  = {[1 3 −2] [2 1 4] [0 1 −1]}.

(c) Creating the matrix whose columns are the vectors in , we get ⎡ ⎤ ⎡ ⎤ 3 1 3 −1 1 0 0 1 ⎣ −2 2 −2 −10 ⎦ , which row reduces to ⎣ 0 1 0 −4 ⎦  2 −1 7 6 0 0 1 0

Since columns 1, 2 and 3 are pivot columns, we choose the 1st, 2nd and 3rd columns of the original matrix for our basis for span(). Hence,  = {[3 −2 2] [1 2 −1] [3 −2 7]}.

(5) (a) First, we convert the polynomials in  into vectors in R4 : (3 − 82 + 1) → [1 −8 0 1], (33 − 22 + ) → [3 −2 1 0], (43 + 2 − 10) → [4 0 2 −10], (3 − 202 −  + 12) → [1 −20 −1 12], (3 + 242 + 2 − 13) → [1 24 2 −13], and (3 + 142 − 7 + 18) → [1 14 −7 18]. Next, we use the Independence Test Method. That is, we create a matrix whose columns are the vectors above. We row reduce this matrix and note which columns are pivot columns. The polynomials for the vectors in the corresponding columns of the original matrix constitute a basis for span() which is a subset of . Thus, we row reduce the matrix ⎤ ⎡ ⎡ ⎤ 1 0 0 0 −3 205 1 3 4 1 1 1 ⎢ 0 1 0 0 0 83 ⎥ ⎢ −8 −2 0 −20 24 14 ⎥ ⎥ ⎢ ⎥ to ⎢ ⎥ ⎢ 181 ⎣ 0 1 2 −1 2 −7 ⎦ 1 − 2 ⎦ ⎣ 0 0 1 0 1 0 −10 12 −13 18 0 0 0 1 0 −91 Since columns 1, 2, 3 and 4 are pivot columns, we choose the polynomials corresponding to the 1st, 2nd, 3rd and 4th columns of the original matrix for our basis for span(). Hence,  = {3 − 82 + 1 33 − 22 +  43 + 2 − 10 3 − 202 −  + 12}.

(6) (a) The first vector, [3 1 −2], is nonzero, so we choose that for a basis  for span(). The second vector in , [0 0 0], is the zero vector, which cannot be in a linearly independent set, so we skip over it. The third vector, [6 2 −3], is not a scalar multiple of [3 1 −2] (the only vector in  so far), so we add it to . We have examined all of the vectors in , so we are finished. The desired basis is  = {[3 1 −2] [6 2 −3]}. (d) We use the inspection method described in the exercise. We notice that the 2nd coordinates of the vectors in  are described in terms of their 1st and 3rd coordinates, that is, 2 = −31 + 3 . This inspires us to consider the 1st and 3rd coordinates as “independent variables” and solve for the second coordinate, creating vectors in a manner similar to Step 3 of the Diagonalization Method of Section 3.4. Thus, we let 1 = 1 and 3 = 0 to obtain 2 = (−3)(1) + 0 = −3. This gives us the vector [1 −3 0]. Next, letting 1 = 0 and 3 = 1 yields 2 = (−3)(0) + 1 = 1, producing [0 1 1].

Now the set  = {[1 −3 0] [0 1 1]} is clearly linearly independent, since neither vector is a scalar multiple of the other. We claim that  also spans , and hence is a basis for span(). To do this, we need to show that every vector in  is a linear combination of the vectors in . So, let v = [1  2  3 ] ∈ . Then, 2 = −31 + 3 . Hence v = [1  −31 + 3  3 ] = 1 [1 −3 0] + 3 [0 1 1], and so v is a linear combination of the vectors in .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

136

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.6

(7) (a) Because the set  is infinite, we follow the method described in the exercise. The finite subset 1 = {3  2  } of  clearly has  ⊆ span(1 ), since every polynomial p in  is of the form p() = 3 + 2 + . And so span(1 ) = span() (using part (3) of Theorem 4.5 to get span() ⊆ span(1 ) and Corollary 4.6 to obtain span(1 ) ⊆ span()).

Next, we would normally use the Independence Test Method on 1 to find . However, in this case, 1 is clearly already linearly independent! Hence, no further work is required. The desired basis is  = 1 = {3  2  }. (c) Because the set  is infinite, we follow the method described in the exercise. The finite subset 1 = {3 + 2   1} of  clearly has  ⊆ span(1 ), since every polynomial p in  is of the form p() = 3 + 2 +  +  = (3 + 2 ) +  + (1). And so span(1 ) = span() (using part (3) of Theorem 4.5 to get span() ⊆ span(1 ) and Corollary 4.6 to obtain span(1 ) ⊆ span()).

Next, we would normally use the Independence Test Method on 1 to find . However, in this case, 1 is clearly already linearly independent! Hence, no further work is required. The desired basis is  = 1 = {3 + 2   1}.

(8) (a) Recall that the standard basis for M33 is  = {Ψ | 1 ≤  ≤ 3 1 ≤  ≤ 3}, where Ψ is the 3 × 3 matrix having 1 as its ( ) entry and zeroes elsewhere. (See Example 3 in Section 4.5.)

Note that  is a subset of the given set .  is linearly independent and spans M33 . Now span() ⊇ span() = M33 , and so, since span() can not be bigger than M33 , span() = M33 .

Hence,  is a basis for span(). (c) Because the given set  is infinite, we will use the method described in Exercise 7 in Section 4.6.

The first step is to guess at such a basis. To help us make this guess, we first observe that ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤    1 0 0 0 1 0 0 0 1 ⎣    ⎦ = ⎣ 0 0 0 ⎦ + ⎣ 1 0 0 ⎦ + ⎣ 0 0 0 ⎦    0 0 0 0 0 0 1 0 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 0 0 0 0 0 0 0 0 + ⎣ 0 1 0 ⎦ + ⎣ 0 0 1 ⎦ +  ⎣ 0 0 0 ⎦ 0 0 0 0 1 0 0 0 1 Thus, we choose ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 ⎬ ⎨ 1 0 0  = ⎣ 0 0 0 ⎦⎣ 1 0 0 ⎦ ⎣ 0 0 0 ⎦⎣ 0 1 0 ⎦ ⎣ 0 0 1 ⎦⎣ 0 0 0 ⎦  ⎭ ⎩ 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1

Note that every matrix in  is symmetric. Also note that we have already shown that every matrix in  is in span(), since we expressed every symmetric matrix as a linear combination of matrices in .

Finally, we need to check that  is linearly independent. But this is also clear, because none of the matrices in  can be expressed as a linear combination of the others, since each contains a 1 in at least one entry in which all of the other matrices in  have a 0.

Hence,  spans span() and is linearly independent, and so  is a basis for span().

(10) (a) We use the Enlarging Method.

Step 1: We let  = {e1  e2  e3  e4  e5 } be the standard basis for R5 .

Step 2: Form the spanning set  = {[1 −3 0 1 4] [2 2 1 −3 1] e1  e2  e3  e4  e5 }. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

137

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.6

Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in . We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis  for span() = R5 .

Since the vectors in  are listed first, this basis will contain  . Hence, we row reduce ⎡ 3 ⎤ 1 ⎡ ⎤ 1 0 0 0 0 13 13 1 2 1 0 0 0 0 ⎢ 0 1 0 0 0 −4 1 ⎥ ⎢ −3 ⎢ 13 13 ⎥ 2 0 1 0 0 0 ⎥ ⎢ ⎥ ⎥ ⎢ 7 5 ⎥ ⎢ 0 ⎢ 1 0 0 1 0 0 ⎥ − 13 13 ⎢ ⎥ ⎥ to ⎢ 0 0 1 0 0 ⎢ ⎣ 1 −3 0 0 0 1 0 ⎦ 11 7 ⎥ ⎣ 0 0 0 1 0 13 13 ⎦ 4 1 0 0 0 0 1 4 1 0 0 0 0 1 − 13 13

Since there are pivots in the first 5 columns, the first 5 columns of the original matrix form the desired basis. That is,  = {[1 −3 0 1 4] [2 2 1 −3 1] [1 0 0 0 0] [0 1 0 0 0] [0 0 1 0 0]}.

(c) We use the Enlarging Method.

Step 1: We let  = {e1  e2  e3  e4  e5 } be the standard basis for R5 .

Step 2: Form the spanning set  = {[1 0 −1 0 0] [0 1 −1 1 0] [2 3 −8 −1 0] e1  e2  e3  e4  e5 }.

Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in . We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis  for span() = R5 .

Since the vectors in  are listed first, this basis will contain  . Hence, we row reduce ⎤ ⎡ 5 ⎡ ⎤ 0 1 0 0 0 − 94 −1 4 1 0 2 1 0 0 0 0 ⎢ 0 1 0 0 1 3 0 0 ⎥ ⎢ 0 ⎥ ⎢ 4 4 1 3 0 1 0 0 0 ⎥ ⎢ ⎥ ⎥ ⎢ 1 1 ⎢ −1 −1 −8 0 0 1 0 0 ⎥ to obtain ⎢ 0 0 1 0 ⎥ 0 − 0 4 4 ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎣ 0 7 3 1 −1 0 0 0 1 0 ⎦ 1 − 0 ⎦ ⎣ 0 0 0 1 4 4 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 Since there are pivots in the first 4 columns and in column 8, the first 4 columns and column 8 of the original matrix form the desired basis. That is,  = {[1 0 −1 0 0] [0 1 −1 1 0] [2 3 −8 −1 0] [1 0 0 0 0] [0 0 0 0 1]} (11) (a) First we convert the polynomials in  to vectors in R5 : (3 − 2 ) → [0 1 −1 0 0], and (4 − 33 + 52 − ) → [1 −3 5 −1 0]. Now we use the Enlarging Method on this set of vectors to find a basis for R5 .

Step 1: We let  = {e1  e2  e3  e4  e5 } be the standard basis for R5 .

Step 2: Form the spanning set  = {[0 1 −1 0 0] [1 −3 5 −1 0] e1  e2  e3  e4  e5 }.

Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in . We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis  for span() = R5 .

Since the vectors corresponding to the polynomials in  are listed first, this basis will contain  .

Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 0 1 1 0 0 0 0 1 0 0 0 −1 −5 0 ⎢ 1 −3 0 1 0 0 0 ⎥ ⎢ 0 1 0 0 0 −1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −1 ⎥ to ⎢ 0 0 1 0 5 0 0 1 0 0 0 1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 −1 0 0 0 1 0 ⎦ ⎣ 0 0 0 1 1 2 0 ⎦ 0 0 0 0 0 0 1 0 0 0 0 0 0 1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

138

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.6

Since there are pivots in columns 1, 2, 3, 4 and 7, columns 1, 2, 3, 4, and 7 of the original matrix form the basis {[0 1 −1 0 0] [1 −3 5 −1 0] e1  e2  e5 } for R5 . Converting these 5-vectors back to polynomials in P4 yields the basis  = {3 − 2  4 − 33 + 52 −  4  3  1}.

(c) First we convert the polynomials in  to vectors in R5 : (4 − 3 + 2 −  + 1) → [1 −1 1 −1 1], (3 − 2 +  − 1) → [0 1 −1 1 −1], and (2 −  + 1) → [0 0 1 −1 1]. Now we use the Enlarging Method on this set of vectors to find a basis for R5 .

Step 1: We let  = {e1  e2  e3  e4  e5 } be the standard basis for R5 .

Step 2: Form the spanning set  = {[1 −1 1 −1 1] [0 1 −1 1 −1] [0 0 1 −1 1] e1  e2  e3  e4  e5 }.

Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in . We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis  for span() = R5 .

Since the vectors corresponding to the polynomials in  are listed first, this basis will contain  .

Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 ⎢ −1 ⎢ 0 1 0 1 1 0 0 1 0 0 1 0 0 0 ⎥ 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 −1 ⎥ ⎢ 1 0 0 1 0 0 ⎥ to obtain ⎢ 0 0 1 0 1 0 0 1 ⎥ ⎢ ⎥ ⎣ −1 ⎣ 0 0 0 0 0 1 0 −1 ⎦ 1 −1 0 0 0 1 0 ⎦ 1 −1 1 0 0 0 0 1 0 0 0 0 0 0 1 1 Since there are pivots in columns 1, 2, 3, 6, and 7, columns 1, 2, 3, 6 and 7 of the original matrix form the basis {[1 −1 1 −1 1] [0 1 −1 1 −1] [0 0 1 −1 1] e3  e4 } for R5 . Converting these 5-vectors back to polynomials in P4 yields the basis  = {4 − 3 + 2 −  + 1 3 − 2 +  − 1 2 −  + 1 2  } (12) (a) First we convert the matrices in  to vectors in R6 . This is done by changing each 3 × 2 matrix A into the 6-vector [11  12  21  22  31  32 ]. Doing this for both of the matrices in  yields the vectors [1 −1 −1 1 0 0] and [0 0 1 −1 −1 1]. Now we use the Enlarging Method on this set of vectors to find a basis for R6 .

Step 1: We let  = {e1  e2  e3  e4  e5  e6 } be the standard basis for R6 .

Step 2: Form the spanning set  = {[1 −1 −1 1 0 0] [0 0 1 −1 −1 1] e1  e2  e3  e4  e5  e6 }.

Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in . We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis  for span() = R6 . Since the vectors corresponding to the matrices in  are listed first, this basis will contain  . Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 0 1 0 0 0 0 0 1 0 0 0 0 1 0 1 ⎢ −1 ⎢ 0 1 0 0 0 0 0 1 0 0 0 0 ⎥ 0 0 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −1 ⎥ ⎢ 1 0 0 1 0 0 0 ⎥ 0 0 1 0 0 −1 0 −1 ⎥ ⎢ ⎢ ⎥ ⎢ 1 −1 0 0 0 1 0 0 ⎥ to obtain ⎢ 0 0 0 1 0 1 0 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 −1 0 0 0 0 1 0 ⎦ ⎣ 0 0 0 0 1 1 0 0 ⎦ 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 1 Since there are pivots in columns 1, 2, 3, 4, 5 and 7, columns 1, 2, 3, 4, 5 and 7 of the original matrix form the basis {[1 −1 −1 1 0 0] [0 0 1 −1 −1 1] e1  e2  e3  e5 } for R6 . Converting these

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

139

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.6

6-vectors back to matrices in M32 yields the basis ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ 1 −1 0 0 1 0 0 1 0 0 0 0 ⎬ ⎨ 1 ⎦  ⎣ 1 −1 ⎦  ⎣ 0 0 ⎦  ⎣ 0 0 ⎦  ⎣ 1 0 ⎦  ⎣ 0 0 ⎦   = ⎣ −1 ⎭ ⎩ 0 0 −1 1 0 0 0 0 0 0 1 0

(c) First we convert the matrices in  to vectors in R6 . This is done by changing each 3 × 2 matrix A into the 6-vector [11  12  21  22  31  32 ]. Doing this for the four matrices in  yields the vectors [3 −1 2 −6 5 1], [−1 2 −4 2 0 0], [6 2 −2 −9 10 2] and [3 −4 8 −9 5 1]. Now we use the Enlarging Method on this set of vectors to find a basis for R6 .

Step 1: We let  = {e1  e2  e3  e4  e5  e6 } be the standard basis for R6 .

Step 2: Form the spanning set  = {[3 −1 2 −6 5 1] [−1 2 −4 2 0 0] [6 2 −2 −9 10 2] [3 −4 8 −9 5 1] e1  e2  e3  e4  e5  e6 } Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in . We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis  for span() = R6 . Since the vectors corresponding to the matrices in  are listed first, this basis will contain  . Hence, we row reduce ⎡ ⎤ 3 −1 6 3 1 0 0 0 0 0 ⎢ −1 2 2 −4 0 1 0 0 0 0 ⎥ ⎢ ⎥ ⎢ 2 −4 −2 8 0 0 1 0 0 0 ⎥ ⎢ ⎥ ⎢ −6 2 −9 −9 0 0 0 1 0 0 ⎥ ⎢ ⎥ ⎣ 5 0 10 5 0 0 0 0 1 0 ⎦ 1 0 2 1 0 0 0 0 0 1 ⎤ ⎡ 2 −3 0 10 0 16 1 0 0 0 3 3 ⎢ 0 1 0 0 −1 0 0 0 0 3 ⎥ ⎥ ⎢ ⎢ 0 0 1 0 0 1 0 −1 0 −5 ⎥ ⎥ to obtain ⎢ ⎢ 0 0 0 1 −2 1 0 − 43 0 −5 ⎥ 3 ⎥ ⎢ ⎣ 0 0 0 0 0 0 1 2 0 10 ⎦ 0 0 0 0 0 0 0 0 1 −5 Since there are pivots in columns 1, 2, 3, 4, 7 and 9, columns 1, 2, 3, 4, 7 and 9 of the original matrix form the basis {[3 −1 2 −6 5 1] [−1 2 −4 2 0 0] [6 2 −2 −9 10 2] [3 −4 8 −9 5 1] e3  e5 } for R6 . Converting these 6-vectors back to matrices in M32 yields the basis ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ −1 2 6 2 3 −4 0 0 0 0 ⎬ ⎨ 3 −1  = ⎣ 2 −6 ⎦  ⎣ −4 2 ⎦  ⎣ −2 −9 ⎦  ⎣ 8 −9 ⎦  ⎣ 1 0 ⎦  ⎣ 0 0 ⎦  ⎭ ⎩ 5 1 0 0 10 2 5 1 0 0 1 0

(13) (b) To find a basis for V, we first guess at a basis , and then verify that  is, indeed, a basis for V. Now, recall that the standard basis for M33 is {Ψ | 1 ≤  ≤ 3 1 ≤  ≤ 3}, where Ψ is the 3×3 matrix having 1 as its ( ) entry and zeroes elsewhere. (See Example 3 in Section 4.5.) However, not all of these matrices are in V, since some have nonzero trace. However, the six matrices Ψ with  6=  are all in V. But these 6 are not sufficient to span V, since all have zero entries on c 2016 Elsevier Ltd. All rights reserved.

Copyright °

140

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.6

the main diagonal, while V contains many matrices having nonzero entries on the main diagonal.

Thus, to get our basis, we want to add more matrices to the 6 we have, linearly independent of those 6, and of each other. However, we can not add more than two matrices, or else the basis would have 9 or more elements, contradicting Theorem 4.13, since V 6= M33 . Consider ⎡ ⎤ ⎡ ⎤ 1 0 0 0 0 0 0 ⎦ and C2 = ⎣ 0 1 0 ⎦ C1 = ⎣ 0 0 0 0 −1 0 0 −1 which are clearly in V. We claim that  = {Ψ | 1 ≤  ≤ 3 1 ≤  ≤ 3  6= } ∪ {C1  C2 } is a basis for V. P First we show that  is a linearly independent set. Suppose 6=  Ψ + 1 C1 + 2 C2 = O3 .

Then, ⎡ ⎤ ⎡ ⎤ 1 12 0 0 0 13 ⎣ 21 2 ⎦ = ⎣ 0 0 0 ⎦ 23 31 32 −1 − 2 0 0 0

Hence, each  = 0 for  6=  and 1 = 2 = 0. Therefore,  is linearly independent by definition.

Next, we prove that  spans V. Let Q ∈ V. Then, 33 = −11 − 22 , and so Q has the form ⎤ ⎡ 11 12 13 ⎦ ⎣ 21 22 23 31 32 −11 − 22 P Thus Q can be expressed as Q = 6=  Ψ + 11 C1 + 22 C2 . Hence, Q ∈ span(). This completes the proof that  is a basis. Therefore, dim(V) = || = 8.

Note: We could have omitted the part in which we verified that  spans V. We already had 8 linearly independent vectors. If a basis for V needed 9 vectors, then we would have had that dim(V) = 9 = dim(M33 ). Theorem 4.13 would then imply that V = M33 , which is clearly not the case. So, dim(V) is at most 8. But dim(V) is also at least 8, because V contains the linearly independent set  which has 8 elements (part (2) of Theorem 4.12). So, dim(V) must equal 8.

Therefore, part (2) of Theorem 4.12 implies that  is a basis for V.

(d) To find a basis for V, we first guess at a basis , and then verify that  is, indeed, a basis for V. Consider that the definition of V involves three “independent variables” , , and . Hence, use the polynomials obtained by first setting  = 1,  = 0,  = 0, then  = 0,  = 1,  = 0, and  = 0,  = 0,  = 1. This yields  = {6 + 4 + 2 −  + 3 −5 + 2 − 2 −3 + 2 +  + 16} Clearly,  spans V, since 6 − 5 + 4 − 3 + ( +  + )2 − ( − ) + (3 − 2 + 16) = (6 + 4 + 2 −  + 3) + (−5 + 2 − 2) + (−3 + 2 +  + 16) To show  is also linearly independent, set (6 + 4 + 2 −  + 3) + (−5 + 2 − 2) + (−3 + 2 +  + 16) = 0 Examining the coefficient of 6 yields  = 0. Similarly, looking at the coefficients of 5 and 3 produces  = 0 and  = 0. Hence,  is linearly independent by definition. This completes the proof that  is a basis. Therefore, dim(V) = || = 3. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

141

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.6

(14) (b) Let C = Ψ −Ψ , where Ψ is the × matrix having 1 as its ( ) entry and zeroes elsewhere.

That is, for  6= , C is the  ×  matrix whose ( ) entry is 1, ( ) entry is −1, and all other entries are zero. A basis for the skew-symmetric  ×  matrices is  = {C | 1 ≤    ≤ }. To show this is a basis, we first prove that  is linearly independent. But each C , for   , has its ( ) entry equal to 1, while every other matrix in  has its ( ) entry equal to zero. Hence, no C is a linear combination of the other matrices in . Therefore,  is linearly independent.

Also, if Q is skew-symmetric, then  = − for all   . That is, entries  above the main diagonal are paired with their additive inverse  below the main diagonal. Also, all of the entries of Q on the main diagonal are zero. Thus, X X X X ( Ψ +  Ψ ) = ( Ψ −  Ψ ) =  (Ψ − Ψ ) =  C  Q= 







which shows that  spans the skew-symmetric matrices. This completes the proof that  is a basis. Hence, dim(skew-symmetric matrices) = || = (2 − )2. (The number (2 − )2 is computed as follows: The number of non-diagonal entries in an  ×  matrix is 2 − . Half of these, (2 − )2, are above the main diagonal. Since  contains one matrix corresponding to each of these entries, || = (2 − )2.)

(16) Let V and  be as given in the statement of the theorem. Let  = { | a set  exists with  ⊆ , | | = , and  linearly independent}.

The empty set is linearly independent and is a subset of . Hence 0 ∈ , and  is nonempty.

Suppose  ∈ . Then, every linearly independent subset  of  is also a linearly independent subset of V. Hence, using part (2) of Theorem 4.12 on  shows that  = | | ≤ dim(V).

Suppose  is the largest number in , which must exist, since  is nonempty and all its elements are ≤ dim(V). Let  be a linearly independent subset of  with || = , which exists because  ∈ .

We want to show that  is a basis for V = span(). Now,  is given as linearly independent. We must show that  spans V. Now,  ⊆ , and so span() ⊆ span() (by Corollary 4.6). We need to show that span() ⊆ span(). To do this, we will prove that  ⊆ span(), for then, parts (2) and (3) of Theorem 4.5 will give us that span() ⊆ span().

We prove that  ⊆ span(): Suppose  = {v1      v } and v ∈ . If v ∈ , then, clearly v ∈ span(), and we are finished. So, suppose that v ∈  . Then  ∪ {v} is a subset of  containing  + 1 elements. Therefore, since  is the largest number in ,  ∪ {v} must be linearly dependent.

Hence, there exist scalars 1         such that 1 v1 + · · · +  v + v = 0 with not all coefficients equal to zero. But if  = 0, then 1 v1 + · · · +  v = 0 with not all  = 0, contradicting the fact that  is linearly independent. Hence,  6= 0 and thus  1 v = (− )v1 + · · · + (− )v    This shows that v ∈ span(), completing the proof that  spans V. Hence, the subset  of  is the desired basis for V, finishing the proof of Theorem 4.14.

(17) (b) No; consider the subspace W of R3 given by W = {[ 0 0] |  ∈ R}. No subset of the basis  = {[1 1 0] [1 −1 0] [0 0 1]} for R3 is a basis for W, since none of the vectors in  are in W. (To verify that  is a basis for R3 , note that no vector in  is a linear combination of the other vectors in . Hence,  is a linearly independent subset of R3 with || = 3 = dim(R3 ). Thus,  is a basis for R3 by part (2) of Theorem 4.12.) (c) Yes; consider Y = span( 0 ).  0 is linearly independent because it is a subset of the linearly independent set . Also,  0 clearly spans Y, so  0 is a basis for Y. Finally,  0 ⊆  ⊆ V, and so c 2016 Elsevier Ltd. All rights reserved.

Copyright °

142

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.7

Y = span( 0 ) ⊆ V, by part (3) of Theorem 4.5. Of course, Y is a subspace of V by part (2) of Theorem 4.5.

(18) (b) In R3 , consider W = {[  0] |   ∈ R}. We could let W10 = {[0 0 ] |  ∈ R} or W20 = {[0  ] |  ∈ R}. We need to show that W10 and W20 both satisfy the given conditions.

For W10 : Suppose v = [1  2  3 ] ∈ R3 . Let w = [1  2  0] and w0 = [0 0 3 ]. Then w ∈ W and w0 ∈ W10 , and v = w+w0 , showing existence for the decomposition. To prove uniqueness, suppose u = [1  2  0] ∈ W and u0 = [0 0 03 ] ∈ W10 , and v = u + u0 . Then [1  2  3 ] = [1  2  03 ] ⇒ 1 = 1 , 2 = 2 , and 3 = 03 ⇒ u = w, and u0 = w0 .

For W20 : Suppose v = [1  2  3 ] ∈ R3 . Let w = [1  2 − 3  0] and w0 = [0 3  3 ]. Then w ∈ W and w0 ∈ W20 , and v = w + w0 , showing existence. To prove uniqueness, suppose u = [1  2  0] ∈ W and u0 = [0 03  03 ] ∈ W20 , with v = u +u0 . Then [1  2  3 ] = [1  2 + 03  03 ].

Hence, 3 = 03 , so u0 = w0 . Now, w + w0 = v = u + u0 . Subtracting w0 = u0 from both sides of this equation yields u = w, finishing the proof.

(22) (a) True. This is the statement of Theorem 4.14. (b) True. This is the statement of Theorem 4.15.

(c) False. The Simplified Span Method produces a basis  for span(), but  is not necessarily a subset of . Instead,  contains vectors of a simpler form that those in . For example, applying the Simplified Span Method to the spanning set  = {[1 1] [1 −2]} for R2 produces the standard basis  = {i j} for R2 . (Try it.) Notice that  is not a subset of .

(d) True. The Independence Test Method uses row reduction to select a linearly independent subset  of  that has the same span as  by removing redundant vectors from . Hence,  is a basis for span(). (e) False. The basis  created by the Enlarging Method satisfies  ⊇  rather than  ⊆  . (This can be seen in Example 6 in Section 4.6.) (f) False. The Enlarging Method uses the Independence Test Method, not the Simplified Span Method. (g) False. The statement has the two methods reversed. The Simplified Span Method actually places the vectors of the given spanning set  as rows in a matrix, while the Independence Test Method places the vectors in  as columns.

Section 4.7 (1) (a) We use the Coordinatization Method.

Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors in the basis , and then find its corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 5 0 ¯¯ 2 1 0 0 ¯¯ 7 [A|v] = ⎣ −4 −7 −4 ¯¯ −1 ⎦ −→ ⎣ 0 1 0 ¯¯ −1 ⎦ 1 2 1 ¯ 0 0 0 1 ¯ −5

Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: [v] = [7 −1 −5], the last column of the row reduced matrix.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

143

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.7

(c) We use the Coordinatization Method.

Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors in the basis , and then find its corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 2 4 1 ¯¯ 7 1 0 0 ¯¯ −2 ⎢ 3 ⎢ 0 1 0 ¯ 4 ⎥ 3 2 ¯¯ −4 ⎥ ⎢ ⎥ ⎢ ¯ ⎥ ⎢ ¯ ⎥ ¯ ⎥ 3 1 ¯ 5 ⎥ −→ ⎢ [A|v] = ⎢ 1 ⎢ 0 0 1 ¯ −5 ⎥  ¯ ¯ ⎣ −2 ⎦ ⎣ 1 −1 ¯ 13 0 0 0 ¯ 0 ⎦ ¯ 2 −1 1 −13 0 0 0 ¯ 0

Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: We eliminate the last two rows of the row reduced matrix, since they contain all zeroes.

Then [v] = [−2 4 −5], the remaining entries of the last column.

(e) We convert the given polynomials to vectors in R3 by changing 2 +  +  to [  ]. Then we use the Coordinatization Method.

Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the polynomials in the basis . Then find its corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 3 1 2 ¯¯ 13 1 0 0 ¯¯ 4 2 3 ¯¯ −5 ⎦ −→ ⎣ 0 1 0 ¯¯ −5 ⎦  [A|v] = ⎣ −1 2 −3 −1 ¯ 20 0 0 1 ¯ 3

Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: [v] = [4 −5 3], the entries of the last column of the row reduced matrix. (Note that, even though we began the problem with polynomials for vectors, the answer to the problem, the coordinatization of the polynomial v, is a vector in R3 .)

(g) We convert the given polynomials to vectors in R4 by changing 3 + 2 +  +  to [   ].

Then we use the Coordinatization Method.

Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the polynomials in the basis . Then find its corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 2 1 −3 ¯¯ 8 1 0 0 ¯¯ −1 ⎢ −1 ⎢ ¯ ⎥ 2 −1 ¯¯ 11 ⎥ ⎥ −→ ⎢ 0 1 0 ¯ 4 ⎥  [A|v] = ⎢ ¯ ¯ ⎣ 3 −1 ⎦ ⎣ 1 ¯ −9 0 0 1 ¯ −2 ⎦ ¯ −1 3 1 11 0 0 0 ¯ 0

Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: We eliminate the last row of the row reduced matrix, since it contains all zeroes. Then, [v] = [−1 4 −2], the remaining entries of the last column of the row reduced matrix. (Note that, even though we began the problem with polynomials for vectors, the answer to the problem, the coordinatization of the polynomial v, is a vector in R3 .) ∙ ¸   4 to [   ]. Then (h) We convert the given matrices in M22 to vectors in R by changing   we use the Coordinatization Method.

Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors obtained

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

144

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.7

by converting the matrices in the basis . Then find its corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 2 1 ¯¯ −3 1 0 0 ¯¯ 2 ⎢ −2 −1 −1 ¯ −2 ⎥ ⎢ ¯ ⎥ ¯ ⎥ −→ ⎢ 0 1 0 ¯ −3 ⎥  [A|v] = ⎢ ¯ ¯ ⎣ 0 ⎦ ⎣ 1 3 ¯ 0 0 0 1 ¯ 1 ⎦ ¯ 1 0 1 3 0 0 0 ¯ 0

Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: We eliminate the last row of the row reduced matrix, since it contains all zeroes. Then, [v] = [2 −3 1], the remaining entries of the last column of the row reduced matrix. (Note that, even though we began the problem with matrices for vectors, the answer to the problem, the coordinatization of the matrix v, is a vector in R3 .) ∙ ¸    6 to [      ]. (j) We convert the given matrices in M32 to vectors in R by changing    Then we use the Coordinatization Method.

Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the matrices in the basis . Then find its corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 −3 ¯¯ 11 1 0 ¯¯ 5 ⎢ 3 ⎢ 0 1 ¯ −2 ⎥ 1 ¯¯ 13 ⎥ ⎢ ⎥ ⎢ ¯ ⎥ ⎢ −1 ¯ −19 ⎥ ⎢ 0 0 ¯ 0 ⎥ 7 ⎢ ¯ ⎥ ⎢ ¯ ⎥ [A|v] = ⎢ −→ ⎢ ¯ ⎥ 1 ¯¯ 8 ⎥ ⎢ 2 ⎥ ⎢ 0 0 ¯ 0 ⎥ ⎣ 1 ⎣ 0 0 ¯ 0 ⎦ 2 ¯¯ 1 ⎦ ¯ 4 5 ¯ 10 0 0 ¯ 0

Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: We eliminate the last four rows of the row reduced matrix, since they contain all zeroes.

Then, [v] = [5 −2], the remaining entries of the last column of the row reduced matrix. (Note that, even though we began the problem with matrices for vectors, the answer to the problem, the coordinatization of the matrix v, is a vector in R2 .)

(2) (a) We use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis , and the columns of C are the vectors in the basis . Then, we find the reduced row echelon form for this matrix: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 1 1 ¯¯ 1 0 0 1 0 0 ¯¯ −102 20 3 ⎣ 5 6 3 ¯¯ 0 1 0 ⎦ −→ ⎣ 0 1 0 ¯¯ 67 −13 −2 ⎦  1 −6 14 ¯ 0 0 1 0 0 1 ¯ 36 −7 −1 The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ −102 20 3 67 −13 −2 ⎦  P=⎣ 36 −7 −1

(c) We convert the given polynomials to vectors in R3 by changing 2 +  +  to [  ]. Then we use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the c 2016 Elsevier Ltd. All rights reserved.

Copyright °

145

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.7

columns of B are the vectors converted from the basis , and the columns of C are the vectors converted from the basis . Then, we find the reduced row echelon form for this matrix: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 3 10 ¯¯ 2 8 1 1 0 0 ¯¯ 20 −30 −69 ⎣ 3 4 17 ¯ 3 1 0 ⎦ −→ ⎣ 0 1 0 ¯ 24 −24 −80 ⎦  ¯ ¯ 1 1 5 ¯ −1 1 6 0 0 1 ¯ −9 11 31

The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ 20 −30 −69 P = ⎣ 24 −24 −80 ⎦  −9 11 31 (d) We convert the given matrices in M22 to vectors in R4 by changing

∙

   

¸

to [   ]. Then

we use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors converted from the basis , and the columns of C are the vectors converted from the basis . Then, we find the reduced row echelon form for this matrix: ¯ ¯ ⎡ ⎤ ⎡ ⎤ −1 1 3 1 ¯¯ 1 2 3 0 1 0 0 0 ¯¯ −1 −4 2 −9 ⎢ 1 0 −4 −1 ¯ 3 1 1 ⎢ 0 1 0 0 ¯ 4 2 ⎥ 5 1 3 ⎥ ⎢ ¯ ⎥ ⎢ ¯ ⎥ ⎣ 3 0 −7 −2 ¯ 5 0 1 −4 ⎦ −→ ⎣ 0 0 1 0 ¯ 0 ⎦ 2 −3 1 ¯ ¯ −1 1 4 1 ¯ 1 4 0 1 0 0 0 1 ¯ −4 −13 13 −15

The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎤ ⎡ −1 −4 2 −9 ⎢ 4 1 3 ⎥ 5 ⎥ P=⎢ ⎣ 0 2 −3 1 ⎦ −4 −13 13 −15

(f) We convert the given polynomials to vectors in R5 by changing 4 + 3 + 2 +  +  to [    ]. Then we use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors converted from the basis , and the columns of C are the vectors converted from the basis . Then, we find the reduced row echelon form for this matrix: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 2 2 ¯¯ 6 1 0 1 0 0 ¯¯ 6 1 2 ⎢ 3 7 ⎢ 0 1 0 ¯ 1 5 ¯¯ 20 5 5 ⎥ 1 2 ⎥ ⎢ ⎥ ⎢ ¯ ⎥ ⎢ 0 4 −3 ¯ 7 ⎥ ⎢ ¯ 7 17 ⎥ −→ ⎢ 0 0 1 ¯ −1 −1 −3 ⎥ ⎢ ¯ ⎥ ⎣ 4 3 ⎣ 0 0 0 ¯ 0 ⎦ 8 ¯¯ 19 −1 −10 ⎦ 0 0 ¯ ¯ ¯ −2 1 −7 −4 6 19 0 0 0 0 0 0 Now, we eliminate the last two rows of the row reduced matrix, since it contains all zeroes. Then, the transition matrix consists of the remaining entries in the matrix on the right side of the row reduced matrix. That is, ⎤ ⎡ 2 6 1 1 2 ⎦ P=⎣ 1 −1 −1 −3

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

146

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.7

(4) (a) In each subpart, we use the Transition Matrix Method. In order to find the transition matrix from basis  to basis  , we form the augmented matrix [Y|X], in which the columns of X are the vectors in the basis , and the columns of Y are the vectors in the basis  . Then, in each case, we convert this matrix to reduced row echelon form. Since, in this problem, there are no rows of all zeroes in any of the row reduced matrices obtained, the matrices to the right of the augmentation bar are the desired transition matrices. (i) To find the transition matrix from  to , we row reduce ¯ ¯ ∙ ¸ ∙ ¸ ∙ ¸ 3 2 ¯¯ 3 7 1 0 ¯¯ 13 31 13 31 , obtaining . Hence, P =  7 5 ¯ 1 2 0 1 ¯ −18 −43 −18 −43 (ii) To find the transition matrix from  to , we row reduce ¯ ¯ ∙ ¸ ∙ ¸ ∙ ¸ 5 2 ¯¯ 3 2 1 0 ¯¯ −11 −8 −11 −8 , obtaining . Hence, Q =  2 1 ¯ 7 5 0 1 ¯ 29 21 29 21 (iii) To find the transition matrix from  to , we row reduce ¯ ¯ ∙ ¸ ∙ ¸ ∙ ¸ 5 2 ¯¯ 3 7 1 0 ¯¯ 1 3 1 3 , obtaining . Hence, T =  2 1 ¯ 1 2 0 1 ¯ −1 −4 −1 −4 A straightforward matrix multiplication verifies that T = QP.

(c) First, we convert the polynomials in each of the given bases to vectors in R3 by changing 2 ++ to [  ]. Then, in each subpart, we use the Transition Matrix Method. Given bases of converted vectors  and  , in order to find the transition matrix from  to  , we form the augmented matrix [Y|X], in which the columns of X are the vectors in the basis , and the columns of Y are the vectors in the basis  . Then, in each case, we convert this matrix to reduced row echelon form. Since, in this problem, there are no rows of all zeroes in any of the row reduced matrices obtained, the matrices to the right of the augmentation bar are the desired transition matrices. (i) To find the transition matrix from  to , we row reduce ¯ ¯ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ¯ 2 8 13 2 8 13 1 2 1 ¯¯ 1 3 3 ¯ ⎣ 4 1 0 ¯ 2 7 9 ⎦ to ⎣ I3 ¯ −6 −25 −43 ⎦ . Hence, P = ⎣ −6 −25 −43 ⎦  ¯ ¯ ¯ 11 45 76 11 45 76 1 0 0 ¯ 2 8 13 (ii) To find the transition matrix from  to , we row reduce ¯ ¯ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ¯ −24 −2 1 −24 −2 1 7 1 1 ¯¯ 1 2 1 ¯ ⎣ −3 3 −1 ⎦ . Hence, Q = ⎣ 30 3 −1 ⎦  7 −2 ¯¯ 4 1 0 ⎦ to ⎣ I3 ¯¯ 30 ¯ 139 13 −5 139 13 −5 2 −3 1 ¯ 1 0 0

(iii) To find the transition matrix from  to , we row reduce ¯ ¯ ⎤ ⎡ ⎤ ⎡ ¯ −25 −97 −150 7 1 1 ¯¯ 1 3 3 ¯ ⎣ −3 185 ⎦  7 −2 ¯¯ 2 7 9 ⎦ to ⎣ I3 ¯¯ 31 120 ¯ 145 562 868 2 −3 1 ¯ 2 8 13 Hence,

⎡

⎤ −25 −97 −150 185 ⎦  T = ⎣ 31 120 145 562 868

A straightforward matrix multiplication verifies that T = QP. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

147

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.7

(5) (a) (i) In accordance with the Simplified Span Method, we row reduce ⎡ ⎤ ⎡ ⎤ 1 −4 1 2 1 1 −4 0 −2 0 ⎣ 6 −24 5 8 3 ⎦ to ⎣ 0 0 1 4 0 ⎦ 3 −12 3 6 2 0 0 0 0 1 Hence,  = ([1 −4 0 −2 0] [0 0 1 4 0] [0 0 0 0 1]).

(ii) We use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis , and the columns of C are the vectors in the basis . Then, we convert this matrix to reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎤ ⎡ 1 0 0 ¯¯ 1 6 3 1 0 0 ¯¯ 1 6 3 ⎢ −4 0 0 ¯ −4 −24 −12 ⎥ ⎢ 0 1 0 ¯ 1 5 3 ⎥ ⎢ ⎥ ¯ ⎥ ⎢ ¯ ⎢ 0 1 0 ¯ 1 ⎥ ⎥ ¯ 5 3 ⎥ −→ ⎢ ⎢ ¯ ⎢ 0 0 1 ¯ 1 3 2 ⎥ ¯ ⎣ −2 4 0 ¯ 2 ⎦ ⎣ 8 6 0 0 0 ¯ 0 0 0 ⎦ ¯ ¯ 0 0 1 1 3 2 0 0 0 ¯ 0 0 0 Now, we eliminate the last two rows of the row reduced matrix, since it contains all zeroes. Then, the transition matrix consists of the remaining entries in the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ 1 6 3 P = ⎣ 1 5 3 ⎦ 1 3 2 (iii) According to Theorem 4.19,

⎡

⎤ 1 −3 3 0 ⎦ Q = P−1 = ⎣ 1 −1 −2 3 −1

where we have computed P−1 by row reducing [P|I3 ] to [I3 |P−1 ].

(iv) We use the Coordinatization Method to find [v] . To do this, we form the augmented matrix [B|v], where the columns of B are the vectors in the basis , and then find the corresponding reduced row echelon form of this matrix: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 6 3 ¯¯ 2 1 0 0 ¯¯ 17 ⎢ −4 −24 −12 ¯ −8 ⎥ ⎢ 0 1 0 ¯ 4 ⎥ ⎢ ¯ ⎥ ⎢ ¯ ⎥ ¯ −2 ⎥ −→ ⎢ 0 0 1 ¯ −13 ⎥  1 5 3 [B|v] = ⎢ ⎢ ¯ ⎥ ⎢ ¯ ⎥ ⎣ 2 ⎣ 0 0 0 ¯ 8 6 ¯¯ −12 ⎦ 0 ⎦ ¯ 1 3 2 ¯ 3 0 0 0 ¯ 0

We eliminate the last two rows of the row reduced matrix, since they contain all zeroes. Then, [v] = [17 4 −13], the remaining entries of the last column.

Next we compute [v] . We form the augmented matrix [C|v], where the columns of C are the vectors in the basis , and then find the corresponding reduced row echelon form of this matrix: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 0 0 ¯¯ 2 1 0 0 ¯¯ 2 ⎢ −4 0 0 ¯ −8 ⎥ ⎢ 0 1 0 ¯ −2 ⎥ ⎢ ¯ ⎥ ⎢ ¯ ⎥ ¯ −2 ⎥ −→ ⎢ 0 0 1 ¯ 3 ⎥  0 1 0 [C|v] = ⎢ ⎢ ¯ ⎥ ⎢ ¯ ⎥ ⎣ −2 4 0 ¯ −12 ⎦ ⎣ 0 0 0 ¯ 0 ⎦ ¯ ¯ 0 0 1 ¯ 3 0 0 0 ¯ 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

148

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.7

We eliminate the last two rows of the row reduced matrix, since they contain all zeroes. Then, [v] = [2 −2 3], the remaining entries of the last column.

(v) Since Q is the transition matrix from  to , Theorem 4.17 implies that Q[v] = [v] . This can be verified for the answers computed above using matrix multiplication.. (c) (i) In accordance with the Simplified Span Method, we row reduce ⎡ ⎤ 3 −1 4 6 ⎢ 6 7 −3 −2 ⎥ ⎢ ⎥ to obtain I4  ⎣ −4 −3 3 4 ⎦ −2 0 1 2

Hence,  = ([1 0 0 0] [0 1 0 0] [0 0 1 0] [0 0 0 1]), the standard basis for R4 . (ii) We use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis , and the columns of C are the vectors in the basis . However, since the matrix C equals I4 , [C|B] is already in reduced row echelon form.

Therefore, the transition matrix P actually equals B. That is, ⎡

(iii) By Theorem 4.19,

⎤ 3 6 −4 −2 ⎢ −1 7 −3 0 ⎥ ⎥ P=⎢ ⎣ 4 −3 3 1 ⎦ 6 −2 4 2

Q=P

−1

⎡

1

⎢ ⎢ −2 ⎢ =⎢ ⎢ −5 ⎣

−4 −12 9 22

7

⎤

⎥ ⎥ 27 − 31 2 ⎥ ⎥ ⎥ 67 − 77 2 ⎦

5 −23 −71

41

where we have computed P−1 by row reducing [P|I3 ] to [I3 |P−1 ].

(iv) We use the Coordinatization Method to find [v] . To do this, we form the augmented matrix [B|v], where the columns of B are the vectors in the basis , and then find the corresponding reduced row echelon form of this matrix: ¯ ¯ ⎡ ⎤ ⎤ ⎡ ¯ 2 3 6 −4 −2 ¯¯ 10 ¯ ⎢ −1 ¯ ⎥ ⎢ 7 −3 0 ¯¯ 14 ⎥ ⎥ −→ ⎢ I4 ¯ 1 ⎥  [B|v] = ⎢ ¯ ¯ ⎣ 4 −3 ⎦ ⎦ ⎣ 3 1 ¯ 3 ¯ −3 ¯ ¯ 6 −2 4 2 12 7 Thus, [v] = [2 1 −3 7], the last column of the row reduced matrix.

Next we compute [v] . But  is the standard basis for R4 , and so [v] actually equals v. That is, [v] = [10 14 3 12]. (v) Since Q is the transition matrix from  to , Theorem 4.17 implies that Q[v] = [v] . This can be verified for the answers computed above using matrix multiplication.

(7) (a) To compute each of these transition matrices, we use the Transition Matrix Method. To do this, we form the augmented matrix [C |B], for 1 ≤  ≤ 5, in which the columns of B are the vectors c 2016 Elsevier Ltd. All rights reserved.

Copyright °

149

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.7

in the basis , and the columns of C are the vectors in the basis  . Then, we convert each of these matrices to reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ ¯ 0 1 0 3 2 −5 ¯¯ −5 3 2 ¯ 9 ¯¯ 9 −9 −5 ⎦ → ⎣ I3 ¯¯ 0 0 1 ⎦ , [C1 |B] = ⎣ −9 −5 ¯ 1 0 0 2 1 −1 ¯ −1 2 1 ¯ ¯ ⎡ ⎤ ⎡ ⎤ ¯ 0 0 1 2 −5 3 ¯¯ −5 3 2 ¯ 9 −9 ¯¯ 9 −9 −5 ⎦ → ⎣ I3 ¯¯ 1 0 0 ⎦ , [C2 |B] = ⎣ −5 ¯ 0 1 0 1 −1 2 ¯ −1 2 1 ¯ ¯ ⎡ ⎤ ⎡ ⎤ ¯ 1 0 0 −5 2 3 ¯¯ −5 3 2 ¯ [C3 |B] = ⎣ 9 −5 −9 ¯¯ 9 −9 −5 ⎦ → ⎣ I3 ¯¯ 0 0 1 ⎦ , ¯ 0 1 0 −1 1 2 ¯ −1 2 1 ¯ ¯ ⎡ ⎤ ⎡ ⎤ ¯ 0 1 0 3 −5 2 ¯¯ −5 3 2 ¯ 9 −5 ¯¯ 9 −9 −5 ⎦ → ⎣ I3 ¯¯ 1 0 0 ⎦ , [C4 |B] = ⎣ −9 ¯ 0 0 1 2 −1 1 ¯ −1 2 1 ¯ ¯ ⎡ ⎤ ⎡ ⎤ ¯ 0 0 1 2 3 −5 ¯¯ −5 3 2 ¯ 9 ¯¯ 9 −9 −5 ⎦ → ⎣ I3 ¯¯ 0 1 0 ⎦.

[C5 |B] = ⎣ −5 −9 ¯ 1 0 0 1 2 −1 ¯ −1 2 1 Thus, the transition matrices from  to 1 , 2 , 3 , 4 , and 5 , respectively, are ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 ⎣ 0 0 1 ⎦⎣ 1 0 0 ⎦⎣ 0 0 1 ⎦⎣ 1 0 0 ⎦⎣ 0 1 0 ⎦ 1 0 0 0 1 0 0 1 0 0 0 1 1 0 0

(10) According to Exercise 9, as suggested in the hint, if P is the matrix whose columns are the vectors in , and Q is the matrix whose columns are the vectors in , then Q−1 P is the transition matrix from  to  Now suppose A is the transition matrix from  to  given in the problem. Then, A = Q−1 P, implying QA = P. This yields Q = PA−1 . Hence, we first compute A−1 by row reducing [A|I3 ] to [I3 |A−1 ]. Then we use the result obtained from that computation to find Q, as follows: ⎡ ⎤⎡ ⎤ ⎡ ⎤ −2 1 −13 49 19 86 −142 −53 −246 5 ⎦ ⎣ −5 −2 −9 ⎦ = ⎣ 64 24 111 ⎦  Q = PA−1 = ⎣ 1 0 3 2 10 3 1 5 167 63 290 So,  = ([−142 64 167] [−53 24 63] [−246 111 290]), the columns of Q.

(11) (b) First,

⎡

⎤⎡ ⎤ ⎡ ⎤ 14 −15 −30 1 14 Av = ⎣ 6 −7 −12 ⎦ ⎣ 4 ⎦ = ⎣ 2 ⎦  3 −3 −7 −2 5

We use the Coordinatization Method to find [Av] . To do this, we form the augmented matrix [B|Av], where the columns of B are the vectors in the basis , and then find its corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎤ ⎡ ¯ 2 5 1 2 ¯¯ 14 ¯ [B|Av] = ⎣ 2 1 0 ¯¯ 2 ⎦ −→ ⎣ I3 ¯¯ −2 ⎦  ¯ 3 1 0 1 ¯ 5

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

150

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 4.7

Thus, [Av] = [2 −2 3], the last column of the row reduced matrix.

Next, we use the Coordinatization Method to find [v] . To do this, we form the augmented matrix [B|v], and find its corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎤ ⎡ ¯ 1 5 1 2 ¯¯ 1 ¯ [B|v] = ⎣ 2 1 0 ¯¯ 4 ⎦ −→ ⎣ I3 ¯¯ 2 ⎦  ¯ −3 1 0 1 ¯ −2

Thus, [v] = [1 2 −3], the last column of the row reduced matrix.

Finally, we compute ⎡ ⎤⎡ ⎤ ⎡ ⎤ 2 0 0 1 2 0 ⎦ ⎣ 2 ⎦ = ⎣ −2 ⎦  D[v] = ⎣ 0 −1 0 0 −1 −3 3 And so, D[v] = [Av] = [2 −2 3].

(13) Let V, , and the  ’s and w’s be as given in the statement of the theorem.

Proof of Part (1): Suppose that [w1 ] = [1       ] and [w2 ] = [1       ]. Then, by definition, w1 = 1 v1 + · · · +  v and w2 = 1 v1 + · · · +  v . Hence, w1 + w2

= 1 v1 + · · · +  v + 1 v1 + · · · +  v = (1 + 1 )v1 + · · · + ( +  )v 

implying [w1 + w2 ] = [(1 + 1 )     ( +  )] which equals [w1 ] + [w2 ] .

Proof of Part (2): Suppose that [w1 ] = [1       ]. Then, by definition, w1 = 1 v1 + · · · +  v .

Hence, 1 w1 = 1 (1 v1 + · · · +  v ) = 1 1 v1 + · · · + 1  v  implying [1 w1 ] = [1 1      1  ], which equals 1 [1       ].

Proof of Part (3): Use induction on .

Base Step ( = 1): This is just part (2).

Inductive Step: Assume true for any linear combination of  vectors, and prove true for a linear combination of  + 1 vectors. Now, [1 w1 + · · · +  w + +1 w+1 ]

= [1 w1 + · · · +  w ] + [+1 w+1 ] (by part (1)) = [1 w1 + · · · +  w ] + +1 [w+1 ] (by part (2)) = 1 [w1 ] + · · · +  [w ] + +1 [w+1 ] ,

by the inductive hypothesis.

(14) Let v ∈ V. Then, applying Theorem 4.17 twice yields (QP)[v] = Q[v] = [v] . Therefore, (QP)[v] = [v] for all v ∈ V, and so Theorem 4.17 implies that QP is the (unique) transition matrix from  to .

(17) (a) False. Typically, changing the order that the vectors appear in an ordered basis also changes the order of the coordinates in the corresponding coordinatization vectors. For example, if v = [1 2 3] ∈ R3 , then since v = 1i + 2j + 3k, [v] = [1 2 3]. However, since v = 2j + 3k + 1i, [v] = [2 3 1]. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

151

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 4 Review

(b) True. Now b = 0b1 + · · · + 0b−1 + 1b + 0b+1 + · · · + 0b . Thus, [b ] is the -vector having the coefficients 0     0 1 0    0, respectively, as its entries. Hence, [b ] = e . (c) True. This is seen by reversing the roles of  and  in the definition of the transition matrix from  to . (d) False. This is close to the statement of Theorem 4.17, except that the positions of [v] and [v] are reversed. The correct conclusion is that P[v] = [v] . For a specific counterexample to the statement given in the problem, consider the ordered bases ∙  = ([1 ¸ 1] [1 −1]) and  = 1 1 2 ([1 0] [0 1]) for R . The transition matrix from  to  is P = , and, for v = [2 6], 1 −1 ∙ ¸ 8 [v] = [4 −2] and [v] = [2 6]. However, P[v] = 6= [v] . −4

(e) False. This is close to the statement of Theorem 4.18, except that the transition matrix T from  to  is QP, not PQ. For a specific counterexample to the statement given in the problem, consider the ordered bases , , and  for R2 given in Exercise 4(a) of this section, whose solution ∙ ¸ ∙ ¸ 13 31 −11 −8 appears above. Then, as computed in Exercise 4(a), P = , Q= , −18 −43 29 21 ∙ ¸ ∙ ¸ 1 3 756 547 and T = . But PQ = 6= T. −1 −4 −1049 −759

(f) True. This is part of Theorem 4.19. (g) False. The matrix P−1 , not P, is the transition matrix from standard coordinates to a basis of fundamental eigenvectors for A, where P is the matrix from the Diagonalization Method whose columns are fundamental eigenvectors for A. This is stated in part (a) of Exercise 8 and mentioned in Example 13 in the textbook. Using A and P from that example gives a specific counterexample to the incorrect given statement. For, if P were the transition matrix from standard coordinates  to an ordered basis  = (b1  b2  b3 ) of eigenvectors for A, then P−1 would be the transition matrix from  to . Hence, we would have b1 = P−1 [b1 ] . But, [b1 ] = e1 (see Exercise 6). However, b1 is the first column of P−1 , which is [1 −2 −1], as shown in Example 13. But [1 −2 −1] is not an eigenvector for A, since Av = [74 32 16], which is not a scalar multiple of [1 −2 −1].

Chapter 4 Review Exercises (2) By part (2) of Theorem 4.1, the zero vector 0 = 0 ¯ [ ] = [0 + 4(0) − 4 0 − 5(0) + 5] = [−4 5] Similarly, by part (3) of Theorem 4.1, the additive inverse of [ ] is (−1) ¯ [ ] = [(−1) + 4(−1) − 4 (−1) − 5(−1) + 5] = [− − 8 −  + 10] (3) (a) The set  = {[3 2 − 1 −4] |  ∈ R} is not a subspace. This is because the zero vector, [0 0 0] is not an element of . For [0 0 0] to be of the form [3 2 − 1 −4], we must have 3 = 0, implying  = 0. But then 2 − 1 = −1 6= 0, which gives a contradiction. ½∙ ¸ ¯ ¾ ¯ 2 +  −4 − 5 ¯   ∈ R is a subspace of M22 . First, letting  =  = 0 (c) The set V = ¯ 0  − 2 ∙ ¸ 0 0 shows that ∈ V, and so V is nonempty. By Theorem 4.2 it is enough to prove that V is 0 0 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

152

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 4 Review

closed under addition and scalar multiplication. ∙ ¸ ∙ ¸ 2 +  −4 − 5 2 +  −4 − 5 For addition, let , ∈ V. Then 0  − 2 0  − 2 ∙ ¸ ∙ ¸ 2 +  −4 − 5 2 +  −4 − 5 + 0  − 2 0  − 2 ∙ ¸ 2( + ) + ( + ) −4( + ) − 5( + ) =  0 ( + ) − 2( + ) which is of the correct form to be in V, with  =  +  and  =  + . ∙ ¸ 2 +  −4 − 5 For scalar multiplication, let ∈ V and  ∈ R. Then 0  − 2 ∙ ¸ ∙ ¸ 2 +  −4 − 5 (2 + ) (−4 − 5)  = 0  − 2 0 ( − 2) ∙ ¸ 2() + () −4() − 5() =  0 () − 2() which is of the correct form to be in V, with  =  and  = .

(d) The set  of all matrices in M22 that are both singular and symmetric is not a subspace of M22 because it is not closed under addition. For example, ∙ ¸ ∙ ¸ 1 0 0 0  ∈  0 0 0 1 but

∙

1 0 0 0

¸

+

∙

0 0 0 1

¸

=

∙

1 0 0 1

¸

∈  

(f) The set  of all polynomials in P4 whose highest order nonzero term has even degree is not a subspace of P4 ¡because ¢it is¡ not closed ¢under addition. For example, 4 − 3 and −4 + 23 are  . both in , but 4 − 3 + −4 + 23 = 3 ∈

(4) (a) With  = {[3 3 −2 4] [3 4 0 3] [5 6 −1 6] [4 4 −3 5]}, we follow the Simplified Span Method: ⎡ ⎤ 3 3 −2 4 ⎢ 3 4 0 3 ⎥ ⎥ Step 1: Form the matrix A = ⎢ ⎣ 5 6 −1 6 ⎦ .

4 4 −3 5 ⎡ ⎤ 1 0 0 5 ⎢ 0 1 0 −3 ⎥ ⎥.

Step 2: A row reduces to C = ⎢ ⎣ 0 0 1 1 ⎦ 0 0 0 0 Step 3: span() = {[1 0 0 5] + [0 1 0 −3] + [0 0 1 1] |    ∈ R} = {[   5 − 3 + ] |    ∈ R}.

(b) The set  = {[1 0 0 5] [0 1 0 −3] [0 0 1 1]} is a basis for span(). First,  clearly spans span(), as can be seen from the description of span() in part (a). Also,  is linearly independent because each vector in  has a 1 in a coordinate in which every other vector has a zero. Since  has 3 elements, dim(span()) = 3. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

153

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 4 Review

(7) (a) With  = {[3 5 −3] [−2 −4 3] [1 2 −1]}, we follow the Independence Test Method. ⎡ ⎤ 3 −2 1 2 ⎦ , whose columns are the vectors in .

Step 1: Form the matrix A = ⎣ 5 −4 −3 3 −1 Step 2: A row reduces to B = I3 .

Step 3: Since there is a pivot in every column,  is linearly independent. (b) Because the full set  is linearly independent,  is a basis for span(). Also,  spans R3 by part (1) of Theorem 4.12. (c) Because  is linearly independent, Theorem 4.9 guarantees that there is only one way to express v = [11 20 −12] as a linear combination of the vectors in .

(8) (a) First, we convert the polynomials in ª ©  = −53 + 22 + 5 − 2 23 − 2 − 2 + 1 3 − 22 −  + 2 −23 + 22 + 3 − 5 into vectors in R4 :

−53 + 22 + 5 − 2 23 − 2 − 2 + 1 3 − 22 −  + 2 and − 23 + 22 + 3 − 5

→ → → →

[−5 2 5 −2] [2 −1 −2 1] [1 −2 −1 2] [−2 2 3 −5]

We follow the Independence Test Method. ⎡ ⎤ −5 2 1 −2 ⎢ 2 −1 −2 2 ⎥ ⎥ , whose columns are the -vectors we found.

Step 1: Form the matrix A = ⎢ ⎣ 5 −2 −1 3 ⎦ −2 1 2 −5 ⎡ ⎤ 1 0 3 0 ⎢ 0 1 8 0 ⎥ ⎥ Step 2: A row reduces to B = ⎢ ⎣ 0 0 0 1 ⎦.

0 0 0 0 Step 3: Since there is no pivot in column 3,  is linearly dependent.

Using the entries in the third column of the matrix B, we see that the third vector equals 3 times the first vector plus 8 times the second vector; that is, ¡ ¢ ¡ ¢ 3 − 22 −  + 2 = 3 −53 + 22 + 5 − 2 + 8 23 − 2 − 2 + 1 

(b) Using the Independence Test Method and the work from part (a), we can find a basis for span() by choosing the vectors in  that correspond to the pivot columns in the row reduced matrix B that we computed. That is, the set {−53 + 22 + 5 − 2 23 − 2 − 2 + 1 − 23 + 22 + 3 − 5} is a basis for span(). From this basis, we see that dim(span()) = 3. Therefore,  does not span P3 since dim(span())  dim(P3 ). c 2016 Elsevier Ltd. All rights reserved.

Copyright °

154

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 4 Review

(c) Yes, there is an alternate linear combination by part (b) of Exercise 21 in Section 4.4. One such alternate linear combination is ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ −5 −53 + 22 + 5 − 2 −5 23 − 2 − 2 + 1 +1 3 − 22 −  + 2 −1 −23 + 22 + 3 − 5  This was found by row reducing the matrix ¯ ⎡ ⎤ ⎡ −5 2 1 −2 ¯¯ 18 1 ⎢ 2 −1 −2 ⎢ 2 ¯¯ −9 ⎥ ⎢ ⎥ , which yields ⎢ 0 ⎣ 5 −2 −1 ⎣ 0 3 ¯¯ −19 ⎦ ¯ −2 1 2 −5 12 0

0 1 0 0

3 8 0 0

¯ ⎤ 0 ¯¯ −2 0 ¯¯ 3 ⎥ ⎥ 1 ¯¯ −1 ⎦ 0 ¯ 0

and taking the particular solution to the system that uses the value 1 for the independent variable corresponding to the third column. ⎡ ⎤ −2 3 −1 4 ⎢ 3 −3 2 −4 ⎥ ⎥ row reduces (12) (a) Using the Simplified Span Method, we find that the matrix ⎢ ⎣ −2 2 −1 3 ⎦ 3 −5 0 −7 to I4 , so the given set of 4 vectors spans R4 . By part (1) of Theorem 4.12, the given set is a basis for R4 .

(13) (a) We will prove that W is a subspace of R4 using Theorem 4.2 by showing that W is nonempty and that it is closed under addition and scalar multiplication. First we show that W is nonempty: 0 ∈ W because A0 = O. Closure under addition: If X1  X2 ∈ W, then A(X1 + X2 ) = AX1 + AX2 = O + O = O. Closure under scalar multiplication: If X ∈ W, then A(X) = AX = O = O. ⎡ ⎤ 1 −3 0 2 0 1 −1 ⎦. We use this to solve the homogeneous system AX = (b) A row reduces to ⎣ 0 0 0 0 0 O. Because there are no pivots in columns 2 and 4, those columns correspond to independent variables. Letting those variables equal 1 and 0, respectively, yields the solution [3 1 0 0] for the homogeneous system. Setting these variables equal to 0 and 1, respectively, produces the solution [−2 0 1 1]. Hence, we obtain the following basis for W: {[3 1 0 0] [−2 0 1 1]} (c) Because the basis from part (b) has two elements, dim(W) = 2. Also, the reduced row echelon form for A from part (b) shows that rank(A) = 2. Therefore, dim(W)+rank(A) = 2+2 = 4 = dim(R4 ).

(14) (a) First, we use direct computation to check that every polynomial in  is in V: If p() = 3 − 3, then p0 () = 32 − 3, and so p0 (1) = 3 − 3 = 0.

If p() = 2 − 2, then p0 () = 2 − 2, and so p0 (1) = 2 − 2 = 0.

If p() = 1, then p0 () = 0, and so p0 (1) = 0.

Thus, every polynomial in  is actually in V.

Next, we convert the polynomials in  to 4-vectors, and use the Independence Test Method. Now, ⎡ ⎤ ⎡ ⎤ 1 0 0 1 0 0 ⎢ 0 ⎢ 0 1 0 ⎥ 1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −3 −2 0 ⎦ clearly row reduces to ⎣ 0 0 1 ⎦ , so  is linearly independent. Finally, since 0 0 1 0 0 0

the polynomial  ∈  V, dim(V)  dim(P3 ) = 4 by Theorem 4.13. But, part (2) of Theorem 4.12 shows that || ≤ dim(V). Hence, 3 = || ≤ dim(V)  dim(P3 ) = 4, and so dim(V) = 3. Part (2) of Theorem 4.12 then implies that  is a basis for V. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

155

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 4 Review

(b) If p() = 3 + 2 +  + , then p0 () = 32 + 2 + , and p00 () = 6 + 2. Setting p00 (1) = 0 produces the equation 6 + 2 = 0, or  = −3. Setting p0 (1) = 0 yields 3 + 2 +  = 0.

Combining this with  = −3 gives us  = −3 − 2(−3) = 3. Note that  is arbitrary. Setting  = 1 and  = 0 produces the polynomial 3 − 32 + 3. Setting  = 0 and  = 1 yields the constant polynomial 1. Note that  = {1 3 − 32 + 3} is a linearly independent subset of W. Now dim(W)  dim(V) = 3 (since W is a proper subset of V). Thus  is a basis for W by an argument similar to that used for  in part (a), and dim(W) = 2.

(15) We follow the method described in Exercise 6 in Section 4.6: Choose v1 = [2 3 0 1]. Skip [−6 9 0 −3], since it equals −3v1 . Choose v2 = [4 3 0 4], since it is not a scalar multiple of v1 . Next, [8 −3 0 6] = 2v1 + v2 , so skip it. Finally, choose v3 = [1 0 2 1], because it is not a linear combination of v1 and v2 , since it does not have a zero in its third coordinate. Thus,  = {[2 −3 0 1] [4 3 0 4] [1 0 2 1]} is a basis for V (17) To perform the Enlarging Method, we first add the standard basis for R4 to  to form a spanning set for R4 . We then use the Independence Test Method to reduce this spanning set to a basis for R4 .

Thus, we row reduce ⎤ ⎡ 1 ⎡ ⎤ 1 0 25 0 0 10 2 1 1 0 0 0 ⎥ ⎢ ⎢ 1 −2 0 1 0 0 ⎥ ⎢ 0 1 15 0 0 − 15 ⎥ ⎥ ⎢ ⎥ to obtain ⎢ ⎢ 0 0 0 1 0 −1 ⎥  ⎣ −1 2 0 0 1 0 ⎦ ⎣ 2 ⎦ 2 −4 0 0 0 1 1 0 0 0 0 1 2 Since the 1st, 2nd, 4th, and 5th columns are pivot columns, we choose the vectors that correspond to those columns. Hence, the desired basis for R4 is {[2 1 −1 2] [1 −2 2 −4] [0 1 0 0] [0 0 1 0]} (20) (a) We use the Coordinatization Method.

Steps 1 and 2: We form the augmented matrix [A|v], where the columns of A are the vectors in the basis , and then find the corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 2 5 −6 ¯¯ 1 1 0 0 ¯¯ −3 2 ¯¯ −7 ⎦ −→ ⎣ 0 1 0 ¯¯ −1 ⎦  [A|v] = ⎣ 1 0 2 1 1 ¯ −9 0 0 1 ¯ −2

Step 3: The row reduced matrix contains no rows having all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: [v] = [−3 −1 −2], the last column of the row reduced matrix. ⎡ ⎤   (c) We convert the given matrices in M23 to vectors in R6 by changing ⎣   ⎦ to [      ].  

Then we use the Coordinatization Method.

Steps 1 and 2: We form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the matrices in the basis . Then we find the corresponding reduced row

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

156

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 4 Review

echelon form: ¯ ⎤ ⎡ −3 −10 2 ¯¯ −43 1 0 ⎢ 3 ¯ −5 ⎥ ⎢ 0 1 3 11 ⎢ ¯ ⎥ ⎢ ⎢ 11 ⎢ 0 0 28 10 ¯¯ 97 ⎥ ⎢ ⎥ ⎢ [A|v] = ⎢ ¯ −9 ⎥ −→ ⎢ 0 0 5 4 14 ⎢ ¯ ⎥ ⎢ ⎣ −2 −6 −16 ¯ −8 ⎦ ⎣ 0 0 ¯ 2 0 3 ¯ −9 0 0 ⎡

¯ ⎤ 0 ¯¯ −3 0 ¯¯ 5 ⎥ ⎥ 1 ¯¯ −1 ⎥ ⎥ 0 ¯¯ 0 ⎥ ⎥ 0 ¯¯ 0 ⎦ 0 ¯ 0

Step 3: The row reduced matrix contains no rows having all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: We eliminate the last three rows of the row reduced matrix, since they contain all zeroes.

Then, [v] = [−3 5 −1], the remaining entries of the last column of the row reduced matrix.

(Note that, even though v and the elements of  are matrices, the coordinatization of v is a vector in R3 .)

(21) (a) First, to find [v] , we use the Coordinatization Method.

Steps 1 and 2: We form the augmented matrix [A|v], where the columns of A are the vectors in the basis , and then find the corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 26 9 −3 ¯¯ 126 1 0 0 ¯¯ 27 [A|v] = ⎣ −47 −16 10 ¯¯ −217 ⎦ −→ ⎣ 0 1 0 ¯¯ −62 ⎦  −10 −1 37 ¯ 14 0 0 1 ¯ 6 Step 3: The row reduced matrix contains no rows having all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: [v] = [27 −62 6], the last column of the row reduced matrix.

We use the Transition Matrix Method to find the transition matrix from  to . To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis , and the columns of C are the vectors in the basis . Then, we find the reduced row echelon form for this matrix: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 2 −3 5 ¯¯ 26 9 −3 1 0 0 ¯¯ 4 2 5 ⎣ −3 5 −10 ¯¯ −47 −16 10 ⎦ −→ ⎣ 0 1 0 ¯¯ −1 0 1 ⎦ 4 −1 −9 ¯ −10 −1 37 0 0 1 ¯ 3 1 −2

The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ 4 2 5 1 ⎦ P = ⎣ −1 0 3 1 −2 Finally, we determine [v] by computing P[v] . Now, ⎡ ⎤⎡ ⎤ ⎡ ⎤ 4 2 5 27 14 1 ⎦ ⎣ −62 ⎦ = ⎣ −21 ⎦  P[v] = ⎣ −1 0 3 1 −2 6 7

Hence, [v] = [14 −21 7].

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

157

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 4 Review

(b) First, we solve for [v] . We convert the given polynomials in  to vectors in R3 by changing 2 +  +  to [  ]. Then we use the Coordinatization Method.

Steps 1 and 2: We form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the polynomials in the basis . Then we find the corresponding reduced row echelon form: ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 3 −2 ¯¯ −13 1 0 0 ¯¯ −4 4 ¯¯ −11 ⎦ −→ ⎣ 0 1 0 ¯¯ −1 ⎦  [A|v] = ⎣ 3 11 1 5 4 ¯ 3 0 0 1 ¯ 3 Step 3: The row reduced matrix contains no rows having all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4.

Step 4: Then, [v] = [−4 −1 3], the entries of the last column of the row reduced matrix. (Note that, even though v and the elements of  are polynomials, the coordinatization of v is a vector in R3 .)

Next, we use the Transition Method to find the transition matrix from  to . We convert the polynomials in  to vectors in R3 in the same manner as we did for the polynomials in .

Then, we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis , and the columns of C are the vectors in the basis . Then, we find the reduced row echelon form for this matrix: ¯ ¯ ⎡ ⎤ ⎡ ⎤ −7 13 −16 ¯¯ 1 3 −2 1 0 0 ¯¯ 4 1 −2 ⎣ 7 −7 18 ¯¯ 3 11 4 ⎦ −→ ⎣ 0 1 0 ¯¯ 1 2 0 ⎦ ¯ ¯ 9 −13 22 1 5 4 0 0 1 −1 1 1

The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ 4 1 −2 0 ⎦ P=⎣ 1 2 −1 1 1 Finally, we determine [v] by computing P[v] . Now, ⎡ ⎤⎡ ⎤ ⎡ ⎤ 4 1 −2 −4 −23 0 ⎦ ⎣ −1 ⎦ = ⎣ −6 ⎦  P[v] = ⎣ 1 2 −1 1 1 3 6

Hence, [v] = [−23 −6 6].

(23) (c) In part (a), your computations likely produced the following set of three fundamental eigenvectors: v1 = [4 4 13], v2 = [−3 2 0] and v3 = [3 0 4]. According to Example 13 in Section 4.7, the transition matrix from the ordered basis  = (v1  v2  v3 ) to standard coordinates is the matrix P from the Diagonalization Method; that is, it is the matrix whose columns are the vectors v1 , 8 in Section 4.7.) Hence, the transition matrix from  to v2 , and v3 . (Also see part (b) of ⎡ ⎤ 4 −3 3 2 0 ⎦. standard coordinates is ⎣ 4 13 0 4 (26) (a) True. To be a vector space, a set with given operations must satisfy all 10 properties in all cases.

So, if we can find a single counterexample to any one of the properties, then the set can not be a vector space.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

158

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 4 Review

(b) True. We will use Theorem 4.2 to prove that V is a subspace of R by showing that V is nonempty and that it is closed under addition and scalar multiplication. First we show that V is nonempty: 0 ∈ V because A0 = 0. Closure under addition: If X1  X2 ∈ V, then A(X1 +X2 ) = AX1 +AX2 = 0 + 0 = 0. Closure under scalar multiplication: If X ∈ V, then A(X) = AX = 0 = 0.

Therefore, since V is a subspace of R , V is itself a vector space. (Note: This was proved in Exercise 14 in Section 4.2 and is merely a generalized version of the proof for part (a) of Exercise 13 in the Chapter 4 Review Exercises.) (c) False. The integers¡ are under scalar multiplication. For example, using the scalar 12 ¢ not closed 1 1 and the integer 1, 2 (1) = 2 , which is not in the set of integers.

(d) True. This is because every subspace is, itself, a vector space. Thus, any subspace must include the zero vector by property (3) from the definition of a vector space. (e) False. For example, let W1 = {[ 0] |  ∈ R} and W2 = {[0 ] |  ∈ R}, both of which are subspaces of R2 . Then W1 ∪ W2 is the set of 2-vectors with a zero in at least one coordinate. However, this union is not closed under vector addition. For example, [1 0] + [0 1] = [1 1] ∈  W1 ∪ W2 .

(f) True. By Theorem 4.5, span() is a subspace of V. If v is a nonzero vector in , then all scalar multiples of v are in span(), since span() is closed under scalar multiplication. But for each scalar , v is a different vector. (If v = v, then ( − )v = 0, so part (4) of Theorem 4.1 shows that  −  = 0, or  = .) Since the number of different scalars in R is infinite, the number of different vectors in span() is infinite.

(g) True. See Exercise 27 in Section 4.3. (h) False. For a counterexample, let 1 = {[1 0]} and 2 = {[2 0]} in R2 . Then span(1 ) = span(2 ) = {[ 0] |  ∈ R}. (i) False. The vectors found using the Simplified Span Method will usually be simpler in form than the original vectors. For example, applying the Simplified Span Method to the basis {[1 1] [1 −1]} for R2 produces the standard basis {[1 0] [0 1]}. In fact, applying the Simplified Span Method to any basis for R produces the standard basis for R .

(j) True. Since  is already a basis, it is linearly independent. The Independence Test Method only removes redundant vectors that are linear combinations of the other vectors in the set. Because  is linearly independent, there are no such redundant vectors in this case, and so no vectors are removed from  . (k) False. In Example 7 in Section 4.1, the “vector” 1 was shown to be the additive identity (that is, the zero vector) in this vector space. No set containing the additive identity is linearly independent. (l) False. The eigenvectors corresponding to a particular eigenvalue are linear combinations of the fundamental eigenvectors for that eigenvalue. In particular, the set consisting of any two distinct nonzero multiples of the same fundamental eigenvector is a linearly dependent set.

Alternately, consider the matrix I2 . Then every nonzero vector in R2 is an eigenvector for I2 corresponding to the eigenvalue 1, because I2 v = v for all v ∈ R2 . Hence, every subset of R2 of nonzero vectors is a set of distinct eigenvectors for I2 . In particular, {[1 0] [0 1] [1 1]} is one such subset that is not linearly independent. (m) True. Exercise 10 in Section 4.4 proves this for 3 × 3 matrices. A similar proof works for  ×  matrices. A different proof is suggested in part (d) of Exercise 15 in Section 4.5, in which you are asked to prove that the rows of a nonsingular  ×  matrix form a basis for R , and are therefore linearly independent. (n) True. See part (b) of Exercise 23 in Section 4.4. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

159

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 4 Review

(o) False. For a specific counterexample, consider the linearly dependent set {[1 0] [2 0]} in R2 . Note that 2[1 0] − 1[2 0] = [0 0].

(p) False. The problem here is that v1 could be the zero vector. For a specific counterexample, consider the linearly dependent set {[0 0] [1 1]}, and note that there does not exist a scalar  such that [1 1] = [0 0]. (q) True. For a set to be a basis for a vector space, it must both span the vector space and be linearly independent. We are given that  is linearly independent, and it spans span( ) by the definition of span( ). (r) False. The dimension of the trivial vector space is zero. The empty set, which contains no vectors, is a basis for the trivial vector space. This is discussed in Example 10 in Section 4.5. (s) True. Using both parts of Theorem 4.12, we get || = | | ≤ dim(V) ≤ ||. Hence, both || and | | must equal dim(V). Therefore, again using both parts of Theorem 4.12, both  and  are bases for V. (t) True. This is the contrapositive of the first two sentences of Theorem 4.13.

(u) True. See part (a) of Exercise 14 in Section 4.6. (v) True. Since  is a basis for W, it is linearly independent. Theorem 4.15 then shows that there is a basis for V that contains .

(w) False. Theorem 4.19 states that P−1 , not P , is the transition matrix from  to . The vector space and bases discussed in Examples 11 and 12 in Section 4.7 provide a specific counterexample, since in that case, P−1 6= P . (x) True. If dim(V) = , then P is an  ×  matrix. To see why, notice that the columns of P are the vectors in  expressed in -coordinates. Thus, the number of columns of P equals the number of vectors in , which is . Also, since each of these columns is the -coordinatization of a vector in V, it has  entries. Therefore, P must have  rows as well.

(y) True. See part (b) of Exercise 8 in Section 4.7.

(z) False. After the row reduction, any rows consisting of all zeros must first be removed. See Example 9 in Section 4.7.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

160

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.1

Chapter 5 Section 5.1 (1) (a) The function  is a linear transformation. To prove this, we show that it satisfies the two properties in the definition of a linear transformation.

Property (1):  ([ ] + [ ]) = = = =

 ([ +   + ]) [3( + ) − 4( + ) −( + ) + 2( + )] [(3 − 4) + (3 − 4) (− + 2) + (− + 2)] [3 − 4 − + 2] + [3 − 4 − + 2] =  ([ ]) +  ([ ])

Property (2):  ([ ]) =  ([ ]) = [3() − 4() −() + 2()] = [(3 − 4) (− + 2)] = [3 − 4 − + 2] =  ([ ]) Also,  is a linear operator because it is a linear transformation whose domain equals its codomain. (b) The function  is not a linear transformation. If it were, then part (1) of Theorem 5.1 would imply (0) = 0. However, ([0 0 0 0]) = [2 −1 0 −3] 6= 0, so  can not be a linear transformation.

Because  is not a linear transformation, it can not be a linear operator. (d) The function  is a linear transformation. To prove this, we show that it satisfies the two properties in the definition of a linear transformation.

Property (1): µ∙ ¸ ∙ ¸¶ µ∙ ¸¶     + +  + =      + + ¸ ∙ 3( +  ) ( + ) − 2( + ) + ( + ) = −4( + ) ( +  ) + ( + ) − 3( + ) ∙ ¸ ( − 2 + ) + ( − 2 + ) 3 + 3 = −4 − 4 ( +  − 3) + ( +  − 3) ∙ ¸ ∙ ¸  − 2 +  3  − 2 +  3 = + −4  +  − 3 −4  +  − 3 µ∙ ¸¶ µ∙ ¸¶     =  +     

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

161

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.1

Property (2): µ ∙ ¸¶ µ∙ ¸¶       =      ∙ ¸ () − 2() + () 3() = −4() () + () − 3() ∙ ¸ ( − 2 + ) (3) = (−4) ( +  − 3) ∙ ¸  − 2 +  3 =  −4  +  − 3 µ∙ ¸¶   =     Also,  is a linear operator because it is a linear transformation whose domain equals its codomain. (f) The function  is not a linear transformation since Property (2) of the definition fails. For example, (83 ) = 22 , but 8(3 ) = 8(2 ). Note that Property (1) also fails, but once we have shown that one of the properties fails it is not necessary to show that the other fails in order to prove that  is not a linear transformation. Also,  is not a linear operator because it is not a linear transformation. (h) The function  is a linear transformation. To prove this, we show that it satisfies the two properties in the definition of a linear transformation.

Property (1): ¶ µ (3 + 2 +  + ) = (( + )3 + ( +  )2 + ( + ) + ( + ))  + (3 +  2 +  + ) = ( + ) + ( +  ) + ( + ) + ( + ) = ( +  +  + ) + ( +  +  + ) = (3 + 2 +  + ) + (3 +  2 +  + )

Property (2): ((3 + 2 +  + )) = = = =

(()3 + ()2 + () + ()) () + () + () + () ( +  +  + ) (3 + 2 +  + )

Finally,  is not a linear operator because its domain, P3 , does not equal its codomain, R.

(j) The function  is not a linear transformation because Property (1) of the definition fails to hold.

For example, (2 + ( + 1)) = (2 +  + 1) = 1, but (2 ) + ( + 1) = 0 + 0 = 0. Note that Property (2) also fails, but once we have shown that one of the properties fails it is not necessary to show that the other fails in order to prove that  is not a linear transformation. Also,  is not a linear operator because it is not a linear transformation. (k) The function  is a linear transformation. To prove this, we show that it satisfies the two properties in the definition of a linear transformation.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

162

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.1

Property (1): ⎛⎡

⎤ ⎡ ⎤⎞ ⎛⎡ ⎤⎞     + +  ⎝⎣   ⎦ + ⎣   ⎦⎠ =  ⎝⎣  +   +  ⎦⎠     +  +

= ( + )4 − ( + )2 + ( + ) = (4 − 2 + ) + (4 − 2 + ) ⎛⎡ ⎤⎞ ⎛⎡ ⎤⎞     =  ⎝⎣   ⎦⎠ +  ⎝⎣   ⎦⎠     

Property (2): ⎛ ⎡

⎤⎞ ⎛⎡ ⎤⎞      ⎝ ⎣   ⎦⎠ =  ⎝⎣   ⎦⎠    

= ()4 − ()2 + () = (4 − 2 + ) ⎛⎡ ⎤⎞   =  ⎝⎣   ⎦⎠   

(l) The function  is not a linear transformation √ because Property (1) of the definition fails to hold.

32 + 42 = 5, but ([3 0]) + ([0 4]) = For example, ([3 0] + [0 4]) = ([3 4]) = √ √ 2 2 2 2 3 + 0 + 0 + 4 = 3 + 4 = 7. Note that Property (2) also fails (try a negative scalar), but once we have shown that one of the properties fails it is not necessary to show that the other fails in order to prove that  is not a linear transformation. Also,  is not a linear operator because it is not a linear transformation.

(10) (c) Since the positive -axis rotates into the positive -axis here, we insert the 2 × 2 matrix from Exercise 9 in the third and first rows and columns to perform the rotation in the -plane, and keep the second row and second column the same as the identity matrix to indicate that the -axis ⎡ ⎤ sin  0 cos  ⎦.

1 0 stays fixed. Thus we get the matrix A = ⎣ 0 cos  0 − sin 

(26) We first need to express i and j as linear combinations of i + j and −2i + 3j. Hence, suppose ½ (i + j) + (−2i + 3j) = i  (i + j) + (−2i + 3j) = j

Setting the first and second coordinates of each side of these two equations equal to each other produces the linear systems ½ ½  − 2 = 1  − 2 = 0 and   + 3 = 0  + 3 = 1 Solving these systems yields  = 35 ,  = − 15 ,  = 25 , and  = 15 .

Therefore, i = 35 (i + j) − 15 (−2i + 3j), and so

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

163

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

¡3

¢

¡

¢

¡

Section 5.1

¢

1 3 1 5 (i + j) − 5 (−2i + 3j) =  5 (i + j) +  − 5 (−2i + 3j) = 35 (i + j) − 15 (−2i + 3j) = 35 (i − 3j) − 15 (−4i + 2j) = 75 i − 11 5 j 2 1 Similarly, j = 5 (i + j) + 5 (−2i + 3j), and so ¢ ¡ ¢ ¡ ¢ ¡ (j) =  25 (i + j) + 15 (−2i + 3j) =  25 (i + j) +  15 (−2i + 3j) = 25 (i + j) + 15 (−2i + 3j) = 25 (i − 3j) + 15 (−4i + 2j) = − 25 i − 45 j.

(i) = 

(29) Suppose part (3) of Theorem 5.1 is true for some . We prove it true for  + 1. (1 v1 + · · · +  v + +1 v+1 ) = (1 v1 + · · · +  v ) + (+1 v+1 ) (by property (1) of a linear transformation) = (1 (v1 ) + · · · +  (v )) + (+1 v+1 ) (by the inductive hypothesis) = 1 (v1 ) + · · · +  (v ) + +1 (v+1 ) (by property (2) of a linear transformation), and we are done.

(30) (b) The converse to the statement in part (a) is: If  = {v1      v } is a linearly independent set of vectors in V, then  = {(v1 )     (v )} is a linearly independent set of vectors in W. For a counterexample, consider  = {v1 }, where v1 is any nonzero vector in V. (For instance, we could have V = W = R2 , with v1 = e1 .) Then  is linearly independent by definition. Let  be the zero linear transformation defined by (v) = 0, for all v ∈ V. Then  = {0}, a linearly dependent subset of W. (31) (2 ◦ 1 )(v) = 2 (1 (v)) = 2 (1 (v)) = 2 (1 (v)) = (2 ◦ 1 )(v).

(33) First, we must show that −1 (W 0 ) is nonempty. Now, 0W ∈ W 0 , so 0V ∈ −1 ({0W }) ⊆ −1 (W 0 ).

Hence, −1 (W 0 ) is nonempty.

Next, if x y ∈ −1 (W 0 ) then we must show that x + y ∈ −1 (W 0 ). But (x) (y) ∈ W 0 by definition of −1 (W 0 ). Hence, (x) + (y) ∈ W 0 , since W 0 is closed under addition. Now  is a linear transformation, and so (x) + (y) = (x + y). Therefore, (x + y) ∈ W 0 , which implies that x + y ∈ −1 (W 0 ).

Finally, if x ∈ −1 (W 0 ) then we must show that x ∈ −1 (W 0 ) for all  ∈ R. Now, (x) ∈ W 0 .

Hence, (x) ∈ W 0 (for any  ∈ R), since W 0 is closed under scalar multiplication. This implies that (x) ∈ W 0 . Therefore, x ∈ −1 (W 0 ). Hence −1 (W 0 ) is a subspace by Theorem 4.2.

(36) (a) False. We are only given that property (2) of the definition of a linear transformation holds.

Property (1) is also required to hold in order for  to be a linear transformation. For a specific example of a function between vector spaces for which property (2) holds but property (1) fails, consider  : R2 → R defined by  ([ ]) =  whenever  6= 0 and  ([0 ]) = 0. To show that property (2) holds for  we consider 3 cases. First, if  6= 0 and  6= 0,  ([ ]) =  ([ ]) =  (since  6= 0) =  ([ ]). Next, if  = 0, then  ([ ]) =  ([0 0]) = 0 = 0 ([ ]). Finally, if  = 0, then  ([0 ]) =  ([0 ]) = 0 = (0) =  ([0 ]). Also, property (1) does not hold because  ([1 1]) +  ([−1 1]) = 1 + 1 = 2, but  ([1 1] + [−1 1]) =  ([0 2]) = 0.1 1 Note that property (1) alone also is insufficient to show that a function between real vector spaces is a linear transformation.

However, describing an appropriate counterexample is quite complicated and beyond the level of this course. What makes constructing such a counterexample difficult is that property (1) for a function  does imply that  (v) =  (v) for every rational number , but not for every real number.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

164

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.2

(b) True. This is proven in Example 4 in Section 5.1 of the textbook. (c) False.  is not a linear transformation, and hence, not a linear operator. Since (0) = [1 −2 3] 6= 0,  does not satisfy part (1) of Theorem 5.1, and so can not be a linear transformation. (One could also show, instead, that  satisfies neither of the two properties in the definition of a linear transformation.) (d) False. If A is a 4 × 3 matrix, then (v) = Av is a linear transformation from R3 to R4 , not from R4 to R3 . (e) True. This is part (1) of Theorem 5.1. (f) False. In order for the composition of functions 1 ◦ 2 to be defined, the range of 2 must be a subset of the domain of 1 . However, in this case, the range of 2 is a subspace of X , while the domain of 1 is V. For a specific counterexample, consider 1 : M23 → M32 given by 1 (A) = A , and 2 : M32 → R given by 2 (A) = 12 . Clearly here, 1 ◦ 2 is not defined.

(Note that, in general, the composition 2 ◦ 1 is a well-defined linear transformation, since the codomain of 1 and the domain of 2 both equal W.)

(g) True. This is part (1) of Theorem 5.3.

(h) True. Since {0W } is a subspace of W, this statement follows directly from part (2) of Theorem 5.3.

Section 5.2 (2) In each part, we will use Theorem 5.5, which states that the th column of A equals [(v )] , where  is an ordered basis for the domain of ,  is an ordered basis for the codomain of , and v is the th basis element in . As directed, we will assume here that  and  are the standard bases in each case. (a) Substituting each standard basis vector into the given formula for  shows that (e1 ) = [−6 −2 3], (e2 ) = [4 3 −1], and (e3 ) = [−1 −5 7]. Using each of these vectors as columns produces the ⎡ ⎤ −6 4 −1 3 −5 ⎦. desired matrix for : ⎣ −2 3 −1 7 (c) Substituting each standard basis vector into the given formula for  shows that (3 ) = [4 1 −2], (2 ) = [−1 3 −7], () = [3 −1 5] and (1) = [3 5 −1]. Using each of these vectors as columns ⎡ ⎤ 4 −1 3 3 3 −1 5 ⎦. produces the desired matrix for : ⎣ 1 −2 −7 5 −1

(3) In each part, we will use Theorem 5.5, which states that the th column of A equals [(v )] , where  is the given ordered basis for the domain of ,  is the given ordered basis for the codomain of , and v is the th basis element in . (a) Substituting each vector in  into the given formula for  shows that ([1 −3 2]) = [4 −7], ([−4 13 −3]) = [−1 25], and ([2 −3 20]) = [56 −24]. Next, we must express each of these answers in -coordinates. We use the Coordinatization Method from Section 4.7. Thus, for each vector v that we need to express in -coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors in , and whose column to the right of the

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

165

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.2

bar is the vector v. But since this process essentially involves solving a linear system, and we have three such systems to solve, we can solve all three simultaneously. Therefore, we row reduce ¯ ¯ ∙ ¸ ∙ ¸ −2 5 ¯¯ 4 −1 56 1 0 ¯¯ −47 128 −288 to  −1 3 ¯ −7 25 −24 0 1 ¯ −18 51 −104

Hence, [([1 −3 2])] = [−47 −18], [([−4 13 −3])] = [128 51], and [([2 −3 20])] = ∙ ¸ −47 128 −288 . [−288 −104]. Using these coordinatizations as columns produces A = −18 51 −104 (Note that A is just the matrix to the right of the bar in the row reduced matrix, above. This is typical, but in general, any rows of all zeroes must be eliminated first, as in the Coordinatization Method.)

(c) Substituting each vector in  into the given formula for  shows that ([5 3]) = 102 + 12 + 6 and ([3 2]) = 72 + 7 + 4. Next, we must express each of these answers in -coordinates. We use the Coordinatization Method from Section 4.7. To do this, we must first convert the above two polynomials, and the three polynomials in  into vectors in R3 by converting each polynomial of the form 2 +  +  to [  ]. Then, for each converted vector v that we need to express in -coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors converted from , and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have two such systems to solve, we can solve both systems simultaneously. Therefore, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 3 −2 1 ¯¯ 10 7 1 0 0 ¯¯ 22 14 ⎣ −2 2 −1 ¯¯ 12 7 ⎦ to ⎣ 0 1 0 ¯¯ 62 39 ⎦  0 −1 1 ¯ 6 4 0 0 1 ¯ 68 43 Hence, [([5 3])] = [22 62 ⎡ 68], and⎤[([3 2])] = [14 39 43]. Using these coordinatizations as 22 14 columns produces A = ⎣ 62 39 ⎦. (Note that A is just the matrix to the right of the bar 68 43 in the row reduced matrix, above. This is typical, but in general, any rows of all zeroes must be eliminated first, as in the Coordinatization Method.)

(e) Substituting each vector in  into the given formula for  shows that (−52 −  − 1) = ∙ ¸ ∙ ¸ ∙ ¸ 6 7 −19 0 5 −1 5 −3 −14 2 and (2 + 1) = . Next, , (−6 + 3 + 1) = −3 1 14 2 1 1 −6 −1 10

we must express each of these answers in -coordinates. We use the Coordinatization Method from Section 4.7. To do this, we must first convert the above matrices, and the six matrices in  into vectors in R6 by converting each matrix A to [11  12  13  21  22  23 ]. Then, for each converted vector v that we need to express in -coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors converted from , and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have 3 such systems to solve, we can solve all 3 systems simultaneously. Therefore, we row reduce

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

166

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

¯ ⎤ 1 0 0 0 0 0 ¯¯ 5 6 0 ⎢ 0 −1 1 0 0 0 ¯¯ −3 7 5 ⎥ ⎢ ⎥ ⎢ 0 ¯ 0 1 0 0 0 ¯ −14 −19 −1 ⎥ ⎢ ⎥ ⎢ 0 0 0 −1 0 0 ¯¯ −6 −3 2 ⎥ ⎢ ⎥ ⎣ 0 0 0 0 1 0 ¯¯ −1 1 1 ⎦ 0 0 0 0 1 1 ¯ 10 14 1 ⎡

⎡ to obtain

⎢ ⎢ ⎢ ⎢ I6 ⎢ ⎢ ⎣

Section 5.2

¯ ⎤ ¯ 5 6 0 ¯ ¯ −11 −26 −6 ⎥ ¯ ⎥ ¯ −14 −19 −1 ⎥ ¯ ⎥ ¯ 6 3 −2 ⎥ ¯ ⎥ ¯ −1 1 1 ⎦ ¯ ¯ 11 13 0

Hence, [(−52 −  − 1)] = [5 −11 −14 6 −1 11], [(−62 + 3 + 1)] = [6 −26 −19 3 1 13], and [(2+1)] = [0 −6 −1 −2 1 0]. Using these coordinatizations as columns shows that A is just the 6 × 3 matrix to the right of the bar in the reduced matrix, above. (This is typical, but in general, any rows of all zeroes must be eliminated first, as in the Coordinatization Method.)

(4) (a) First we find the matrix for  with respect to the standard basis  for R3 . Substituting each vector in  into the given formula for  shows that (e1 ) = [−2 0 1], (e2 ) = [1 −1 0], and ⎡ ⎤ −2 1 0 (e3 ) = [0 −1 3]. Using each of these vectors as columns produces A = ⎣ 0 −1 −1 ⎦.

1 0 3

Next, we use Theorem 5.6 to compute A . To do this, we need the inverse of the transition matrix P from  to  and the transition matrix Q from  to . Now P is the inverse of the matrix whose columns are the vectors in , and Q is the inverse of the matrix whose columns ⎡ ⎤ 15 2 3 are the vectors in  (see Exercise 8(a) in Section 4.7). Hence, P−1 = ⎣ −6 0 −1 ⎦ and 4 1 1 ⎡ ⎡ ⎤ ⎤ 1 0 2 1 −2 −6 ¢ ¡ −1 3 −2 ⎦ . Using row reduction yields Q = Q−1 Q−1 = ⎣ −3 = ⎣ 1 −1 −4 ⎦ .

1 −1 1 0 1 3 ⎡ ⎤ −202 −32 −43 Finally, using Theorem 5.6 produces A = QA P−1 = ⎣ −146 −23 −31 ⎦.

83 14 18

(b) First we find the matrix for  with respect to the standard basis  for M22 and the standard basis  for R2 . Recall that the standard basis  for M22 is {Ψ | 1 ≤  ≤ 2 1 ≤  ≤ 2}, where Ψ is the 2 × 2 matrix having 1 as its ( ) entry and zeroes elsewhere. (See Example 3 in Section 4.5.) Substituting each vector in  into the given formula for  shows that (Ψ11 ) = [6 −2], (Ψ12 ) = [−1 3], (Ψ21 ) = [3 −1] and (Ψ22 ) = [−2 4]. Using each of these vectors as columns ∙ ¸ 6 −1 3 −2 . produces A = −2 3 −1 4

Next, we use Theorem 5.6 to compute A . To do this, we need the inverse of the transition matrix P from  to  and the transition matrix Q from  to . To compute P (or P−1 ), we need to convert the matrices in  into vectors in R4 . This is done by converting each 2 × 2 matrix A to [11  12  21  22 ]. After doing this, P is the inverse of the matrix whose columns are the resulting vectors obtained from , and Q is the inverse of the matrix whose columns are the vectors in ⎡ ⎤ 2 0 1 1 ∙ ¸ ⎢ 1 2 1 1 ⎥ −2 −1 −1 −1 ⎢ ⎥ .  (see Exercise 8(a) in Section 4.7). Hence, P = ⎣ and Q = 0 1 2 1 ⎦ 5 2 1 1 1 1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

167

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

¢−1 ¡ Using Theorem 2.14 yields Q = Q−1 =

∙

2 1 −5 −2 ∙ ¸ 21 7 21 16 −1 A = QA P = . −51 −13 −51 −38

¸

Section 5.2

. Finally, using Theorem 5.6 produces

(6) (a) First we use the method of Theorem 5.5. Substituting each vector in  into the given formula for  shows that ([4 −1]) = [9 7] and ([−7 2]) = [−16 −13]. Next, we must express each of these answers in -coordinates. We use the Coordinatization Method from Section 4.7. Thus, for each vector v that we need to express in -coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors in , and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have 2 such systems to solve, we can solve both systems simultaneously. Therefore, we row reduce ¯ ¯ ∙ ¸ ∙ ¸ 4 −7 ¯¯ 9 −16 1 0 ¯¯ 67 −123 yielding  −1 2 ¯ 7 −13 0 1 ¯ 37 −68

Hence, [([4 −1])] = [67 37] and [([−7 2])] = [−123 −68]. Using these coordinatizations as ∙ ¸ 67 −123 . columns produces A = 37 −68 Next, we use the method of Theorem 5.6. We begin by finding the matrix for  with respect to the standard basis  for R2 . Substituting each vector in  into the given formula for  shows that (i) = [2 1] and (j) = [−1 −3]. Using each of these vectors as columns produces ∙ ¸ 2 −1 . Next, we use Theorem 5.6 to compute A . To do this, we need the A = 1 −3 transition matrix P from  to . Now P is the inverse of the matrix whose columns are the ∙ ¸ 4 −7 . By Theorem 2.14, vectors in  (see Exercise 8(a) in Section 4.7). Hence, P−1 = −1 2 ∙ ¸ ¢−1 ¡ 2 7 P = P−1 = . Finally, using Theorem 5.6 produces A = PA P−1 = 1 4 ∙ ¸ 67 −123 .

37 −68

(b) First we use the method of Theorem 5.5. Substituting each vector in  into the given formula for  shows that (22 + 2 − 1) = 42 + 3 + 1, () = 2 − 1, and (−32 − 2 + 1) = −42 − 5 − 2.

Next, we must express each of these answers in -coordinates. We use the Coordinatization Method from Section 4.7. To do this, we must first convert the above polynomials, and the three polynomials in  into vectors in R3 by converting each polynomial of the form 2 +  +  to [  ]. Then, for each converted vector v that we need to express in -coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors obtained from , and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have 3 such systems to solve, we can solve all 3 systems simultaneously. Therefore, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 2 0 −3 ¯¯ 4 1 −4 1 0 0 ¯¯ −7 2 10 ⎣ 2 1 −2 ¯ 3 0 −5 ⎦ to ⎣ 0 1 0 ¯¯ 5 −2 −9 ⎦  ¯ −1 0 1 ¯ 1 −1 −2 0 0 1 ¯ −6 1 8 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

168

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.2

Hence, [(22 +2−1)] = [−7 5 −6], [()] = [2 −2 1], and [(−32 −2+1)] = [10 −9 8]. ⎡ ⎤ −7 2 10 Using these coordinatizations as columns produces A = ⎣ 5 −2 −9 ⎦. −6 1 8

Next, we use the method of Theorem 5.6. We begin by finding the matrix for  with respect to the standard basis  for P2 . Substituting each vector in  into the given formula for  shows that (2 ) = 2 + 1, () = 2 − 1, and (1) = −22 +  − 1. We convert these results to vectors ⎡ ⎤ 0 1 −2 0 1 ⎦. in R3 . Then, using each of these vectors as columns produces A = ⎣ 2 1 −1 −1

Next, we use Theorem 5.6 to compute A . To do this, we need the transition matrix P from  to . Now P is the inverse of the matrix whose columns are the polynomials in , converted ⎡ ⎤ 2 0 −3 into vectors in R3 (see Exercise 8(a) in Section 4.7). Hence, P−1 = ⎣ 2 1 −2 ⎦ . Using −1 0 1 ⎡ ⎤ −1 0 −3 ¡ ¢−1 2 ⎦. Finally, using Theorem 5.6 produces row reduction yields P = P−1 = ⎣ 0 1 ⎡ ⎤ −1 0 −2 −7 2 10 A = PA P−1 = ⎣ 5 −2 −9 ⎦. −6 1 8

(7) (a) Computing the derivative of each polynomial in the standard basis  for P3 shows that (3 ) = 32 , (2 ) = 2, () = 1, and (1) = 0. We convert these resulting polynomials in P2 to vectors in R3 . Then, using each of these vectors as columns produces ⎡ ⎤ 3 0 0 0 A = ⎣ 0 2 0 0 ⎦, where  is the standard basis for P2 . Now, by Theorem 5.5, 0 0 1 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 4 3 0 0 0 ⎢ 12 ⎥ −5 ⎥ ⎣ ⎦ [(43 − 52 + 6 − 7)] = A [43 − 52 + 6 − 7] = ⎣ 0 2 0 0 ⎦ ⎢ ⎣ 6 ⎦ = −10 .

0 0 1 0 6 −7 Converting back from -coordinates to polynomials yields (43 − 52 + 6 − 7)0 = 122 − 10 + 6.

(8) (a) To do this, we need to calculate (i) and (j). Rotating the vector i counterclockwise through an √ angle of 6 produces the vector [ 23  12 ]. Rotating j counterclockwise through an angle of 6 yields √ [− 12  23 ]. By Theorem 5.5, the matrix for  with respect to the standard basis is the matrix " √ # 3 − 12 2 √ whose columns are (i) and (j); that is, .

3 1 2

2

(9) (b) First we need to compute (b ) for each matrix b ∈ . Since  is just the transpose transformation, the results of these computations are, respectively, ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 0 0 0 0 −1 0 0 0 0 ⎣ 0 0 ⎦⎣ 1 0 ⎦⎣ 1 0 ⎦⎣ 0 0 ⎦  ⎣ 0 −1 ⎦  and ⎣ 0 0 ⎦  0 0 −1 0 0 0 0 0 0 −1 0 1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

169

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.2

Next, we must express each of these answers in -coordinates. We use the Coordinatization Method from Section 4.7. To do this, we must first convert the above matrices, and the 6 matrices in  into vectors in R6 by changing each 3 × 2 matrix A to the vector [11  12  21  22  31  32 ].

Then, for each resulting vector v that we need to express in -coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors converted from , and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have 6 such systems to solve, we can solve all 6 systems simultaneously. Therefore, we row reduce ¯ ⎡ ⎤ 1 1 0 0 0 0 ¯¯ 1 0 0 0 0 0 ⎢ 1 −1 0 0 0 0 ¯¯ 0 0 0 −1 0 0 ⎥ ⎢ ⎥ ⎢ 0 ¯ 0 1 1 0 0 ¯ 0 1 1 0 0 0 ⎥ ⎢ ⎥ ⎢ 0 0 1 −1 0 0 ¯¯ 0 0 0 0 −1 0 ⎥ ⎢ ⎥ ⎣ 0 0 0 0 1 1 ¯¯ 0 −1 0 0 0 0 ⎦ 0 0 0 0 1 −1 ¯ 0 0 0 0 −1 1

to obtain

⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ I6 ⎢ ⎢ ⎢ ⎣

¯ 1 ¯ 0 ¯ 2 ¯ 1 0 ¯ 2 ¯ 1 ¯ 0 ¯ 2 ¯ 1 ¯ 0 2 ¯ ¯ ¯ 0 − 12 ¯ ¯ 0 −1 2

0 − 12

0

0

1 2

1 2 1 2

0 − 12

0

0

0

0

0

0 1 2 − 12 1 2

0

⎤

⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ ⎥ 0 ⎥ ⎥ 1 ⎥ 2 ⎦

− 12

The columns of the matrix to the right of the bar gives the -coordinatizations of the images of the 6 matrices in  under . Hence, that 6 × 6 matrix is A .

(10) According to Theorem 5.7, the matrix A for the composition 2 ◦1 is the matrix product A A . ⎡ ⎤ ⎡ ⎤ ¸ −12 12 −2 4 −1 ∙ −2 3 −1 6 −2 ⎦.

0 ⎦ = ⎣ −4 Hence, A = ⎣ 2 4 0 −2 −10 −3 7 −1 −3

(13) In each part, we use the fact that, by Theorem 5.5, the th column of A is the -coordinatization of (v ). (a) For each , (v ) = v . But [v ] = e . (See the comment just after Example 4 in Section 4.7.)

Hence, the th column of A is e , and so A = I . (c) For each , (v ) = v . But [v ] = [v ] (by part (2) of Theorem 4.16) = e (see the comment just after Example 4 in Section 4.7). Hence, the th column of A is e , and so A = I . (e) Now (v1 ) = v , and for each   1, (v ) = v−1 . But [v ] = e . (See the comment just after Example 4 in Section 4.7.) Hence, the 1st column of A is e , and the th column of A for   1 is e−1 . Thus, A is the  ×  matrix whose columns are e , e1 , e2 ,   , e−1 , respectively.

(15) First, note that [(2 ◦1 )(v)] = [2 (1 (v))] = A [1 (v)] = A (A [v] ) = (A A )[v] .

Therefore, by the uniqueness condition in Theorem 5.5, A A is the matrix for 2 ◦1 with respect to  and .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

170

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ¯ ¯  − 89 ¯ ¯ (18) (a) A () = |I3 −A | = ¯ − 29 ¯ ¯ −2 9 and algebraic simplification.

− 29

 − 59 4 9

Section 5.2

¯ ¯ ¯ ¯ 4 ¯ = (−1)2 , where we have used basketweaving 9 ¯ − 5 ¯ − 29

9

(b) For 1 = 1, we row reduce [I3 − A | 0] = ¯ ⎡ 1 ⎤ − 29 − 29 ¯¯ 0 9 ⎢ 2 ⎥ 4 4 ¯ to obtain ⎣ −9 9 9 ¯¯ 0 ⎦ 2 4 4 ¯ 0 −9 9 9

¯ ⎤ 1 −2 −2 ¯¯ 0 ⎣ 0 0 0 ¯¯ 0 ⎦  0 0 0 ¯ 0 ⎡

There are two independent variables, and hence Step 3 of the Diagonalization Method of Section 3.4 produces two fundamental eigenvectors for 1 . Following that method, the basis for 1 = {[2 1 0] [2 0 1]}.

For 2 = 0, we row reduce [0I3 − A | 0] = [−A | 0] = ¯ ⎡ 8 ⎤ ⎡ ⎤ 1 ¯¯ − 9 − 29 − 29 ¯¯ 0 1 0 2 ¯ 0 ⎢ 2 ⎥ 4 ¯ to obtain ⎣ 0 1 −1 ¯¯ 0 ⎦  ⎣ − 9 − 59 9 ¯¯ 0 ⎦ 4 0 0 0 ¯ 0 −2 −5 ¯ 0 9

9

9

There is one independent variable. Setting it equal to 2 to eliminate fractions produces one fundamental eigenvector for 2 . Following the Diagonalization Method, the basis for 0 = {[−1 2 2]}.

Combining these bases for 1 and 0 yields  = ([2 1 0] [2 0 1] [−1 2 2]) (Note: the answer to part (c) will vary if the vectors in  are ordered differently). (c) The transition matrix P from  to  is the matrix whose columns are the vectors in  (see ⎡ ⎤ 2 2 −1 2 ⎦.

Exercise 8(b) in Section 4.7). Hence, P = ⎣ 1 0 0 1 2

(23) Let : V −→ W be a linear transformation such that (v1 ) = w1  (v2 ) = w2    , (v ) = w  with {v1  v2      v } a basis for V. If v ∈ V, then v = 1 v1 + 2 v2 + · · · +  v  for unique 1  2       ∈ R (by Theorem 4.9). But then (v) = (1 v1 + 2 v2 + · · · +  v ) = 1 (v1 ) + 2 (v2 ) + · · · +  (v ) = 1 w1 + 2 w2 + · · · +  w  Hence, for each v ∈ V, (v) is completely determined by the given values for w1      w and the unique values of 1       . Therefore,  is uniquely determined.

(24) (a) True. This is shown in Theorem 5.4. (b) True. This follows from Theorem 5.4. (c) False. Bases for the domain and codomain must be chosen. Different choices of bases typically produce different matrices for . For instance, Examples 3 and 4 in Section 5.2 of the textbook show two different matrices for the same linear transformation. (d) False. This statement has the correct equation scrambled. The correct equation from Theorem 5.5 states that A [v] = [(v)] . For a specific counterexample to the scrambled equation in the problem statement, consider : R2 → R given by ([ ]) =  and let  = (i j),  = (1) and v = [1 2]. Then A = [1 0], a 1 × 2 matrix, [v] = [1 2], and [(v)] = [1] = [1], a 1-vector.

Note that A [(v)] is not even defined, because the matrix and vector are of incompatible sizes. Hence, the scrambled equation can not be correct. (e) True. This is part of the statement of Theorem 5.5. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

171

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.3

(f) False. The given matrix is the matrix (in standard coordinates) for the projection onto the ⎡ ⎤ 1 0 0 plane. The matrix for the projection  onto the -plane (in standard coordinates) is ⎣ 0 0 0 ⎦, 0 0 1 since (i) = i, (j) = 0, and (k) = k. (g) True. To see this, take the equation A = QA P−1 from Theorem 5.6, and multiply both sides on the right by P. (h) True. This is shown just before Example 6 in Section 5.2. (i) True. This follows from Exercise 6 in Section 3.4. (j) False. According to Theorem 5.7, the matrix for 2 ◦ 1 with respect to the standard basis is ∙ ¸∙ ¸ ∙ ¸ ∙ ¸∙ ¸ ∙ ¸ 0 1 1 2 3 4 1 2 0 1 2 1 = , not = .

1 0 3 4 1 2 3 4 1 0 4 3

Section 5.3 (1) In each part, let A represent the given 3 × 3 matrix such that (v) = Av. (a) Yes, because ([1 −2 3]) = [0 0 0].

(c) No. If v = [2 −1 4] were in range(), then there would exist a vector x = [1  2  3 ] such that Ax = v. This is equivalent to the system ⎧ 2 ⎨ 51 + 2 − 3 = + 3 = −1  −31 ⎩ 4 1 − 2 − 3 = which has no solutions, since

¯ ⎤ 5 1 −1 ¯¯ 2 ⎣ −3 0 1 ¯¯ −1 ⎦ 1 −1 −1 ¯ 4 ⎡

row reduces to

¯ ⎤ 1 0 − 13 ¯¯ 13 ⎢ 2 ¯ 1 ⎥ ⎣ 0 1 3 ¯¯ 3 ⎦  0 0 0 ¯ 4 ⎡

Note: in the above, we stopped the row reduction process at the augmentation bar. If you use a calculator to perform the row reduction, it may continue row reducing into the column beyond the bar, making that column equal e3 .

(2) (a) No, since (3 − 52 + 3 − 6) = 63 + 4 − 9 6= 0.

(c) Yes. To show this, we need to find p() = 3 + 2 +  +  such that (p) = 83 −  − 1.

Equating the coefficients of (p) to those of 83 −  − 1 gives the system ⎧ 2 = 8 ← 3 terms ⎨ − −  = −1 ←  terms ⎩ − +  = −1 ← constant terms One solution is  = 1,  = 0,  = 4,  = 3. And so, (3 + 4 + 3) = 83 −  − 1, showing that 83 −  − 1 ∈ range().

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

172

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.3

(3) In each part, let A represent the given matrix such that (v) = Av. ⎡ ⎤ 1 0 2 (a) The reduced row echelon form of A is B = ⎣ 0 1 −3 ⎦. Then, by the Kernel Method, a basis 0 0 0 for ker() is found by considering the solution to the system BX = 0 computed by setting the one independent variable in the system equal to 1. This yields the basis {[−2 3 1]} for ker().

Since ker() has a basis with 1 element, dim(ker()) = 1. Next, by the Range Method, a basis for range() consists of the first 2 columns of A, since those columns in  contain nonzero pivots. Thus a basis for range() = {[1 −2 3] [−1 3 −3]} and dim(range()) = 2. Note that dim(ker()) + dim(range()) = 1 + 2 = 3 = dim(R3 ). ⎡ ⎤ 1 0 −1 1 ⎢ 0 1 3 −2 ⎥ ⎢ ⎥ ⎢ 0 0 ⎥ (d) The reduced row echelon form of A is B = ⎢ 0 0 ⎥. Then, by the Kernel Method, ⎣ 0 0 0 0 ⎦ 0 0 0 0 a basis for ker() is found by considering certain particular solutions to the system BX = 0.

We note that 3 and 4 are independent variables in the system. Letting 3 = 1, 4 = 0 produces the vector [1 −3 1 0]. Using 3 = 0, 4 = 1 gives us [−1 2 0 1]. This yields the basis {[1 −3 1 0] [−1 2 0 1]} for ker(). Since ker() has a basis with 2 elements, dim(ker()) = 2. Next, by the Range Method, a basis for range() consists of the first 2 columns of A, since those columns in  contain nonzero pivots. Thus a basis for range() = {[−14 −4 −6 3 4] [−8 −1 2 −7 2]} and dim(range()) = 2. Note that dim(ker()) + dim(range()) = 2 + 2 = 4 = dim(R4 ).

(4) We will solve each part by inspection (mostly). (a) Now, ker() is the set of vectors sent to 0 by . But, ([1  2  3 ]) = [0 2 ] = [0 0] precisely when 2 = 0, and so ker() = {[1  0 3 ] | 1  3 ∈ R} = {[ 0 ] |   ∈ R}, which has basis {[1 0 0], [0 0 1]}. Thus, dim(ker()) = 2. Also, range() = {[0 2 ] | 2 ∈ R}, the set of images obtained using , which can also be expressed as {[0 ] |  ∈ R}. Hence, a basis for range() = {[0 1]} and dim(range()) = 1. Finally, note that dim(ker()) + dim(range()) = 2 + 1 = 3 = dim(R3 ). (d) Now, ker() is the set of vectors sent to 0 by . But (4 +3 +2 ++) = 2 ++ = 0 precisely when  =  =  = 0, and so ker() = {4 + 3 |   ∈ R}, which has basis {4  3 }.

Thus, dim(ker()) = 2. Also, range() = {2 +  +  |    ∈ R}, the set of images obtained using . This is precisely P2 . Hence, a basis for range() = {2   1} and dim(range()) = 3.

Finally, note that dim(ker()) + dim(range()) = 2 + 3 = 5 = dim(P4 ). (f) Now, ker() is the set of vectors sent to 0 by . But, ([1  2  3 ]) = [1  0 1 −2 +3 ] = [0 0 0] precisely when 1 = 0 and 1 − 2 + 3 = 0. That is when 1 = 0 and 2 = 3 . Hence ker() = {[0 2  2 ] | 2 ∈ R}, which can also be expressed as {[0  ] |  ∈ R}. Thus, ker() has basis {[0 1 1]}, and so dim(ker()) = 1. Also, range() = {[1  0 1 − 2 + 3 ] | 1  2  3 ∈ R}, the set of images obtained using . Hence, range() = {1 [1 0 1] + 2 [0 0 −1] + 3 [0 0 1] | 1  2  3 ∈ R} and so {[1 0 1] [0 0 −1] [0 0 1]} spans range(). However, this set is not linearly independent, since [0 0 1] = (−1)[0 0 −1]. So, if we eliminate the last vector, we get the basis {[1 0 1] [0 0 −1]} c 2016 Elsevier Ltd. All rights reserved.

Copyright °

173

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.3

for range(). (A simpler basis for range() can be found using the Simplified Span Method, yielding {[1 0 0] [0 0 1]}.) And so, dim(range()) = 2. Finally, note that dim(ker())+dim(range()) = 1 + 2 = 3 = dim(R3 ). (g) Now, ker() is the set of vectors sent to O22 by . But (A) = A = O22 precisely when A = O22 . Hence, ker() = {O22 }, having basis { } (the empty set), and dim(ker()) = 0. Also, every 2 × 2 matrix is the image of its transpose (since (A ) = A), proving range() = M22 .

Therefore, dim(range()) = 4, and a basis for range() = standard basis for M22 . Notice that dim(ker()) + dim(range()) = 0 + 4 = 4 = dim(M22 ). (i) Now, ker() is the set of vectors sent to 0 by . Note that (p) = [p(1) p0 (1)] = [0 0] precisely when p(1) = 0 and p0 (1) = 0. But p(1) = 0 implies that ( − 1) is a factor of p. Thus, p = (−1)(+). Next, p0 (1) = 0 implies ((1) +)(1) +((1) −1)() = 0 (using the product rule and substituting  = 1), which gives  = −. Thus, p = (−1)(−) = (2 −2+1). Therefore, a basis for ker() is {2 −2+ 1}, and dim(ker()) = 1. A spanning set for range() can be found by computing the images of a basis for P2 . Now (2 ) = [1 2], () = [1 1], and (1) = [1 0].

The first two of these form a linearly independent set, and the fact that dim(R2 ) = 2 shows that the third vector is redundant. Hence, a basis for range() is {[1 2] [1 1]} and dim(range()) = 2.

Since dim(range()) = dim(R2 ), range() = R2 , and so a simpler basis for range() = standard basis for R2 . Notice that dim(ker()) + dim(range()) = 1 + 2 = 3 = dim(P2 ).

(6) First we show that  is a linear transformation by verifying the two properties.

P3 Property (1): (A + B) = trace(A + B) = =1 ( +  ) P3 P3 = =1  + =1  = trace(A) + trace(B) = (A) + (B).

P P Property (2): (A) = trace(A) = 3=1  =  3=1  = (trace(A)) = (A).

Next, ker() is the set of all 3 × 3 matrices whose trace is 0. Thus, ⎧⎡ ⎫ ⎤¯ ¯  ⎨   ⎬ ¯ ⎦ ¯         ∈ R   ker() = ⎣   ¯ ⎩ ⎭   − −  ¯

A detailed computation for a basis for ker() is given in the solution to Exercise 13(b) in Section 4.6, found earlier in this manual. The basis found there has eight elements, and so dim(ker()) = 8. Also, it is possible to obtain any real number  as the trace of some 3×3 matrix since trace( 3 I3 ) = . Hence, range() = R, and so dim(range()) = 1.

(8) First, ker() is the set of polynomials p such that 2 p() = 0 for all . But since 2 is 0 only when  = 0, p() must be 0 for all other values of . However, no nonzero polynomial can have an infinite number of roots. Hence, p = 0. Thus, ker() = {0}, and so dim(ker()) = 0. Next, the images under  are precisely those having 2 as a factor. Thus, range() = {4 + 3 + 2 |    ∈ R}. Clearly, a basis for range() is {4  3  2 }, and so dim(range()) = 3.

(10) First, let us consider the case  ≤ . Taking the th derivative of a polynomial reduces its degree by , and so the polynomials of degree less than  are precisely those that have th derivative equal to 0.

Hence, when  ≤ , ker() = {all polynomials of degree less than } = P−1 . Therefore, dim(ker()) = . Also, computing a th antiderivative for a polynomial p of degree ≤ ( − ) will yield a pre-image under  for p. Hence, range() = P− , and dim(range()) =  −  + 1.

When   ,  sends all of P to 0, and so ker() = P , dim(ker()) =  + 1, range() = {0}, and dim(range()) = 0.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

174

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.3

(12) First, ker() = {0} because (X) = 0 =⇒ AX = 0 =⇒ A−1 AX = A−1 0 =⇒ X = 0. Similarly, every vector X is in the range of  since (A−1 X) = A(A−1 X) = X. Thus, range() = R . µ∙ ¸¶ ∙ ¸∙ ¸ ∙ ¸ ∙ ¸  1 −1  1 −1 1 −1 (16) Consider  = . Since row reduces to , the Kernel  1 −1  1 −1 0 0 Method and the Range Method show that ker() = range() = {[ ] |  ∈ R}.

(17) Let , A, , , and B be as given in the statement of the problem. (a) Let P be the transition matrix from the standard basis to , and let Q be the transition matrix from the standard basis to . Note that both P and Q are nonsingular by Theorem 4.19. Then,

by Theorem 5.6, B = QAP−1 . Hence, rank(B) = rank(QAP−1 ) = rank(AP−1 ) (by part (a) of Review Exercise 16 in Chapter 2) = rank(A) (by part (b) of Review Exercise 16 in Chapter 2). (b) The comments prior to Theorem 5.9 in the text show that parts (a) and (b) of the theorem are true if A is the matrix for  with respect to the standard bases. However, part (a) of this exercise shows that if B is the matrix for  with respect to any bases, rank(B) = rank(A). Hence, rank(B) can be substituted for any occurrence of rank(A) in the statement of Theorem 5.9. Thus, parts (a) and (b) of Theorem 5.9 hold when the matrix for  with respect to any bases is used. Part (c) of Theorem 5.9 follows immediately from parts (a) and (b).

(20) (a) False. By definition, the set {(v) | v ∈ V} is range(), not ker(). These sets are usually unequal.

For specific counterexamples, consider Examples 1, 2, and 3 in Section 5.3 of the textbook. (b) False. Range(), in general, is a subspace of the codomain W, not of the domain V. A linear transformation such as that given in Example 4 in Section 5.3 provides a specific counterexample. (c) True. This is a direct consequence of the Dimension Theorem. (d) False. The Dimension Theorem relates the dimensions of ker() and the domain V of  with dim(range()), not with the dimension of the codomain W. For a specific counterexample, consider : R5 → R3 given by (v) = 0 for all v ∈ R5 . Then ker() = R5 , and so dim(ker()) = 5 6= 2. (e) True. By part (2) of Theorem 5.9, dim(ker()) =  − rank(A). Now, rank(A) is the number of nonzero rows in the reduced row echelon form for A. Hence, it gives the number of pivots obtained from A, which is the number of pivot columns for A. But since A has  columns in all, it must have  − rank(A) nonpivot columns. (f) False. By part (1) of Theorem 5.9, dim(range()) = rank(A). See Example 5 in Section 5.3 for a specific counterexample. In that example, rank(A) = 3, and dim(range()) = 3 6= 5 − rank(A).

(g) False. By Corollary 5.11, rank(A) = rank(A ). Hence, if rank(A) = 2, then rank(A ) = 2 as well. (h) False. Even though Corollary 5.11 assures us that dim(row space of A) = dim(column space of A), the subspaces themselves are not necessarily the same. In fact, unless A is square, they are not even subspaces of the same vector space. See the comments after Example 7 in Section 5.3, which indicate that counterexamples to the statement are provided by Examples 4, 5 and 7 in Section 5.3.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

175

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.4

Section 5.4 (1) In each part, we will check whether  is one-to-one using part (1) of Theorem 5.12, which states that  is one-to-one if and only if ker() = {0}. (a)  is not one-to-one because ([1 0 0]) = [0 0 0 0]. Thus, [1 0 0] ∈ ker(), and so ker() 6= {0}.

Also,  is not onto because [0 0 0 1] is not in range(). This is because all vectors in range() have a zero in their fourth coordinate. (c)  is one-to-one because ([  ]) = [0 0 0] implies that [2  +  +  −] = [0 0 0], which is equivalent to the system ⎧ = 0 ⎨ 2  +  +  = 0  ⎩ − = 0

A quick inspection shows that  =  =  = 0 is the only solution, and hence ker() = {0}.

Next,  is onto, because every vector [  ] can be expressed as [2  +  +  −], where  = 2 ,  = −, and  =  − 2 + .

(e)  is one-to-one because (2 +  + ) = 0 implies that  +  =  +  =  +  = 0. Thus,  +  =  + , and so  = . Using this and  +  = 0 gives  =  = 0. Then  +  = 0 implies  = 0. Hence, ker() = {0}. Next,  is onto because every polynomial 2 +  +  can be expressed as ( + )2 + ( + ) + ( + ), where  = ( −  + )2,  = ( +  − )2, and  = (− +  + )2. µ∙ ¸¶ ∙ ¸ ∙ ¸ 0 1 0 0 0 0 1 0 (g)  is not one-to-one because  = . Hence, ∈ ker(), 1 0 −1 0 0 1 0 −1 ∙ ¸   can be and so ker() 6= {O23 }. However,  is onto because every 2 × 2 matrix   ∙ ¸  − expressed as , where  = ,  = −,  = 2,  = , and  = 0.

2  +  ∙ ¸ 0 0 (h)  is one-to-one because (2 +  + ) = implies that  +  =  −  = −3 = 0, which 0 0 ∙ ¸ 0 1 gives  =  =  = 0. Hence ker() = {0}. Next,  is not onto, because is not in 0 0 range(), since every matrix in range() has a zero for its (1 2) entry.

(2) In each part, let A represent the given matrix such that (v) = Av. (a) The matrix A row reduces to I2 . Hence, rank(A) = 2. Therefore, Theorem 5.9 implies that dim(ker()) = 0 and dim(range()) = 2. Since dim(ker()) = 0,  is one-to-one. Also, dim(range()) = 2 = dim(R2 ) implies that  is onto. ⎤ ⎡ 1 0 − 25 ⎥ ⎢ (c) The matrix A row reduces to ⎣ 0 1 − 65 ⎦. Hence, rank(A) = 2. Therefore, Theorem 5.9 0 0

0

implies that dim(ker()) = 1 and dim(range()) = 2. Since dim(ker()) 6= 0,  is not one-to-one.

Also, dim(range()) = 2  3 = dim(R3 ) implies that  is not onto.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

176

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.4

(3) In each part, let A represent the given matrix for . (a) The matrix A row reduces to I3 . Hence, rank(A) = 3. Therefore, Theorem 5.9 implies that dim(ker()) = 0 and dim(range()) = 3. Since dim(ker()) = 0,  is one-to-one. Also, dim(range()) = 3 = dim(P2 ) implies that  is onto. ⎡ ⎤ 19 1 0 − 10 11 11 ⎢ 3 9 ⎥ ⎢ 0 1 ⎥ − 11 11 ⎢ ⎥. Hence, rank(A) = 2. Therefore, Theorem (c) The matrix A row reduces to ⎢ 0 0 ⎥ ⎣ 0 0 ⎦ 0 0 0 0 5.9 implies that dim(ker()) = 2 and dim(range()) = 2. Since dim(ker()) 6= 0,  is not one-to-one. Also, dim(range()) = 2  4 = dim(P3 ) implies that  is not onto.

(7) (a) Because dim(R6 ) = dim(P5 ), Corollary 5.13 tells us that if  were one-to-one,  would also be onto. But since  is not onto,  is also not one-to-one. (b) Because dim(M22 ) = dim(P3 ), Corollary 5.13 tells us that if  were onto,  would also be one-to-one. But since  is not one-to-one,  is also not onto.

(12) (a) False. The second half of the given statement is the converse of what it means for a function to be one-to-one. That is,  is one-to one if and only if (v1 ) = (v2 ) implies v1 = v2 . Any linear transformation that is not one-to-one, such as Example 2 in Section 5.4 of the textbook, provides a specific counterexample to the given statement. Note that the statement “v1 = v2 implies (v1 ) = (v2 )” is actually true for every function, not just one-to-one functions. (b) False. The correct definition of what it means for : V → W to be onto is that for every w ∈ W there is a v ∈ V such that (v) = w. The statement “for all v ∈ V, there is some w ∈ W such that (v) = w” is true for every function : V → W, not just those that are onto. Thus, any linear transformation that is not onto provides a counterexample. One such function is described just after Example 2 in Section 5.4 of the textbook. (c) True. This is one direction of the “if and only if” statement in part (1) of Theorem 5.12. (d) True. This follows from part (2) of Theorem 5.14. (See the discussion directly after the proof of Theorem 5.14 in the textbook.) Also, you are asked to prove this in part (b) of Exercise 11 in Section 5.4. (e) True. This is the contrapositive of one half of the if and only if statement in part (1) of Theorem 5.12. (f) True. This is the contrapositive of one half of the if and only if statement in part (2) of Theorem 5.12. (g) True. This follows from the contrapositive of one half of Theorem 5.13. (h) False. For a specific counterexample, consider the linear transformation : R3 → R3 given by (v) = 0 for all v ∈ R3 . The linearly independent set  = {i j} has image ( ) = {0}, which is linearly dependent. (i) False. For a specific counterexample, consider the linear transformation : R3 → R2 given by ([  ]) = [ ]. Let  = {[1 0 0] [0 1 0]}, which does not span R3 . However, () = {[1 0] [0 1]}, which does span R2 .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

177

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.5

Section 5.5 (1) In each part, let A represent the given matrix for 1 and let B represent the given matrix for 2 . (a) (i) By Theorem 5.16, 1 and 2 are isomorphisms if and only if A and B are nonsingular. But cofactor expansions of each along the third column shows that |A| = 1 and |B| = 3. Since neither of these are zero, both matrices are nonsingular. −1 −1 v and −1 v, where (ii) By Theorem 5.16, −1 1 (v) = A 2 (v) = B ⎤ ⎡ ⎤ ⎡ 0 0 1 1 0 0 A−1 = ⎣ 0 −1 0 ⎦ and B−1 = ⎣ 0 0 − 13 ⎦ 2 1 0 1 −2 0

are obtained by row reduction.

(iii) By Theorem 5.7, (2 ◦ 1 )(v) = (BA)v. Computing BA produces ⎛⎡ ⎤⎞ ⎡ ⎤⎡ ⎤ 1 0 −2 1 1 4 −2 ⎦ ⎣ 2 ⎦  (2 ◦ 1 ) ⎝⎣ 2 ⎦⎠ = ⎣ 1 3 3 0 3 0

(iv) By Theorem 5.16, (2 ◦ 1 )−1 (v) = (BA)−1 (v) Using row reduction to compute the inverse of the matrix we found in part (iii) gives ⎛⎡ ⎤⎞ ⎡ 2 1 0 ⎤ ⎡ ⎤ 1 1 ⎥ ⎢ (2 ◦ 1 )−1 ⎝⎣ 2 ⎦⎠ = ⎣ 0 0 13 ⎦ ⎣ 2 ⎦  3 3 1 0 2 3

−1 −1 −1 B )v. Using the matrices A−1 and B−1 we (v) By Theorem 5.7, (−1 1 ◦ 2 )(v) = (A computed in part (ii) yields ⎤ ⎡ 2 1 0 ⎥ ⎢ A−1 B−1 = ⎣ 0 0 13 ⎦ 

1 0

2 3

−1 −1 Hence, (−1 , since parts (iv) and (v) show that 1 ◦ 2 ) is the same linear operator as (2 ◦ 1 ) they have the same matrix with respect to the standard basis.

(c) (i) By Theorem 5.16, 1 and 2 are isomorphisms if and only if A and B are nonsingular. But using basketweaving on each shows that |A| = −1 and |B| = 1. Since neither of these are zero, both matrices are nonsingular. −1 −1 v and −1 v, where (ii) Theorem 5.16, −1 1 (v) = A 2 (v) = B ⎡ ⎤ ⎡ ⎤ 2 −4 −1 1 0 −1 1 −3 ⎦ A−1 = ⎣ 7 −13 −3 ⎦ and B−1 = ⎣ 3 5 −10 −3 −1 −2 2

are obtained by row reduction.

(iii) By Theorem 5.7, (2 ◦ 1 )(v) = (BA)v. Computing BA produces ⎤⎞ ⎡ ⎤⎡ ⎤ ⎛⎡ 29 −6 −4 1 1 (2 ◦ 1 ) ⎝⎣ 2 ⎦⎠ = ⎣ 21 −5 −2 ⎦ ⎣ 2 ⎦  3 3 38 −8 −5 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

178

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.5

(iv) By Theorem 5.16, (2 ◦ 1 )−1 (v) = (BA)−1 (v) Using row reduction to compute the inverse of the matrix we found in part (iii) gives ⎛⎡ ⎤⎞ ⎡ ⎤⎡ ⎤ 1 −9 −2 8 1 (2 ◦ 1 )−1 ⎝⎣ 2 ⎦⎠ = ⎣ −29 −7 26 ⎦ ⎣ 2 ⎦  3 3 −22 −4 19

−1 −1 −1 B )v. Using the matrices A−1 and B−1 we (v) By Theorem 5.7, (−1 1 ◦ 2 )(v) = (A computed in part (ii) yields ⎡ ⎤ −9 −2 8 A−1 B−1 = ⎣ −29 −7 26 ⎦  −22 −4 19

−1 −1 , since parts (iv) and (v) show that Hence, (−1 1 ◦ 2 ) is the same linear operator as (2 ◦ 1 ) they have the same matrix with respect to the standard basis.

(5) (a) Now (i) = j and (j) = i. Since the columns of the matrix for  are the images of i and j, ∙ ¸ 0 1 respectively, the matrix is .

1 0 (8) Suppose dim(V) =   0 and dim(W) =   0. −1 −1 Suppose A is nonsingular, then A is square,  = , A−1  exists, and A A = A A = −1 I . Let  be the linear operator from W to V whose matrix with respect to  and  is A . Then, for all v ∈ V, [( ◦ )(v)] = A−1  A [v] = I [v] = [v] . Therefore, ( ◦ )(v) = v for all v ∈ V, since coordinatizations are unique. Similarly, for all w ∈ W, [( ◦ )(w)] = A A−1  [w] = I [w] = [w] . Hence ( ◦ )(w) = w for all w ∈ W. Therefore,  acts as an inverse for . Hence,  is an isomorphism by Theorem 5.15.

(15) We want to use this exercise as part of the proof of the Dimension Theorem, so we will not use the Dimension Theorem in this proof. (Notice that we can use both Theorem 5.14 and part (a) of Exercise 11 because they were previously proven without invoking the Dimension Theorem.)

Because Y is a subspace of the finite dimensional vector space V, Y is also finite dimensional (Theorem 4.13). Suppose {y1      y } is a basis for Y. Because  is one-to-one, the vectors (y1 )     (y ) are distinct. By part (1) of Theorem 5.14, {(y1 )     (y )} = ({y1      y }) is linearly independent. Part (a) of Exercise 11 in Section 5.4 shows that span(({y1      y })) = (span({y1      y })) = (Y). Therefore,{(y1 )     (y )} is a -element basis for (Y). Hence, dim((Y)) =  = dim(Y).

(16) (c) Let v ∈ 1 (ker( ◦ 1 )). Thus, there is an x ∈ ker( ◦ 1 ) such that 1 (x) = v. Then,  (v) =  (1 (x)) = ( ◦ 1 )(x) = 0W . Thus, v ∈ ker( ). Therefore, 1 (ker( ◦ 1 )) ⊆ ker( ).

Now suppose that v ∈ ker( ). Then  (v) = 0W . Thus, ( ◦ 1 )(1−1 (v)) = 0W as well.

Hence, 1−1 (v) ∈ ker( ◦ 1 ). Therefore, v = 1 (1−1 (v)) ∈ 1 (ker( ◦ 1 )), implying ker( ) ⊆ 1 (ker( ◦ 1 )), completing the proof.

(e) Suppose y ∈ range(2 ◦  ). Hence, there is a v ∈ V such that (2 ◦  )(v) = y. Thus, y = 2 ( (v)). Clearly,  (v) ∈ range( ), and so y ∈ 2 (range( )). This proves that range(2 ◦  ) ⊆ 2 (range( )).

Next, suppose that y ∈ 2 (range( )). Then there is some w ∈ range( ) such that 2 (w) = y.

Since w ∈ range( ), there is a v ∈ V such that  (v) = w. Therefore, (2 ◦  )(v) = 2 ( (v)) = 2 (w) = y. This establishes that 2 (range( )) ⊆ range(2 ◦  ), finishing the proof.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

179

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.6

(17) (a) Part (c) of Exercise 16¡ states (ker 2 ◦  and ¡ that 1−1 ¢¢ ( ◦ 1 )) = ker ( ). Substituting  = −1 −1 ker  = ker ( yields  ◦  ◦  ◦ ).

Now  =  ◦  ◦  1 = −1 2 2 2 1 1 1 1 . Hence, −1 1 (ker ( )) = ker (2 ◦ ). (b) Part (a) of Exercise 16 shows that ker (2 ◦ ) = ker (). This combined with part (a) of this exercise shows that −1 1 (ker ( )) = ker (). −1 (c) From part (b), dim(ker ()) = dim(−1 1 (ker ( ))). Apply Exercise 15 with  replaced by 1 −1 and Y = ker( ) to show that dim(1 (ker ( ))) = dim(ker( )), completing the proof.

(20) (a) True. This is part of Theorem 5.15. (b) True. This is part of Theorem 5.15. (c) False. In fact, this statement contradicts Theorem 5.16, which implies that  is an isomorphism if and only if A is nonsingular. But A is nonsingular if and only if |A | 6= 0. Any linear operator on a nontrivial finite dimensional vector space provides a counterexample for the statement given in the problem. (d) False. While Corollary 5.13 shows this is true if V and W are finite dimensional vector spaces having the same dimension, the statement is not necessarily true otherwise. For example, the linear transformation of Example 2 in Section 5.4 of the textbook is onto, but is not one-to-one.

Also, Exercise 1(h) in Section 5.4, whose solution appears in this manual, gives an example of a linear transformation that is one-to-one but not onto. (e) False. The problem is that ker() is a subspace of V while ker( ◦  ) is a subspace of X . For a particular counterexample, let  be the zero linear operator on R3 and let  : P2 → R3 be given by  (2 +  + ) = [  ]. Note that ker() = R3 while ker( ◦  ) = P2 . (Notice, however, that if the vector spaces are finite dimensional, then dim(ker()) = dim(ker( ◦  )).) (f) True. This follows from part (b) of Exercise 16.

(g) True. By Theorem 5.15, −1 is a linear transformation. The given statement then follows from part (3) of Theorem 5.1. (h) True. Since dim(R28 ) = dim(P27 ) = dim(M74 ) = 28, Corollary 5.20 shows that each of these vector spaces is isomorphic to the other two.

Section 5.6 (1) In all parts, we just need to perform Step 2 of the Generalized Diagonalization Method. However, this corresponds precisely with the Diagonalization Method from Section 3.4. Also, in each part, let A represent the given matrix. ¯ ¯ ¯  − 2 −1 ¯ ¯ ¯ = ( − 2)2 . Hence,  = 2 is the only eigenvalue for , and the (a) First, A () = ¯ 0 −2 ¯ algebraic multiplicity for  is 2. To find a basis for 2 , we row reduce ¯ ¯ ∙ ¸ ∙ ¸ 0 −1 ¯¯ 0 0 1 ¯¯ 0 [ 2I2 − A| 0] = to obtain  0 0 ¯ 0 0 0 ¯ 0

Setting the independent variable 1 of the associated system equal to 1 produces the basis {[1 0]} for 2 . Since dim(2 ) = 1, the geometric multiplicity of  is 1.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

180

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.6

¯ ¯ ¯  −1 −1 ¯¯ ¯ 1 ¯¯ = 3 −22 −+2 (using basketweaving) = (−1)(+1)(−2). (c) First, A () = ¯¯ 1  − 4 ¯ 1 −5 +2 ¯ Hence, the eigenvalues for  are 1 = 1, 2 = −1, and 3 = 2, each having algebraic multiplicity 1. Next, we solve for a basis for each eigenspace.

For 1 = 1, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 −1 −1 ¯¯ 0 1 0 −2 ¯¯ 0 1 ¯¯ 0 ⎦ to obtain ⎣ 0 1 −1 ¯¯ 0 ⎦  [ 1I3 − A| 0] = ⎣ 1 −3 1 −5 3 ¯ 0 0 0 0 ¯ 0

Setting the independent variable 3 of the associated system equal to 1 produces the basis {[2 1 1]} for 1 .

For 2 = −1, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ −1 −1 −1 ¯¯ 0 1 0 1 ¯¯ 0 1 ¯¯ 0 ⎦ to obtain ⎣ 0 1 0 ¯¯ 0 ⎦  [ −1I3 − A| 0] = ⎣ 1 −5 1 −5 1 ¯ 0 0 0 0 ¯ 0

Setting the independent variable 3 of the associated system equal to 1 produces the basis {[−1 0 1]} for −1 .

For 3 = 2, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 2 −1 −1 ¯¯ 0 1 0 −1 ¯¯ 0 1 ¯¯ 0 ⎦ to obtain ⎣ 0 1 −1 ¯¯ 0 ⎦  [ 2I3 − A| 0] = ⎣ 1 −2 1 −5 4 ¯ 0 0 0 0 ¯ 0

Setting the independent variable 3 of the associated system equal to 1 produces the basis {[1 1 1]} for 2 .

Since each eigenspace is one-dimensional, each eigenvalue has geometric multiplicity 1. ¯ ¯ ¯ −2 0 0 ¯¯ ³ ´ ¯ 6 ¯¯ = ( − 2) ( + 3)( − 8) + 30 (using cofactor expansion along (d) A () = ¯¯ −4  + 3 ¯ 4 −5  − 8 ¯ the 1st row) = ( − 2)2 ( − 3). Hence, the eigenvalues for  are 1 = 2 and 2 = 3. The algebraic multiplicities of these eigenvalues are 2 and 1, respectively. Next, we solve for a basis for each eigenspace.

For 1 = 2, we row reduce ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 − 54 − 32 ¯¯ 0 0 0 0 ¯¯ 0 ⎢ ¯ ⎥ 5 6 ¯¯ 0 ⎦ to obtain ⎣ 0 0 0 ¯ 0 ⎦ [ 2I3 − A| 0] = ⎣ −4 ¯ 4 −5 −6 ¯ 0 0 0 0 ¯ 0

Setting the independent variables 2 = 4 (to avoid fractions) and 3 = 0 produces the fundamental eigenvector [5 4 0]. Setting 2 = 0 and 3 = 2 yields [3 0 2]. Hence, {[5 4 0] [3 0 2]} is a basis for 2 , and the geometric multiplicity of 1 is 2.

For 2 = 3, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 0 0 ¯¯ 0 1 0 0 ¯¯ 0 6 6 ¯¯ 0 ⎦ to obtain ⎣ 0 1 1 ¯¯ 0 ⎦  [ 3I3 − A| 0] = ⎣ −4 4 −5 −5 ¯ 0 0 0 0 ¯ 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

181

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.6

Setting the independent variable 3 equal to 1 produces the basis {[0 −1 1]} for 3 , and the geometric multiplicity of 2 is 1.

(2) (b) Step 1: We let  = (2   1), the standard basis for P2 . Then, since (2 ) = 22 −2, () = −1, ⎡ ⎤ 2 0 0 1 0 ⎦ for  with respect to . and (1) = 0, we get the matrix A = ⎣ −2 0 −1 0 ¯ ¯ ¯ −2 0 0 ¯¯ ¯  − 1 0 ¯¯ = ( − 2)( − 1) (since the corresponding matrix is Step 2: First, A () = ¯¯ 2 ¯ 0 1  ¯ lower triangular). Hence, the eigenvalues for  are 1 = 2, 2 = 1 and 3 = 0. Next, we solve for a basis for each eigenspace.

For 1 = 2, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 0 0 0 ¯¯ 0 1 0 −1 ¯¯ 0 2 ¯¯ 0 ⎦  [ 2I3 − A| 0] = ⎣ 2 1 0 ¯¯ 0 ⎦ to obtain ⎣ 0 1 ¯ 0 1 2 0 0 0 0 ¯ 0

Setting the independent variable 3 of the associated system equal to 1 produces the basis {[1 −2 1]} for 2 in R3 .

For 2 = 1, we row reduce ⎡

¯ ⎤ −1 0 0 ¯¯ 0 [ 1I3 − A| 0] = ⎣ 2 0 0 ¯¯ 0 ⎦ 0 1 1 ¯ 0

to obtain

¯ ⎤ 1 0 0 ¯¯ 0 ⎣ 0 1 1 ¯ 0 ⎦ ¯ 0 0 0 ¯ 0 ⎡

Setting the independent variable 3 of the associated system equal to 1 produces the basis {[0 −1 1]} for 1 in R3 .

For 3 = 0, we row reduce

¯ ⎤ −2 0 0 ¯¯ 0 [ 0I3 − A| 0] = [ −A| 0] = ⎣ 2 −1 0 ¯¯ 0 ⎦ 0 1 0 ¯ 0 ⎡

to obtain

¯ ⎤ 1 0 0 ¯¯ 0 ⎣ 0 1 0 ¯ 0 ⎦ ¯ 0 0 0 ¯ 0 ⎡

Setting the independent variable 3 of the associated system equal to 1 produces the basis {[0 0 1]} for 0 in R3 .

Since the union of the three bases is  = {[1 −2 1] [0 −1 1] [0 0 1]} which has 3 elements,  is diagonalizable.

Step 3: Converting the vectors in  into polynomials produces the basis  = (2 −2+1 −+1 1) for P2 . The matrix for  with respect to  is the diagonal matrix having the eigenvalues for  ⎡ ⎤ ⎡ ⎤ 2 0 0 1 0 0 on its main diagonal; that is, D = ⎣ 0 1 0 ⎦. Let P = ⎣ −2 −1 0 ⎦, the matrix whose 0 0 0 1 1 1

columns are the vectors in . Then row reduction yields P−1 = P. (This does not occur in general.) Matrix multiplication verifies that D = P−1 AP. (Try it!)

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

182

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.6

(d) Step 1: We let  = (2   1), the standard basis for P2 . Then, since (2 ) = −2 − 12 + 18, ⎡ ⎤ −1 0 0 0 ⎦. () = −4, and (1) = −5, the matrix for  with respect to  is A = ⎣ −12 −4 18 0 −5 ¯ ¯ ¯ +1 0 0 ¯¯ ¯ 0 ¯¯ = ( + 1)( + 4)( + 5) (since the corresponding Step 2: First, A () = ¯¯ 12  + 4 ¯ −18 0 +5 ¯

matrix is lower triangular). Hence, the eigenvalues for  are 1 = −1, 2 = −4 and 3 = −5.

Next, we solve for a basis for each eigenspace.

For 1 = −1, we row reduce ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 0 − 29 ¯¯ 0 0 0 0 ¯¯ 0 ⎢ ⎥ 8 ¯ [ −1I3 − A| 0] = ⎣ 12 3 0 ¯¯ 0 ⎦ to obtain ⎣ 0 1 9 ¯¯ 0 ⎦  −18 0 4 ¯ 0 0 0 0 ¯ 0

Setting the independent variable 3 of the associated system equal to 9 (to eliminate fractions) produces the basis {[2 −8 9]} for −1 in R3 .

For 2 = −4, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ −3 0 0 ¯¯ 0 1 0 0 ¯¯ 0 [ −4I3 − A| 0] = ⎣ 12 0 0 ¯¯ 0 ⎦ to obtain ⎣ 0 0 1 ¯¯ 0 ⎦  −18 0 1 ¯ 0 0 0 0 ¯ 0

Setting the independent variable 2 of the associated system equal to 1 produces the basis {[0 1 0]} for −4 in R3 .

For 3 = −5, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ −4 0 0 ¯¯ 0 1 0 0 ¯¯ 0 [ −5I3 − A| 0] = ⎣ 12 −1 0 ¯¯ 0 ⎦ to obtain ⎣ 0 1 0 ¯¯ 0 ⎦  −18 0 0 ¯ 0 0 0 0 ¯ 0

Setting the independent variable 3 of the associated system equal to 1 produces the basis {[0 0 1]} for −5 in R3 .

Since the union of the three bases is  = {[2 −8 9] [0 1 0] [0 0 1]} which has 3 elements,  is diagonalizable.

Step 3: Converting the vectors in  into polynomials produces the basis  = (22 − 8 + 9  1) for P2 . The matrix for  with respect to  is the diagonal matrix having the eigenvalues for ⎡ ⎤ ⎡ ⎤ −1 0 0 2 0 0 0 ⎦. Let P = ⎣ −8 1 0 ⎦ , the matrix  on its main diagonal; that is, D = ⎣ 0 −4 0 0 −5 9 0 1 ⎤ ⎡ 1 0 0 2 ⎥ ⎢ whose columns are the vectors in . Then row reduction yields P−1 = ⎣ 4 1 0 ⎦. Matrix − 92 0 1 multiplication verifies that D = P−1 AP. (Try it!) √

(e) Step 1: We let  = (i j), the standard basis for R2 . Then, since (i) = [ 12  23 ], and (j) = " √ # 1 3 √ − 2 [− 23  12 ], we get the matrix A = √23 for  with respect to .

1 2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

2

183

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ¯ ¯ − 1 ¯ √2 Step 2: First, A () = ¯ ¯ − 3 2

Section 5.6

¯ ¯ ¯ ¯ = 2 −  + 1, which has no real roots. Hence,  has no 1 ¯ − √ 3 2

2

eigenvalues, and so is not diagonalizable. (h) Step 1: Let  be the standard basis for M22 ; that is,  = {Ψ | 1 ≤  ≤ 2 1 ≤  ≤ 2}, where Ψ is the 2 × 2 matrix having 1 in its ( ) entry and zeroes elsewhere. (See Example 3 ∙ ¸ ∙ ¸ ∙ ¸ −4 0 0 −4 3 0 , (Ψ12 ) = , (Ψ21 ) = , and in Section 4.5.) Now (Ψ11 ) = −10 0 0 −10 7 0 ⎡ ⎤ −4 0 3 0 ∙ ¸ ⎢ 0 3 0 −4 0 3 ⎥ ⎥ (Ψ22 ) = , and so the matrix for  with respect to  is A = ⎢ ⎣ 0 7 −10 0 7 0 ⎦ 0 −10 0 7 ¯ ¯ ¯ +4 ¯ 0 −3 0 ¯ ¯ ¯ 0 +4 0 −3 ¯¯ ¯ Step 2: First, A () = ¯ . Using cofactor expansion along the fourth 0 −7 0 ¯¯ ¯ 10 ¯ 0 10 0 −7 ¯ row, and cofactor expansion on the second row for each of the resulting 3 × 3 determinants yields A () = 10 (3 (( + 4)( − 7) + 30)) + ( − 7)( + 4) (( + 4)( − 7) + 30) = (( + 4)( − 7) + 30)2 = ( − 1)2 ( − 2)2 

Hence, the eigenvalues for  are 1 = 1, and 2 = 2. Next, we solve for a basis for each eigenspace.

For 1 = 1, we row reduce ¯ ⎡ ⎤ 5 0 −3 0 ¯¯ 0 ⎢ 0 5 0 −3 ¯¯ 0 ⎥ ⎥ [ 1I3 − A| 0] = ⎢ ⎣ 10 0 −6 0 ¯¯ 0 ⎦ 0 10 0 −6 ¯ 0

to obtain

⎡

¯ ⎤ 0 ¯¯ 0 ¯ ⎥ 0 − 35 ¯ 0 ⎥ ¯ ⎥ 0 0 ¯¯ 0 ⎦ 0 0 ¯ 0

1 0 − 35

⎢ ⎢ 0 1 ⎢ ⎣ 0 0 0 0

Setting the independent variables 3 = 5 (to avoid fractions) and 4 = 0 produces the fundamental eigenvector [3 0 5 0]. Setting 3 = 0 and 4 = 5 yields [0 3 0 5]. Hence, {[3 0 5 0] [0 3 0 5]} is a basis for 1 in R4 .

For 2 = 2, we row reduce ¯ ⎤ ⎡ ¯ ⎡ ⎤ 0 ¯¯ 0 1 0 − 12 6 0 −3 0 ¯¯ 0 ¯ ⎥ ⎢ ⎢ 0 6 0 −3 ¯¯ 0 ⎥ 0 1 0 − 12 ¯ 0 ⎥ ⎥ to obtain ⎢ [ 2I3 − A| 0] = ⎢ ¯ ⎥ ⎢ ⎣ 10 0 −5 0 ¯¯ 0 ⎦ ⎣ 0 0 0 0 ¯¯ 0 ⎦ 0 10 0 −5 ¯ 0 0 0 0 0 ¯ 0

Setting the independent variables 3 = 2 (to avoid fractions) and 4 = 0 produces the fundamental eigenvector [1 0 2 0]. Setting 3 = 0 and 4 = 2 yields [0 1 0 2]. Hence, {[1 0 2 0] [0 1 0 2]} is a basis for 2 in R4 .

The union of the three bases is  = {[3 0 5 0] [0 3 0 5] [1 0 2 0] [0 1 0 2]}. Since || = 4,  is diagonalizable.

Step 3: Converting the vectors in  into matrices produces the basis µ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸¶ 3 0 0 3 1 0 0 1 =    5 0 0 5 2 0 0 2 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

184

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.6

for M22 . The matrix for  with respect to  is the diagonal matrix having the eigenvalues for ⎡ ⎤ ⎡ ⎤ 1 0 0 0 3 0 1 0 ⎢ 0 1 0 0 ⎥ ⎢ 0 3 0 1 ⎥ ⎥ ⎢ ⎥  on its main diagonal; that is, D = ⎢ ⎣ 0 0 2 0 ⎦. Let P = ⎣ 5 0 2 0 ⎦ , the matrix 0 0 0 2 0 5 0 2 ⎡ ⎤ 2 0 −1 0 ⎢ 0 2 0 −1 ⎥ ⎥. whose columns are the vectors in . Then row reduction yields P−1 = ⎢ ⎣ −5 0 3 0 ⎦ 0 −5 0 3

Matrix multiplication verifies that D = P−1 AP. (Try it!)

(4) (a) Consider  = (2   1), the standard basis for P2 . Since  = 1 in this part, (2 +  +  ) = ( + 1)2 + ( + 1) +  . Therefore, (2 ) = ( + 1)2 = 2 + 2 + 1, () =  + 1, and (1) = 1. ⎡ ⎤ 1 0 0 Hence, we get the matrix A = ⎣ 2 1 0 ⎦ for  with respect to .

1 1 1 ¯ ¯ ¯ −1 0 0 ¯¯ ¯ 0 ¯¯ = ( − 1)3 (since the corresponding matrix is lower Now, A () = ¯¯ −2  − 1 ¯ −1 −1  − 1 ¯ triangular). Hence, the only eigenvalue for  is  = 1. Next, we solve for a basis for 1 . We row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 0 0 0 ¯¯ 0 1 0 0 ¯¯ 0 0 0 ¯¯ 0 ⎦ to obtain ⎣ 0 1 0 ¯¯ 0 ⎦  [ 1I3 − A| 0] = ⎣ −2 −1 −1 0 ¯ 0 0 0 0 ¯ 0

Setting the independent variable 3 of the associated system equal to 1 produces the basis {[0 0 1]} for 1 in R3 . Converting the vector in this basis back to polynomial form yields the basis 1 = {1} in P2 . (Note that our process has produced only one fundamental eigenvector for , and so  is not diagonalizable.) ¯ ¯ ⎡ ⎤ ¯ −1 1 1 −1 −1 1 ¯¯ ¯ −1 0 ¯¯ = ( − 1)3 (since the cor(7) (a) Let A = ⎣ 0 1 0 ⎦. Then, A () = ¯¯ 0 ¯ 0 0 1 0 0 −1 ¯

responding matrix is upper triangular). Hence, the only eigenvalue for A is  = 1, which has algebraic multiplicity 3. Next, we solve for a basis for 1 . We row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 0 −1 1 ¯¯ 0 0 1 −1 ¯¯ 0 0 0 ¯¯ 0 ⎦ to obtain ⎣ 0 0 0 ¯¯ 0 ⎦  [ 1I3 − A| 0] = ⎣ 0 0 0 0 ¯ 0 0 0 0 ¯ 0

Setting the independent variables 1 = 1 and 3 = 0 produces the fundamental eigenvector [1 0 0]. Setting 1 = 0 and 3 = 1 yields [0 1 1]. Hence, {[1 0 0] [0 1 1]} is a basis for 1 .

Since this basis has 2 elements,  = 1 has geometric multiplicity 2. ¯ ¯ ⎡ ⎤ ¯ −1 1 0 0 0 0 ¯¯ ¯  − 1 0 ¯¯ = ( − 1)2  (since the corresponding (b) Let A = ⎣ 0 1 0 ⎦. Then, A () = ¯¯ 0 ¯ 0 0 0 0 0  ¯ matrix is upper triangular). Hence,  = 1 is an eigenvalue for A which has algebraic multiplicity

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

185

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 5.6

2. (Zero is also an eigenvalue for A.) ⎡Next, we solve ¯ for ⎤ a basis for 1 . Performing the row op0 0 0 ¯¯ 0 eration h1i ↔ h3i puts [ 1I3 − A| 0] = ⎣ 0 0 0 ¯¯ 0 ⎦ into reduced row echelon form, yielding 0 0 1 ¯ 0 ¯ ⎡ ⎤ ¯ 0 0 1 ¯ 0 ⎣ 0 0 0 ¯ 0 ⎦. Setting the independent variables 1 = 1 and 2 = 0 produces the fundamen¯ 0 0 0 ¯ 0

tal eigenvector [1 0 0]. Setting 1 = 0 and 2 = 1 yields [0 1 0]. Hence, {[1 0 0] [0 1 0]} is a basis for 1 . Since this basis has 2 elements,  = 1 has geometric multiplicity 2.

(15) Suppose v ∈ 1 ∩ 2  Then v ∈ 1 and v ∈ 2 . Now v ∈ 1 =⇒ v ∈ 1 =⇒ (v) = 1 v Similarly, v ∈ 2 =⇒ v ∈ 2 =⇒ (v) = 2 v Hence, 1 v = (v) = 2 v and so (1 −2 )v = 0 Since 1 6=2 , v = 0 But 0 can not be an element of any basis. This contradiction shows that 1 ∩ 2 must be empty.

(16) (a)  is a subspace, hence it is closed under linear combinations. Therefore, since u is a linear combination of basis vectors for  , it must also be in  .

P P (b) First, substitute u for =1  v in the given double sum equation. This proves =1 u = 0. are eigenvectors Now, the set of all nonzero u ’s is linearly independent by Theorem 5.23, since Pthey  u corresponding to distinct eigenvalues. But then the nonzero terms in =1  would give a nontrivial linear combination (of vectors from a linearly independent set) equal to the zero vector.

This contradiction shows that all of the u ’s must equal 0.

P  v , for each . But {v1      v } is linearly (c) Using part (b) and the definition of u , 0 = =1 independent, since it is a basis for  . Hence, for each , 1 = · · · =  = 0. (d) We have shown that any linear combination of vectors in  that produces the zero vector must use all zero coefficients. Therefore  is linearly independent by definition.

(18) (a) False. The correct definition for the eigenspace  is  = {v ∈ V | (v) = v}. Note that the set  = {(v) | v ∈ V} is the set of images of  multiplied by . But since (v) = (v), it is not hard to show that  equals range(). For a specific example in which  6= , consider Example 7 in Section 5.6 of the textbook, using  = 2. In that example,  is an isomorphism, so  = range() = R3 . However, the example shows that  has dimension 1, since the algebraic multiplicity of  = 2 is only 1. (b) True. This is the definition of the characteristic polynomial of a linear operator. (The characteristic polynomial for  is independent of the particular ordered basis chosen for V (because of Theorems 5.5 and 5.6). Therefore, the definition of a characteristic polynomial of a linear operator makes sense as given in the text.) (c) True. This follows directly from Theorem 5.22. (d) False. While one half of this “if and only if” statement is just Theorem 5.23, the converse of that statement (the other half of the “if and only if”) is false. For a specific counterexample, consider the zero linear operator : R2 → R2 . Then  = 0 is the only eigenvalue, and 0 = R2 . Thus, {i j} is a set of linearly independent eigenvectors for the same eigenvalue  = 0. (e) True. This is Theorem 5.25.

(f) True. This follows from Theorem 5.28 (which indicates that a union of the bases for the distinct eigenspaces contains 6 vectors) and Theorem 5.25 (which indicates that the 6 vectors in the union of these bases are linearly independent, and hence form a basis for R6 ). c 2016 Elsevier Ltd. All rights reserved.

Copyright °

186

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 5 Review

(g) True. This is the purpose of the Method. (h) False. For a specific example in which the algebraic multiplicity of an eigenvalue is strictly larger than its geometric multiplicity, see Example 12 in Section 5.6 of the textbook. (Note that by Theorem 5.26, the algebraic multiplicity of an eigenvalue  is always greater than or equal to its geometric multiplicity.) (i) False. According to Theorem 5.28, each geometric multiplicity must also equal its corresponding algebraic multiplicity in order for  to be diagonalizable. For a specific counterexample, consider : R7 → R7 given by (e ) = e+1 , for 1 ≤  ≤ 6, and (e7 ) = 0. Then it can be shown that  () = 7 , and so  = 0 is the only eigenvalue for , having algebraic multiplicity 7. However, it can be shown that 0 has basis {e7 }, and so the geometric multiplicity of  = 0 is 1. Hence, by Theorem 5.28,  is not diagonalizable. You should verify all of the claims made in this example. (j) True. Since A () = (−1)(−4) = (1−)(4−), this follows directly from the Cayley-Hamilton Theorem.

Chapter 5 Review Exercises (1) (a) We present three different proofs that the function  is not a linear transformation.

Solution 1: The function  is not a linear transformation because  ([0 0 0]) = [0 1 0]. The image of the zero vector under a linear transformation must be the zero vector in the codomain (part (1) of Theorem 5.1).

Solution 2: Note that  ([1 0 0]) +  ([0 1 0]) = [0 4 5] + [−1 1 2] = [−1 5 7], but  ([1 0 0] + [0 1 0]) =  ([1 1 0]) = [−1 4 7], and so property (1) in the definition of a linear transformation is not satisfied.

Solution 3: Note that  (2[1 0 0]) =  ([2 0 0]) = [0 7 10] while 2 ([1 0 0]) = 2[0 4 5] = [0 8 10] and so property (2) in the definition of a linear transformation is not satisfied.

(3)  (A1 ) +  (A2 ) = CA1 B−1 + CA2 B−1 = C(A1 B−1 + A2 B−1 ) = C(A1 + A2 )B−1 =  (A1 + A2 );  (A) = C(A)B−1 = CAB−1 =  (A).

(4) Creating the matrix whose columns are the images of i, j, and k, we see that ⎡ ⎤⎡ ⎤ −3 5 −4  0 ⎦ ⎣  ⎦ = [−3 + 5 − 4 2 −  4 + 3 − 2] ([  ]) = ⎣ 2 −1 4 3 −2  Hence,

⎡

⎤⎡ ⎤ ⎡ ⎤ −3 5 −4 6 20 0 ⎦ ⎣ 2 ⎦ = ⎣ 10 ⎦  ([6 2 −7]) = ⎣ 2 −1 4 3 −2 −7 44

(5) (b) By Theorem 5.2, 2 ◦ 1 : V → X is a linear transformation. Thus, by part (2) of Theorem 5.3, (2 ◦ 1 )−1 (X 0 ) is a subspace of V. (6) (a) We will use Theorem 5.5, which states that the th column of A equals [(v )] , where  is the given ordered basis for the domain of ,  is the given ordered basis for the codomain of , and v is the th basis element in .

Substituting each vector in  into the given formula for  shows that ([−5 −3 −2]) = [−13 1], ([3 0 1]) = [2 12], and ([5 2 2]) = [10 6]. Next, we must express each of these c 2016 Elsevier Ltd. All rights reserved.

Copyright °

187

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 5 Review

answers in -coordinates. We use the Coordinatization Method from Section 4.7. Thus, for each vector v that we need to express in -coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors in , and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have three such systems to solve, we can solve all three simultaneously. Therefore, we row reduce ¯ ¯ ∙ ¸ ∙ ¸ 4 −3 ¯¯ −13 2 10 1 0 ¯¯ 29 32 −2 to  3 −2 ¯ 1 12 6 0 1 ¯ 43 42 −6 Hence, [([−5 −3 −2])] = [29 43], [([3 0 1])] = [32 42], and [([5 2 2])] = [−2 −6]. Using ∙ ¸ 29 32 −2 . (Note that A is just these coordinatizations as columns produces A = 43 42 −6 the matrix to the right of the bar in the row reduced matrix, above. This is typical, but in general, any rows of all zeroes must be eliminated first, as in the Coordinatization Method.)

(7) (b) First we find the matrix for  with respect to the standard basis  for P2 and the standard  ∙ basis ¸ 6 0 2 for M22 . Substituting each vector in  into the given formula for  shows that ( ) = , 2 1 ∙ ¸ ∙ ¸ −1 3 −1 2 () = , and (1) = . Converting these matrices to -coordinates creates 0 −5 −4 1 ⎡ ⎤ 6 −1 −1 ⎢ 0 3 2 ⎥ ⎥. Next, we use Theorem 4-vectors, which we use as columns to produce A = ⎢ ⎣ 2 0 −4 ⎦ 1 −5 1

5.6 to compute A . To do this, we need the inverse of the transition matrix P from  to  and the transition matrix Q from  to . To compute P (or P−1 ), we need to convert the polynomials in  into vectors in R3 . After doing this, P is the inverse of the matrix whose columns are the resulting vectors obtained from , and Q is the inverse of the matrix whose ⎡ ⎤ ⎡ ⎤ 3 2 1 4 −5 3 −2 ⎢ 2 1 1 2 ⎥ ⎥ 1 ⎦ and Q−1 = ⎢ columns are the vectors in . Hence, P−1 = ⎣ 2 −1 ⎣ 2 1 2 2 ⎦.

5 −1 3 5 4 4 7 ⎡ ⎤ −1 3 −1 0 ¢−1 ⎢ −4 ¡ 6 −5 2 ⎥ ⎥. Finally, using Theorem 5.6 produces =⎢ So, Q = Q−1 ⎣ 0 −1 1 0 ⎦ 3 −5 3 −1 ⎡ ⎤ 115 −45 59 ⎢ 374 −146 190 ⎥ ⎥.

A = QA P−1 = ⎢ ⎣ −46 15 −25 ⎦ −271 108 −137 ¯ ¯ ¯ ¯  − 18 − 36 − 12 41 41 41 ¯ ¯ ¯ ¯ 36 31 24  + 41 (9) (a) A () = |I3 − A | = ¯ − 41 ¯ = 3 − 2 −  + 1 = ( + 1)( − 1)2 , where 41 ¯ ¯ 12 ¯ − 24  − 49 ¯ 41

41

41

we have used basketweaving and algebraic simplification.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

188

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 5 Review ⎡

1 ⎢ 0 (10) (a) Let A represent the given matrix. The reduced row echelon form of A is B = ⎢ ⎣ 0 0

⎤ 0 −2 3 1 3 −4 ⎥ ⎥.

0 0 0 ⎦ 0 0 0

Then, by the Kernel Method, a basis for ker() is found by considering the solution to the system BX = 0 computed by setting each independent variable in the system equal to 1, while the other is set to 0. This yields the basis {[2 −3 1 0] [−3 4 0 1]} for ker(). Since ker() has a basis with 2 elements, dim(ker()) = 2. Next, by the Range Method, a basis for range() consists of the first 2 columns of A, since those columns in  contain nonzero pivots. Thus a basis for range() = {[3 2 2 1] [1 1 3 4]} and dim(range()) = 2.

(12) (a) If v ∈ ker(1 ), then 1 (v) = 0W . Thus, (2 ◦ 1 )(v) = 2 (1 (v)) = 2 (0W ) = 0X , and so v ∈ ker(2 ◦ 1 ). Therefore, ker(1 ) ⊆ ker(2 ◦ 1 ). Hence, dim(ker(1 )) ≤ dim(ker(2 ◦ 1 )).

(b) Let 1 be the projection onto the -axis (1 ([ ]) = [ 0]) and 2 be the projection onto the -axis (2 ([ ]) = [0 ]). A basis for ker(1 ) is {[0 1]}, while ker(2 ◦ 1 ) = R2 . Thus, dim(ker(1 )) = 1  2 = dim(ker(2 ◦ 1 )).

(15) (a) Let A represent the given matrix. The reduced row echelon form of A is I3 . Then, by the Kernel Method, ker() = {0}, and so dim(ker()) = 0, and  is one-to-one. By Corollary 5.13,  is also onto. Hence dim(range()) = 3.

(18) (a) 1 and 2 are isomorphisms because their (given) matrices are nonsingular (by Theorem 5.16).

Their inverses are given in part (b). (b) The matrix for 2 ◦ 1 with respect to the standard basis is found by multiplying the matrices for 2 and 1 with respect to the standard bases (by Theorem 5.7). In particular: ⎡ ⎤⎡ ⎤ 9 8 5 4 3 6 1 1 ⎢ 9 13 ⎢ 4 7 ⎥ 2 −2 1 ⎥ ⎥⎢ 5 ⎥ matrix for 2 ◦ 1 = ⎢ ⎣ 5 9 2 5 ⎦⎣ 2 1 0 1 ⎦ −5 −2 −2 0 1 −1 −2 −1 ⎡ ⎤ 81 71 −15 18 ⎢ 107 77 −31 19 ⎥ ⎥ = ⎢ ⎣ 69 45 −23 11 ⎦ −29 −36 −1 −9 −1 Also, computing inverses of the matrices for 1 and 2 gives the matrices for −1 1 and 2 (by Theorem 5.16): ⎡ ⎤ 2 −10 19 11 1⎢ 5 −10 −5 ⎥ ⎢ 0 ⎥; = matrix for −1 1 ⎣ 3 −15 26 14 ⎦ 5 −4 15 −23 −17 ⎡ ⎤ −8 26 −30 2 1⎢ 41 −4 ⎥ ⎢ 10 −35 ⎥ = matrix for −1 2 ⎣ 10 −30 34 −2 ⎦ 2 −14 49 −57 6

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

189

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 5 Review

(21) (a) If p = 4 + 3 + 2 , then (p) = p0 + p00 = (43 + 32 + 2) + (122 + 6 + 2) = 43 + (12 + 3)2 + (6 + 2) + 2. Hence, p ∈ ker() if and only if ⎧ 4 = 0 ⎪ ⎪ ⎨ 12 + 3 = 0  6 + 2 = 0 ⎪ ⎪ ⎩ 2 = 0 Since this system has only the trivial solution, ker() = {0}. Therefore, by part (1) of Theorem 5.12,  is one-to-one.

(22) In all parts, we just need to perform Step 2 of the Generalized Diagonalization Method. However, this corresponds precisely with the Diagonalization Method from Section 3.4. Also, in each part, let A represent the given matrix. ¯ ¯ ¯  + 9 −18 16 ¯¯ ¯ (a) (i) First, A () = ¯¯ −32  + 63 −56 ¯¯ = 3 − 32 −  + 3 (using basketweaving) ¯ −44 84  − 75 ¯ = ( − 1)( + 1)( − 3). Hence, the eigenvalues for  are 1 = 1, 2 = −1, and 3 = 3, each having algebraic multiplicity 1. Next, we solve for a basis for each eigenspace. ¯ ⎡ ⎤ 1 ¯ ¯ ⎡ ⎤ 1 0 4 ¯ ¯ 0 10 −18 16 ¯ 0 ⎢ ¯ ⎥ 3 ¯ ⎥ For 1 = 1, we row reduce [ 1I3 − A| 0] = ⎣ −32 64 −56 ¯¯ 0 ⎦ to ⎢ ⎣ 0 1 − 4 ¯ 0 ⎦. ¯ ¯ −44 84 −74 0 0 0 0 ¯ 0 Setting the independent variable 3 of the associated system equal to 1, and then eliminating fractions, produces the basis {[−1 3 4]} for 1 . ¯ ⎡ ⎤ 1 ¯ ¯ ⎡ ⎤ 1 0 5 ¯ ¯ 0 8 −18 16 ¯ 0 ⎢ ¯ ⎥ 4 ¯ ⎥ For 2 = −1, we row reduce [ −1I3 − A| 0] = ⎣ −32 62 −56 ¯¯ 0 ⎦ to ⎢ ⎣ 0 1 − 5 ¯ 0 ⎦. ¯ ¯ −44 84 −76 0 0 0 0 ¯ 0

Setting the independent variable 3 of the associated system equal to 1, and then eliminating fractions, produces the basis {[−1 4 5]} for −1 . ¯ ⎡ ⎤ 2 ¯ ¯ ⎡ ⎤ 1 0 9 ¯ 0 12 −18 16 ¯¯ 0 ⎢ ¯ ⎥ 20 ¯ ⎥ For 3 = 3, we row reduce [ 3I3 − A| 0] = ⎣ −32 66 −56 ¯¯ 0 ⎦ to ⎢ ⎣ 0 1 − 27 ¯ 0 ⎦. ¯ −44 84 −72 ¯ 0 0 0 0 ¯ 0

Setting the independent variable 3 of the associated system equal to 1, and then eliminating fractions, produces the basis {[−6 20 27]} for 3 .

(ii) Since each eigenspace is one-dimensional, each eigenvalue has geometric multiplicity 1, which equals its algebraic multiplicity. Since the sum of the geometric multiplicities is 3, which equals the dimension of R3 ,  is diagonalizable.

(iii) The ordered basis  is the union of the bases of fundamental eigenvectors we found for each eigenspace. Thus,  = {[−1 3 4] [−1 4 5] [−6 20 27]}. The transition matrix P is the matrix whose columns are these vectors, and the diagonal matrix D = P−1 AP is the diagonal matrix ⎡ ⎤ 1 0 0 with the eigenvalues 1, −1, and 3 on the main diagonal. Hence, D = ⎣ 0 −1 0 ⎦, and 0 0 3 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

190

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 5 Review

⎡

⎤ ⎡ ⎤ −1 −1 −6 −8 3 −4 4 20 ⎦. (Note that P−1 = ⎣ 1 3 −2 ⎦.)

P=⎣ 3 4 5 27 1 −1 1 ¯ ¯ ¯  + 97 −20 −12 ¯¯ ¯  − 63 −36 ¯¯ = 3 − 52 + 3 + 9 (using basketweaving) = (c) (i) First, A () = ¯¯ 300 ¯ 300 −60  − 39 ¯

( + 1)( − 3)2 . Hence, the eigenvalues for  are 1 = −1, and 2 = 3, with 1 having algebraic multiplicity 1 and 2 having algebraic multiplicity 2. Next, we solve for a basis for each eigenspace. ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 0 − 13 ¯¯ 0 96 −20 −12 ¯¯ 0 ⎢ ¯ ⎥ For 1 = −1, we row reduce [ −1I3 − A| 0] = ⎣ 300 −64 −36 ¯¯ 0 ⎦ to ⎣ 0 1 −1 ¯ 0 ⎦. ¯ ¯ 300 −60 −40 0 0 0 0 ¯ 0 Setting the independent variable 3 of the associated system equal to 1, and then eliminating fractions, produces the basis {[1 3 3]} for −1  ¯ ⎡ ⎤ ¯ ⎡ ⎤ 3 ¯ 1 − 15 − 25 100 −20 −12 ¯¯ 0 ¯ 0 ⎢ ¯ ⎥ For 2 = 3, we row reduce [ 3I3 − A| 0] = ⎣ 300 −60 −36 ¯¯ 0 ⎦ to ⎣ 0 0 0 ¯¯ 0 ⎦.

300 −60 −36 ¯ 0 0 0 0 ¯ 0

Setting the independent variable 2 of the associated system equal to 1 and 3 equal to 0, and then eliminating fractions, produces the fundamental eigenvector [1 5 0] for 3 . Setting the independent variable 3 of the associated system equal to 1 and 2 equal to 0, and then eliminating fractions, produces the fundamental eigenvector [3 0 25]. This gives us a basis {[1 5 0] [3 0 25]} for 3 . (ii) The eigenspace −1 is one-dimensional, and so 1 = −1 has geometric multiplicity 1, which equals its algebraic multiplicity. The eigenspace 3 is two-dimensional, and so 2 = 3 has geometric multiplicity 2, which also equals its algebraic multiplicity. Since the sum of the geometric multiplicities is 3, which equals the dimension of R3 ,  is diagonalizable. (iii) The ordered basis  is the union of the bases of fundamental eigenvectors we found for each eigenspace. Thus,  = {[1 3 3] [1 5 0] [3 0 25]}. The transition matrix P is the matrix whose columns are these vectors, and the diagonal matrix D = P−1 AP is the diagonal ma⎡ ⎤ −1 0 0 trix with the eigenvalues −1, 3, and 3 on the main diagonal. Hence, D = ⎣ 0 3 0 ⎦, and 0 0 3 ⎡ ⎤ ⎡ ⎤ 125 −25 −15 1 1 3 16 9 ⎦.)

P = ⎣ 3 5 0 ⎦. (Note that P−1 = 15 ⎣ −75 −15 3 2 3 0 25

(25) (a) True. This follows directly from Theorem 5.4.

(b) False. There are many such linear transformations because the value of (k) has not been specified. (Once the value of (k) is known, the linear transformation is determined by Theorem 5.4.)

Choosing either (k) = i or (k) = j would produce two different linear transformations having (i) = j and (j) = i. (c) True. See Example 11 in Section 5.1. Use  = −45◦ .

(d) True. The properties (y1 + y2 ) = (y1 ) + (y2 ) and (y1 ) = (y1 ) hold for all vectors y1 , y2 in Y and scalars  because Y is a subset of V and  is a linear transformation on V. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

191

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 5 Review

(e) True. By Theorem 5.5, the th column of the matrix for the linear transformation is Be . But this is the th column of B. (f) False. Similar matrices must be square. The matrices A and A for the linear transformation are not square if dim(V) 6= dim(W). However, even if dim(V) = dim(W), the matrices A and A might not be similar. For example, consider : R3 → P2 given by ([  ]) = 2 +  + .

The matrix for  with respect to the standard bases is I3 . Note that I3 is the only matrix similar to I3 . However, if we continue to use the standard basis for R3 , but use the basis {2 + 1  + 1 1} ⎡ ⎤ 1 0 0 1 0 ⎦, which is not similar to I3 . for P2 , the matrix for  is ⎣ 0 −1 −1 1

(g) False. By the Dimension Theorem, dim(ker())+dim(range()) = 5. However, two equal integers added together can not equal an odd integer. (h) True. Because dim(row space of A) = rank(A) (by the Simplified Span Method), the given statement is true by part (1) of Theorem 5.9. (i) True. See the discussion in the text just prior to the Range Method in Section 5.3. (j) False. Only range() is guaranteed to be finite dimensional. Counterexample: Define : P → P by (p) = p. The range of  is P , which is finite dimensional, but the codomain is P, which is infinite dimensional. (k) False. ker() is a subspace of the domain, and is never empty, since it must contain the zero vector. Part (1) of Theorem 5.12 says that  is one-to-one if and only if ker() = {0}. The identity linear operator on R3 gives a counterexample to the given statement. (l) True. The Dimension Theorem states that dim(ker()) + dim(range()) = dim(V). Thus, dim(range()) = dim(V) if and only if dim(ker()) = 0. The given statement then follows by part (1) of Theorem 5.12. (m) False. Counterexample: The zero linear operator on R3 is neither one-to-one nor onto. (n) True. The Dimension Theorem states that range() is finite dimensional. But, if  is onto, then range() = W, and so W is finite dimensional. (o) True. This is part (1) of Theorem 5.14.

(p) False.  () = 3 is a one-to-one and onto function from R to R, but is not a linear transformation.

Also, translation functions on R are one-to-one and onto, but are not linear transformations (see the discussion after Example 4 in Section 5.1). (q) False. Counterexample: the zero linear operator on R3 has O3 as its matrix with respect to any bases. O3 is square, but the zero operator is clearly not an isomorphism. (Note, however, that with V, W as given, if  is an isomorphism, then the matrix for  with respect to any bases for V and W is square.) (r) True.  ◦  is the identity operator because every vector will reflect back to itself.

(s) False. Only nontrivial finite dimensional vector spaces are isomorphic to R for some . For example, P is not isomorphic to R for any value of . (t) True. Both W1 and W2 are two dimensional vector spaces, and so, by Corollary 5.20, they are isomorphic to each other.

(u) True. This follows from part (a) of Exercise 16 in Section 5.5 with  in place of 2 and  in place of  . The given statement can be proven directly by showing that each set is a subset of the other. First we show ker() ⊆ ker( ◦ ). Let v ∈ ker(). Then (v) = 0. Thus, c 2016 Elsevier Ltd. All rights reserved.

Copyright °

192

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 5 Review

( ◦)(v) =  ((v)) =  (0) = 0, and so v ∈ ker( ◦). Finally we show ker( ◦) ⊆ ker().

Let v ∈ ker( ◦). Then ( ◦)(v) = 0. Thus, (v) = (( −1 ◦ )◦)(v) =  −1 (( ◦)(v)) =  −1 (0) = 0, and so v ∈ ker().

(v) False. Note that range() is a subspace of W, while range( ◦ ) is a subspace of X . For a specific counterexample, let  be the identity linear operator on P3 (so range() = P3 ), and let  be the coordinatization isomorphism from P3 to R4 (so range( ◦ ) = R4 ). (Note, however, that in general, dim(range()) = dim(range( ◦ )), since  is an isomorphism.)

(w) True. This is merely a rewording of the definition of eigenspace in terms of the kernel of an operator.

(x) True. Each algebraic multiplicity must be at least 1 by Theorem 5.26, and since there are  distinct eigenvalues, the sum of the algebraic multiplicities must be at least . However, from the discussion just before Theorem 5.28, the sum of the algebraic multiplicities can be at most .

Therefore, each algebraic multiplicity must equal 1. (y) True. If 2 is a factor of the characteristic polynomial, then the eigenvalue 0 has algebraic multiplicity at least 2. But, if dim(0 ) = 1, then the geometric multiplicity of the eigenvalue 0 is 1. Hence, by Theorem 5.28,  is not diagonalizable. (z) True. By Theorem 5.25, 1 ∪ 2 ∪ · · · ∪  is a linearly independent subset of V. Therefore, it is a basis for the subspace W = span(1 ∪ 2 ∪ · · · ∪  ) of V.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

193

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.1

Chapter 6 Section 6.1 (1) (a) The vectors√are orthogonal because [3 −2] · [4 6] = 0. The vectors are not orthonormal because k[3 −2]k = 13 6= 1. h i h i 9 (c) The vectors are not orthogonal because √313  − √213 · √110  − √310 = √130 6= 0. Since the vectors are not orthogonal, they are also not orthonormal. (f) The set of vectors is orthogonal because [2 −3 1 2]·[−1 2 8 0] = 0, [2 −3 1 2]·[6 −1 1 −8] = 0, and [−1 2 8 0] · [6 −1 1 −8] = 0. The vectors are not orthonormal because k[2 −3 1 2]k = √ 18 6= 0.

(2) In each part, let A represent the given matrix. (a) Straightforward computation shows that AA = I2 . Therefore, A−1 = A , and so A is an orthogonal matrix. ⎡ ⎤ 109 27 −81 19 −27 ⎦ 6= I3 . Therefore, A is not an orthogonal matrix. (c) AA = ⎣ 27 −81 −27 91

(e) Straightforward computation shows that AA = I4 . Therefore, A−1 = A , and so A is an orthogonal matrix. h √ i h √ i (3) (a) First,  is an orthonormal basis for R2 because it has 2 elements and − 23  12 · 12  23 = 0, °h √ i° °h √ i° ° ° ° ° ° − 23  12 ° = 1, and ° 12  23 ° = 1. So, by Theorem 6.3, h³ h √ i´ ³ h √ i´i h √ i √ [−2 3] · − 23  12  [−2 3] · 12  23 = 2 23+3  3 23−2 . [v] = (c) First,  is an horthonormal basis for Ri4 because it has 4 elements and £1 1 1 1¤ 3 1 1 1 √ √ √ √ 2  − 2  2  2 · 2 3  2 3  − 2 3  − 2 3 = 0, h i i £1 1 1 1¤ £1 1 1 1¤ h √2  √1  √1 √1  √1 · 0 · 0 0 −  −    −   = 0, = 0, 2 2 2 2 2 2 2 2 6 6 6 2 2 i h i h i h i h 3 1 1 1 3 1 1 1 √2  √1  √1 √1  √1 √ √ √ √ √ √ √ √   −  −   −  − · 0 = 0, · 0 0 − = 0, 2 3 2 3 2 3 2 3 6 6 6 2 3 2 3 2 3 2 3 2 2 i h i h °£ ¤° 0 √26  √16  √16 · 0 0 − √12  √12 = 0, ° 12  − 12  12  12 ° = 1, °h °h °h i° i° i° ° ° ° ° ° ° √ 1 1 1 √2  √1  √1 ° = 1, and ° 0 0 − √1  √1 ° = 1. So, by √ √  −  − = 1, 0 ° ° ° 2 3 3  2√ 3 2 3 2 3 6 6 6 2 2 h h¡ i´ £ ¤¢ ³ 3 1 1 1  √  − 2√  − 2√  Theorem 6.3, [v] = [8 4 −3 5] · 12  − 12  12  12  [8 4 −3 5] · 2√ 3 2 3 3 3 i´ ³ h i´i h ³ h √ √ √ i = 3 133 3  5 3 6  4 2 .

[8 4 −3 5] · 0 √26  √16  √16  [8 4 −3 5] · 0 0 − √12  √12

(4) (a) First, let v1 = [5 −1 2]. Next,

¶ [2 −1 −4] · [5 −1 2] [5 −1 2] v2 = [2 −1 −4] − [5 −1 2] · [5 −1 2] ¤ £ 3 9 10 [5 −1 2] = 32  − 10  − 21 = [2 −1 −4] − 30 5  We multiply v2 by 3 to simplify its form, yielding v2 = [5 −3 −14]. Hence, an orthogonal basis for the subspace is {[5 −1 2] [5 −3 −14]}. µ

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

194

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

(c) First, let v1 = [2 1 0 −1]. Next,

Section 6.1

¶ [1 1 1 −1] · [2 1 0 −1] [2 1 0 −1] [2 1 0 −1] · [2 1 0 −1] ¤ £ = [1 1 1 −1] − 46 [2 1 0 −1] = − 13  13  1 − 13  Multiplying by 3 to eliminate fractions yields v2 = [−1 1 3 −1]. Finally, ¶ µ [1 −2 1 1] · [2 1 0 −1] v3 = [1 −2 1 1] − [2 1 0 −1] [2 1 0 −1] · [2 1 0 −1] µ ¶ [1 −2 1 1] · [−1 1 3 −1] − [−1 1 3 −1] [−1 1 3 −1] · [−1 1 3 −1] v2 = [1 1 1 −1] −

µ

£5 7 5 3¤ (−1) = [1 −2 1 1] − (−1) 6 [2 1 0 −1] − 12 [−1 1 3 −1] = 4  − 4  4  4 . Multiplying by 4 to eliminate fractions produces v3 = [5 −7 5 3]. Hence, an orthogonal basis for the subspace is {[2 1 0 −1] [−1 1 3 −1] [5 −7 5 3]}.

(5) (a) First, we must find an ordinary basis for R3 containing [2 2 −3]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the vector we have and use the Independence Test Method. Hence, we row reduce ⎤ ⎡ ⎡ ⎤ 1 0 0 − 13 2 1 0 0 ⎢ 2 ⎥ ⎣ 2 0 1 0 ⎦ to ⎣ 0 1 0 3 ⎦ 2 −3 0 0 1 0 0 1 3

giving the basis {[2 2 −3] [1 0 0] [0 1 0]}. We perform the Gram-Schmidt Process on this set.

Let v1 = [2 2 −3]. Then ¶ µ [1 0 0] · [2 2 −3] v2 = [1 0 0] − [2 2 −3] [2 2 −3] · [2 2 −3] ¤ £ 2 4 6 = [1 0 0] − 17 [2 2 −3] = 13 17  − 17  17  Multiplying by 17 to eliminate fractions yields v2 = [13 −4 6]. Finally, ¶ µ ¶ µ [0 1 0] · [13 −4 6] [0 1 0] · [2 2 −3] v3 = [0 1 0] − [2 2 −3] − [13 −4 6] [2 2 −3] · [2 2 −3] [13 −4 6] · [13 −4 6] £ 9 6¤ 2 13 = [0 1 0] − 17 [2 2 −3] − (−4) 221 [13 −4 6] = 0 13  13 . Multiplying by 3 to simplify the form of 3 the vector gives us v3 = [0 3 2]. Hence, an orthogonal basis for R containing [2 2 −3] is {[2 2 −3] [13 −4 6] [0 3 2]}.

(c) First, we must find an ordinary basis for R3 containing [1 −3 1] and [2 5 13]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the given vector, and use the Independence Test Method. Hence, we row reduce ⎡ ⎡ ⎤ 5 ⎤ 1 0 0 − 13 44 44 1 2 1 0 0 ⎢ 3 ⎥ 1 ⎣ −3 5 0 1 0 ⎦ to ⎣ 0 1 0 44 44 ⎦  1 1 13 0 0 1 0 0 1 − 14 4 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

195

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.1

giving the basis {[1 −3 1] [2 5 13] [1 0 0]}. We perform the Gram-Schmidt Process on this set.

Let v1 = [1 −3 1]. Since we know that the given set of vectors is orthogonal, we may simply use v2 = [2 5 13]. (If the two given vectors were not already orthogonal, we would need to apply Gram-Schmidt to obtain v2 .) Finally, ¶ µ ¶ µ [1 0 0] · [2 5 13] [1 0 0] · [1 −3 1] [1 −3 1] − [2 5 13] v3 = [1 0 0] − [1 −3 1] · [1 −3 1] [2 5 13] · [2 5 13] ¤ £ 1 2 = [1 0 0] − 11 [1 −3 1] − 198 [2 5 13] = 89  29  − 29  Multiplying by 92 to simplify the form of the vector gives v3 = [4 1 −1]. Hence, an orthogonal basis for R3 containing the two given vectors is {[1 −3 1] [2 5 13] [4 1 −1]}.

(e) First, we must find an ordinary basis for R4 containing [2 1 −2 1]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R4 to the given vector and use the Independence Test Method. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 2 1 0 0 0 1 0 0 0 1 ⎢ 1 0 1 0 0 ⎥ ⎢ 0 1 0 0 −2 ⎥ ⎢ ⎥ ⎢ ⎥ to ⎣ −2 0 0 1 0 ⎦ ⎣ 0 0 1 0 −1 ⎦  1 0 0 0 1 0 0 0 1 2

giving us the basis {[2 1 −2 1] [1 0 0 0] [0 1 0 0] [0 0 1 0]}. We perform the Gram-Schmidt Process on this set.

Let v1 = [2 1 −2 1]. Then, ¶ µ [1 0 0 0] · [2 1 −2 1] [2 1 −2 1] v2 = [1 0 0 0] − [2 1 −2 1] · [2 1 −2 1] ¤ £ 2 = [1 0 0 0] − 10 [2 1 −2 1] = 35  − 15  25  − 15  Multiplying by 5 to eliminate fractions yields v2 = [3 −1 2 −1]. Next, ¶ µ [0 1 0 0] · [2 1 −2 1] [2 1 −2 1] v3 = [0 1 0 0] − [2 1 −2 1] · [2 1 −2 1] µ ¶ [0 1 0 0] · [3 −1 2 −1] − [3 −1 2 −1] [3 −1 2 −1] · [3 −1 2 −1] £ 5 1 1¤ 1 = [0 1 0 0] − 10 [2 1 −2 1] − (−1) 15 [3 −1 2 −1] = 0 6  3  − 6  Multiplying by 6 to eliminate fractions gives us v3 = [0 5 2 −1]. Finally, ¶ µ [0 0 1 0] · [2 1 −2 1] [2 1 −2 1] v4 = [0 0 1 0] − [2 1 −2 1] · [2 1 −2 1] µ ¶ [0 0 1 0] · [3 −1 2 −1] − [3 −1 2 −1] [3 −1 2 −1] · [3 −1 2 −1] µ ¶ [0 0 1 0] · [0 5 2 −1] − [0 5 2 −1] [0 5 2 −1] · [0 5 2 −1] ¤ £ 2 2 1 2 = [0 0 1 0] − (−2) 10 [2 1 −2 1] − 15 [3 −1 2 −1] − 30 [0 5 2 −1] = 0 0 5  5  Multiplying by 5 to eliminate fractions produces v4 = [0 0 1 2]. Hence, an orthogonal basis for R4 containing the given vector is {[2 1 −2 1] [3 −1 2 −1] [0 5 2 −1] [0 0 1 2]}.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

196

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.1

(7) In each part, let A represent the given matrix. Then, according to the directions for the problem, a vector in the direction of the axis of rotation is found by solving for an eigenvector for A corresponding to the eigenvalue  = 1. (a) First, we must verify that A is an orthogonal matrix. Calculating A and performing matrix multiplication easily shows that AA = I3 . Therefore, A−1 = A , and so A is orthogonal.

Basketweaving quickly verifies that |A| = 1. Finally, we need to find a fundamental eigenvector for A corresponding to  = 1. To do this, we row reduce ¯ ⎡ ⎤ ⎡ ⎤ 9 6 9 ¯ 1 ¯¯ − 11 1 0 11 11 ¯ 0 3 ¯ 0 ⎢ ⎥ 9 2 ¯ 5 ⎣ 0 1 −1 ¯ 0 ⎦  − 11 [1I3 − A | 0] = ⎣ 11 to ¯ 0 ⎦ 11 ¯ ¯ 5 ¯ 6 7 0 0 0 ¯ 0 − − 0 11

11

11

Setting the independent variable 3 equal to 3 to eliminate fractions produces the fundamental eigenvector [−1 3 3]. This is a vector along the desired axis of rotation.

(c) First, we must verify that A is an orthogonal matrix. Calculating A and performing matrix multiplication easily shows that AA = I3 . Therefore, A−1 = A , and so A is orthogonal.

Basketweaving quickly verifies that |A| = 1. Finally, we need to find a fundamental eigenvector for A corresponding to  = 1. To do this, we row reduce ¯ ⎡ 1 ⎤ ¯ ⎡ ⎤ − 27 − 37 ¯¯ 0 7 1 0 −5 ¯¯ 0 ⎢ 3 ¯ ⎥ 13 2 ⎣ 0 1 −1 ¯ 0 ⎦  [1I3 − A | 0] = ⎣ − 7 to 7 7 ¯¯ 0 ⎦ ¯ 13 ¯ 0 0 0 ¯ 0 0 − 27 − 37 7

Setting the independent variable 3 equal to 1 produces the fundamental eigenvector [5 1 1]. This is a vector along the desired axis of rotation.

(8) (b) No. The vectors might not be unit vectors after scalar multiplication. For example, if we multiply every vector in the orthonormal set {[1 0] [0 1]} by 2, we get {[2 0] [0 2]}. This new set is orthogonal, but it is not orthonormal because each of its elements has length 2.

(11) (a) To prove that A−1 is orthogonal, we must show that its transpose equals its inverse. But, (A−1 )

= (A )−1 = (A−1 )−1

by part (4) of Theorem 2.12 since A is orthogonal.

(b) We must prove that (AB) = (AB)−1 . To do that, we multiply (AB) by (AB) and show that we get I. Now, AB(AB)

= = = = =

AB(B A ) A(BB )A AIA AA I

Hence (AB) = (AB)−1 .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

197

by Theorem 1.18 by part (1) of Theorem 1.16 since B is orthogonal since A is orthogonal.

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.1

(16) (b) By part (1) of Theorem 6.7, a 3×3 matrix is orthogonal if and only if its rows form an orthonormal basis for R3 . Thus, we must find an orthonormal basis containing √16 [1 2 1]. To do this, we will first find an orthogonal basis containing [1 2 1] and then normalize the vectors afterwards.

First, we must find an ordinary basis for R3 containing [1 2 1]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the vector we have and then use the Independence Test Method. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 1 0 0 1 0 0 1 ⎣ 2 0 1 0 ⎦ ⎣ 0 1 0 −1 ⎦  to 1 0 0 1 0 0 1 −2 giving the basis {[1 2 1] [1 0 0] [0 1 0]}.

Next, we perform the Gram-Schmidt Process on this set. Let v1 = [1 2 1]. Then ¶ µ [1 0 0] · [1 2 1] [1 2 1] v2 = [1 0 0] − [1 2 1] · [1 2 1] ¤ £ = [1 0 0]− 16 [1 2 1] = 56  − 13  − 16 . Multiplying by 6 to eliminate fractions yields v2 = [5 −2 −1].

Next, ¶ µ ¶ µ [0 1 0] · [5 −2 −1] [0 1 0] · [1 2 1] [1 2 1] − [5 −2 −1] v3 = [0 1 0] − [1 2 1] · [1 2 1] [5 −2 −1] · [5 −2 −1]

£ 1 2¤ = [0 1 0] − 26 [1 2 1] − (−2) 30 [5 −2 −1] = 0 5  − 5 . Multiplying by 5 to eliminate fractions gives us v3 = [0 1 −2]. Normalizing v1 , v2 , and v3 produces the orthonormal basis ¸ ∙ ¸ ∙ ¸¾ ½∙ 5 1 1 2 1 2 1 2 √  √  √  √  −√  −√  0 √  − √ 6 6 6 30 30 30 5 5 for R3 . Using these vectors as rows for a matrix, rationalizing denominators and simplifying gives us the orthogonal matrix √ √ ⎤ ⎡ √6 6 6 6

⎢ √ ⎢ 30 ⎣ 6 0

3 √ 30 − 15 √ 5 5

whose first row is the given vector √16 [1 2 1].

6 √ 30 − 30 √ −255

⎥ ⎥ ⎦

(17) Proof of the other half of Part (1) of Theorem 6.7: Suppose that the rows of A form an orthonormal basis for R . Then, ( ) entry of AA

= (th row of A) · (th column of A ) ½ 1 = = (th row of A) · (th row of A) = 0 = 6 

since the rows of A form an orthonormal set. Hence AA = I , and A is orthogonal.

Proof of Part (2): A is orthogonal if and only if A is orthogonal (by part (2) of Theorem 6.6) if and only if the rows of A form an orthonormal basis for R (by part (1) of Theorem 6.7) if and only if the columns of A form an orthonormal basis for R . c 2016 Elsevier Ltd. All rights reserved.

Copyright °

198

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.2

(22) (a) False. The remaining vectors in the set must also be perpendicular to each other. For example, the set {[0 0] [1 1] [2 3]} contains the zero vector but is not orthogonal because [1 1] · [2 3] = 5 6= 0. (b) True. This is because e · e = 0 for  6= , and ke k = 1 for all . (c) True. This is part of Theorem 6.3.

(d) False. If the set of vectors {w1      w } is not orthogonal at the outset, adding more vectors will not make the original vectors perpendicular to each other. For example, the linearly independent set {[1 1 0] [1 0 1]} can not be enlarged to an orthogonal basis for R3 , because any such enlargement {[1 1 0] [1 0 1] v3 } will still have [1 1 0] · [1 0 1] = 1 6= 0. (e) True. By Theorem 4.13, W has a finite basis . Also, ||  0, since W is nontrivial. Then, by Theorem 6.4, the Gram-Schmidt Process can be used on the basis  to find an orthogonal basis for W.

(f) True. Since A A = I , Theorem 2.10 tells us that A−1 = A . Hence, A is orthogonal by definition.

(g) True. This follows directly from parts (3) and (1) of Theorem 6.6. (h) False. According to Theorem 6.7, the rows or columns of a matrix must be orthonormal, not just orthogonal, in order to guarantee that the matrix is orthogonal. So, for example, the rows of ∙ ¸ 1 2 A= form an orthogonal set, but A is not orthogonal, since A A = 5I2 6= I2 .

2 −1

(i) True. By part (1) of Theorem 6.7, the rows of A form an orthonormal basis for R . However, the set of rows of (A) is the same as the set of rows of A, only listed in a different order. However, these vectors form an orthonormal basis for R regardless of the order in which they are listed.

Hence, part (1) of Theorem 6.7 asserts that (A) is also an orthogonal matrix.

(j) True. This is the statement of Theorem 6.8.

Section 6.2 (1) If you need assistance combining the Gram-Schmidt Process with the Enlarging Method of Section 4.6, please review Examples 4 and 5 in Section 6.1 of the textbook and Exercise 5 in Section 6.1. Solutions for the starred parts for Exercise 5 appear above in this manual. (a) The vector [2 3], found by inspection, is clearly orthogonal to [3 −2]. Hence [2 3] ∈ W ⊥ . Now,  W ⊥ . Therefore, W ⊥ W ⊥ can not be 2-dimensional, or it would be all of R2 , and yet [3 −2] ∈ ⊥ must be 1-dimensional, and so W = span({[2 3]}). Note that dim(W) + dim(W ⊥ ) = 1 + 1 = 2 = dim(R2 ). (c) First, we find an orthogonal basis for W by performing the Gram-Schmidt Process on its basis {[1 4 −2] [2 1 −1]}. This results in the orthogonal basis  = {[1 4 −2] [34 −11 −5]} for W. Next, we enlarge this set to a basis for R3 using the Enlarging Method of Section 4.6. This results in the basis  = {[1 4 −2] [34 −11 −5] [1 0 0]} for R3 . Now we perform the Gram-Schmidt Process on the basis  to find an orthogonal basis  for R3 . Note that since the first two vectors in  are already perpendicular to each other, they will not be changed by the Gram-Schmidt Process.

Completing the Gram-Schmidt Process produces the basis  = {[1 4 −2] [34 −11 −5] [2 3 7]}.

Since  is an orthogonal basis for W, Theorem 6.12 shows that {[2 3 7]} is an orthogonal basis for W ⊥ . Hence, W ⊥ = span({[2 3 7]}). Note that dim(W) + dim(W ⊥ ) = 2 + 1 = 3 = dim(R3 ).

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

199

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.2

(e) Looking at the given equation for the plane W, we see that W is defined to be {[  ] | [  ] · [−2 5 −1] = 0}. That is, W is the set of vectors orthogonal to [−2 5 −1], and hence orthogonal to the subspace Y = span({[−2 5 −1]}) of R3 . So, by definition of orthogonal ¡ ¢⊥ complement, W = Y ⊥ . Hence, by Corollary 6.14, W ⊥ = Y ⊥ = Y = span({[−2 5 −1]}) Note that dim(W) + dim(W ⊥ ) = 2 + 1 = 3 = dim(R3 ). (f) First, we find an orthogonal basis for W by performing the Gram-Schmidt Process on its basis {[1 −1 0 2] [0 1 2 −1]}. This results in the orthogonal basis  = {[1 −1 0 2] [1 1 4 0]} for W. Next, we enlarge this set to a basis for R4 using the Enlarging Method of Section 4.6.

This results in the basis  = {[1 −1 0 2] [1 1 4 0] [1 0 0 0] [0 1 0 0]} for R4 . Now we perform the Gram-Schmidt Process on the basis  to find an orthogonal basis  for R4 . Note that since the first two vectors in  are already perpendicular to each other, they will not be changed by the Gram-Schmidt Process. Completing the Gram-Schmidt Process produces the basis  = {[1 −1 0 2] [1 1 4 0] [7 1 −2 −3] [0 4 −1 2]}. Since  is an orthogonal basis for W, Theorem 6.12 shows that {[7 1 −2 −3] [0 4 −1 2]} is an orthogonal basis for W ⊥ . Hence, W ⊥ = span({[7 1 −2 −3] [0 4 −1 2]}). Note that dim(W) + dim(W ⊥ ) = 2 + 2 = 4 = dim(R4 ).

(2) (a) First we use the Gram-Schmidt Process on the basis {[1 −2 −1] [3 −1 0]} for W to find the following orthogonal basis for W: {[1 this orthogonal basis for o n −2 −1] [13 4 5]}. Normalizing 1 [13 4 5] . Hence, by definition, W produces the orthonormal basis √16 [1 −2 −1] √210 µ µ ¶¶ µ ¶ 1 1 √ [1 −2 −1] w1 = projW v = [−1 3 2] · √ [1 −2 −1] 6 6 ¶¶ µ ¶ µ µ 1 1 √ [13 4 5] [13 4 5] + [−1 3 2] · √ 210 210 £ 33 111 12 ¤ 1 1 = 6 (−9)[1 −2 −1] + 210 (9)[13 4 5] = − 35  35  7  ¤ £ 2 ¤ £ 111 12 6 2 Finally, w2 = projW ⊥ v = v − projW v = [−1 3 2] − − 33 35  35  7 = − 35  − 35  7 . (b) Looking at the given equation for the plane W, we see that W is defined to be {[  ] | [  ]·[2 −2 1] = 0}. That is, W is the set of vectors orthogonal to [2 −2 1], and hence orthogonal to the subspace Y = span({[2 −2 1]}) of R3 . So, by definition of orthogonal comple¡ ¢⊥ ¡© ª¢ ment, W = Y ⊥ . Now, by Corollary 6.14, W ⊥ = Y ⊥ = Y = span 13 [2 −2 1] , where we have normalized [2 −2 1] to by definition of£ projY v, w¤2 ¡ create an orthonormal ¡ ¢¢basis ¡ 1 for Y. ¢Hence, 1 = projW ⊥ v = projY v = [1 −4 3] · 13 [2 −2 1] [2 −2 1] = (13)[2 −2 1] = 26  − 26  13 .

3 £ 17 109 14 ¤ 9 9 £ 26 269 13 ¤ Finally, w1 = projW v = v − projW ⊥ v = [1 −4 3] − 9  − 9  9 = − 9  − 9  9 . Note that in this problem we found it easier to compute projW ⊥ v first and use it to calculate projW v because it was easier to determine an orthonormal basis for W ⊥ than for W. (4) (a) Let v = [−2 3 1]. By Theorem 6.18, the minimum distance from  to W is kv − projW vk.

Thus, we need to compute projW v. To do this, we need an orthonormal basis for W. We begin with the given basis {[−1 4 4] [2 −1 0]} for W and apply the Gram-Schmidt Process to obtainn {[−1 4 4] [20 −3 8]}. Normalizing the vectors in this basis produces the orthonormal o 1 1 √ √ [−1 4 4] 473 [20 −3 8] for W. Thus, by definition, basis 33 µ µ ¶¶ µ ¶ 1 1 √ [−1 4 4] projW v = [−2 3 1] · √ [−1 4 4] 33 33 ¶¶ µ ¶ µ µ 1 1 √ [20 −3 8] [20 −3 8] + [−2 3 1] · √ 473 473 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

200

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.2

¤ £ 1 1 . Hence, the minimum distance from  = 33 (18)[−1 4 4] + 473 (−41)[20 −3 8] = − 98  105  64 43 43 43 √ ° £ 98 105 64 ¤° °£ 12 24 21 ¤° to W is kv − projW vk = °[−2 3 1] − − 43  43  43 ° = ° 43  43  − 43 ° = 3 43129 

(d) Let v = [−1 4 −2 2]. By Theorem 6.18, the minimum distance from  to W is kv − projW vk = kprojW ⊥ vk. To compute this, we will find an orthonormal basis for W ⊥ . Now W is defined to be {[   ] | [   ]·[2 0 −3 2] = 0}. That is, W is the set of vectors orthogonal to [2 0 −3 2], and hence orthogonal to the subspace Y = span({[2 0 −3 2]}). So, by definition of orthogonal ¡ ¢⊥ ⊥ complement, W = Y ⊥ . Thus, by Corollary 6.14, W ⊥ = Y ⊥ n = Y. Therefore, o a basis for W is {[2 0 −3 2]}. Normalizing, we get the orthonormal basis √117 [2 0 −3 2] for W ⊥ . So, by ³ ³ ´´ ³ ´ √1 [2 0 −3 2] = 1 (8) [2 0 −3 2]. definition, projW ⊥ v = [−1 4 −2 2] · √117 [2 0 −3 2] 17 °178 ° 8√17 ° ° Hence, the minimum distance from  to W is kprojW ⊥ vk = 17 [2 0 −3 2] = 17 .

(5) In each part, let A represent the given matrix. This problem utilizes the methods described in the boxed statements after Examples 8 and 10 in Section 6.2 of the textbook. (a) The first step is to find the eigenvalues of A. Now, ¯ ¯ 3 3 2 ¯ ¯  − 11 11 11 ¯ ¯ ¯ ¯ 3 1  − 10 A () = ¯ ¯ = 3 − 22 +  (by basketweaving) = ( − 1)2  11 11 11 ¯ ¯ 3 1 ¯  − 10 ¯ 11

11

11

Hence, A has eigenvalues 1 = 0 and 2 = 1, as required for  to be an orthogonal projection. ( can not be an orthogonal reflection since neither eigenvalue is −1.) Next, by row reducing [0I3 − A | 0] and [1I3 − A | 0], we find the following bases for the associated eigenspaces: basis for 0 = {[3 1 1]}, basis for 1 = {[−1 3 0] [−1 0 3]}. Since the basis vector [3 1 1] for 0 is perpendicular to both basis vectors for 1 , the operator  whose matrix is A with respect to the standard basis is an orthogonal projection onto the plane determined by 1 . But since 1 = 0⊥ , 1 is the plane consisting of the vectors orthogonal to [3 1 1]. Hence,  is the orthogonal projection onto the plane 3 +  +  = 0.

(d) The first step is to find the eigenvalues of A. Now, ¯ ¯ 7 2 14 ¯ ¯  − 15 15 15 ¯ ¯ ¯ ¯ 7 4  − 14 A () = ¯ ¯ = 3 − 2 −  + 1 (by basketweaving) = ( − 1)2 ( + 1).

15 15 15 ¯ ¯ 4 1 ¯  + 25 ¯ 5 5

Hence, A has eigenvalues 1 = 1 and 2 = −1, as required for  to be an orthogonal reflection. ( can not be an orthogonal projection since neither eigenvalue is 0.) Next, by row reducing [1I3 − A | 0] and [−1I3 − A | 0], we find the following bases for the associated eigenspaces: basis for 1 = {[−1 4 0] [−7 0 4]}, basis for −1 = {[2 1 3]}. However, [2 1 3] · [−1 4 0] = 2 6= 0.

Therefore, the eigenspaces are not orthogonal complements. Hence, the operator  is neither an orthogonal projection nor an orthogonal reflection.

(6) In this problem, we utilize the method described in the boxed statement after Example 8 of Section 6.2 in the textbook.

Let W = {[  ] | 2− +2 = 0} = {[  ] | [  ]·[2 −1 2] = 0}. Then, if Y = span({[2 −1 2]}), W = Y ⊥ , and, by Corollary 6.14, W ⊥ = Y. So, {[2 −1 2]} is a basis for W ⊥ , and by the Inspection Method of Section 4.6, {[1 0 −1] [1 2 0]} is a basis for W. Now W = 1 , the eigenspace corresponding to the eigenvalue 1 = 1, for , since the plane W is left fixed by . Also, W ⊥ = 0 , the eigenspace c 2016 Elsevier Ltd. All rights reserved.

Copyright °

201

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.2

corresponding to the eigenvalue 2 = 0 for , since all vectors perpendicular to W are sent to 0 by . So, we know by the Generalized Diagonalization Method of Section 5.6 that if A is the matrix for  with respect to the standard basis for R3 , then D = P−1 AP, where P is a matrix whose columns are a basis for R3 consisting of fundamental eigenvectors for , and D is a diagonal matrix with the eigenvalues for ⎤on its main diagonal. Note that D = P−1 AP implies that A = ⎡PDP−1 . Setting ⎡ ⎤ 0 0 0 2 1 1 0 2 ⎦.

D = ⎣ 0 1 0 ⎦ and, using the bases for W ⊥ = 0 and W = 1 from above, P = ⎣ −1 0 0 1 2 −1 0 ⎡ ⎤ ⎡ ⎤ 2 −1 2 5 2 −4 2 ⎦.

Row reduction yields P−1 = 19 ⎣ 4 −2 −5 ⎦. Hence, A = PDP−1 = 19 ⎣ 2 8 1 4 1 −4 2 5

(9) (a) The following argument works for any orthogonal projection onto a plane through the origin in R3 .

An orthogonal projection onto a plane in R3 has two eigenvalues, 1 = 1 and 2 = 0. The eigenspace 1 is the given plane, and thus is 2-dimensional. Hence, the geometric multiplicity of 1 is 2. The eigenspace 0 is the orthogonal complement of the plane, since the vectors perpendicular to the plane are sent to 0 by . By Corollary 6.13, dim(0 ) = 3 − dim(1 ) = 1. Hence, the geometric multiplicity of 2 is 1. Since the geometric multiplicities of the eigenvalues add up to 3,  is diagonalizable, and each geometric multiplicity must equal its corresponding algebraic multiplicity. Therefore, the factor ( − 1) appears in  () raised to the second power, and the factor of  appears in  () raised to the first power. Hence,  () = ( − 1)2  = 3 − 22 + . (c) The following argument works for any orthogonal reflection through a plane through the origin in R3 .

An orthogonal reflection through a plane in R3 has two eigenvalues, 1 = 1 and 2 = −1. The eigenspace 1 is the given plane, and thus is 2-dimensional. Hence, the geometric multiplicity of 1 is 2. The eigenspace −1 is the orthogonal complement of the plane, since the vectors perpendicular to the plane are reflected to their opposites by . By Corollary 6.13, dim(0 ) = 3 − dim(1 ) = 1. Hence, the geometric multiplicity of 2 is 1. Since the geometric multiplicities of the eigenvalues add up to 3,  is diagonalizable, and each geometric multiplicity must equal its corresponding algebraic multiplicity. Therefore, the factor ( − 1) appears in  () raised to the second power, and the factor of ( + 1) appears in  () raised to the first power. Hence,  () = ( − 1)2 ( + 1) = 3 − 2 −  + 1.

(10) (a) We will consider three different approaches to this problem. The third method is the easiest, but it involves using the cross product of vectors in R3 , and so can only be used in R3 and when dim(W) = 2.

Method #1: First we find an orthonormal basis for W. To do this, we need an orthogonal basis, which we get by applying the Gram-Schmidt Process to the given basis {[2 −1 1] [1 0 −3]}.

So, let v1 = [2 −1 1]. Then ¶ µ [1 0 −3] · [2 −1 1] [2 −1 1] v2 = [1 0 −3] − [2 −1 1] · [2 −1 1] £ 4 1 17 ¤ = [1 0 −3] − (−1) 6 [2 −1 1] = 3  − 6  − 6 . Multiplying by 6 to eliminate fractions yields v2 = [8 −1 −17]. We normalize these vectors to find the following orthonormal basis for W: ½ ¾ 1 1 √ [2 −1 1]  √ [8 −1 −17]  6 354 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

202

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.2

Now the matrix for  with respect to the standard basis is the 3 × 3 matrix whose columns are (i), (j), and (k). Using the definition of the projection operator we get µ µ ¶¶ µ ¶ 1 1 √ [2 −1 1] (i) = projW i = i · √ [2 −1 1] 6 6 ¶¶ µ ¶ µ µ 1 1 √ [8 −1 −17] [8 −1 −17] + i· √ 354 354 1 4 1 = 3 [2 −1 1] + 177 [8 −1 −17] = 59 [50 −21 −3] µ µ ¶¶ µ ¶ 1 1 √ √ (j) = projW j = j · [2 −1 1] [2 −1 1] 6 6 ¶¶ µ ¶ µ µ 1 1 √ √ [8 −1 −17] [8 −1 −17] + j· 354 354 =

(−1) (−1) 1 6 [2 −1 1] + 354 [8 −1 −17] = 59 [−21 10 −7]

and (k) = projW k =

=

µ µ ¶¶ µ ¶ 1 1 √ [2 −1 1] k · √ [2 −1 1] 6 6 ¶¶ µ ¶ µ µ 1 1 √ [8 −1 −17] [8 −1 −17] + k· √ 354 354

(−17) 1 1 6 [2 −1 1] + 354 [8 −1 −17] = 59 [−3 −7 58]

⎡

⎤ 50 −21 −3 1 ⎣ −21 10 −7 ⎦.

Therefore, the matrix for  with respect to the standard basis is A = 59 −3 −7 58

Method #2: The subspace W is a plane in R3 , and  is the orthogonal projection onto that plane. The plane W is also the eigenspace 1 for , corresponding to the eigenvalue 1 = 1.

The eigenspace 0 , corresponding to the eigenvalue 2 = 0 is W ⊥ . We are given the basis  = {[2 −1 1] [1 0 −3]} for W, but we also need a basis for W ⊥ . To find this basis for W ⊥ , we first expand the basis  for W to a basis  for R3 using the Enlarging Method, producing  = {[2 −1 1] [1 0 −3] [1 0 0]}. Next, apply the Gram-Schmidt Process to , obtaining the orthogonal basis  = {[2 −1 1] [8 −1 −17] [3 7 1]} for R3 . Now {[2 −1 1] [8 −1 −17]} is a basis for W = 1 , and so Theorem 6.12 shows that {[3 7 1]} is a basis for W ⊥ = 0 . Hence each vector in  is an eigenvector for . Thus, the matrix A for  with respect to the standard ⎡ ⎤ ⎡ ⎤ 1 0 0 2 8 3 basis is given by A = PDP−1 , where D = ⎣ 0 1 0 ⎦ and P = ⎣ −1 −1 7 ⎦. Using row 0 0 0 1 −17 1 ⎡ ⎤ 118 −59 59 1 ⎣ 8 −1 −17 ⎦. Performing the matrix reduction to compute P−1 produces P−1 = 354 18 42 6 product PDP−1 gives the matrix A computed above using Method #1.

Method #3: The subspace W is a plane is R3 , and  is the orthogonal projection onto that plane. The plane W is also the eigenspace 1 for , corresponding to the eigenvalue 1 = 1. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

203

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.2

The eigenspace 0 corresponding to the eigenvalue 2 = 0 is W ⊥ . We are given the basis  = {[2 −1 1] [1 0 −3]} for W, but we also need a basis for W ⊥ . In Exercise 8 of Section 3.1 we defined the cross product, v × w, of two vectors in R3 . In that exercise it is proven that v × w is orthogonal to both v and w. Thus, if we take the cross product of the two vectors in the basis  for W, we will get a vector orthogonal to both of these vectors, which places it in W ⊥ . Hence, [2 −1 1] × [1 0 −3] = [3 7 1] ∈ W ⊥ . But since dim(W ⊥ ) = 1 (by Corollary 6.13), {[3 7 1]} must be a basis for W ⊥ . So now we have a basis  = {[3 7 1] [2 −1 1] [1 0 −3]} for R3 consisting of eigenvectors. Thus, the matrix A for  with respect to the standard basis is given ⎡ ⎤ ⎡ ⎤ 0 0 0 3 2 1 0 ⎦. Using row reduction to

by A = PDP−1 , where D = ⎣ 0 1 0 ⎦ and P = ⎣ 7 −1 0 0 1 1 1 −3 ⎡ ⎤ 3 7 1 1 ⎣ 21 −10 7 ⎦. Performing the matrix product PDP−1 compute P−1 produces P−1 = 59 8 −1 −17 gives the matrix A computed above using Method #1.

(c) (Note: Three solutions using three different methods were presented for part (a) above. However, the third method used there will not work in part (c) because the cross product is not defined in R4 . Hence, we will only solve this problem two ways, corresponding to the first two methods presented in part (a).)

Method #1: First we find an orthonormal basis for W. To do this, we need an orthogonal basis, which we get by applying the Gram-Schmidt Process to the given basis {[1 2 1 0] [−1 0 −2 1]}.

So, let v1 = [1 2 1 0]. Then ¶ µ [−1 0 −2 1] · [1 2 1 0] [1 2 1 0] v2 = [−1 0 −2 1] − [1 2 1 0] · [1 2 1 0] £ 1 ¤ 3 = [−1 0 −2 1] − (−3) 6 [1 2 1 0] = − 2  1 − 2  1 . Multiplying by 2 to eliminate fractions yields v2 = [−1 2 −3 2]. We normalize these vectors to find the following orthonormal basis for W: ½ ¾ 1 1 √ [1 2 1 0] √ [−1 2 −3 2]  6 3 2 Now the matrix for  with respect to the standard basis is the 4 × 4 matrix whose columns are (e1 ), (e2 ), (e3 ) and (e4 ). Using the definition of the projection operator we get µ µ ¶¶ µ ¶ 1 1 √ [1 2 1 0] (e1 ) = projW e1 = e1 · √ [1 2 1 0] 6 6 µ ¶¶ µ ¶ µ 1 1 √ [−1 2 −3 2] √ [−1 2 −3 2] + e1 · 3 2 3 2 =

(−1) 1 1 6 [1 2 1 0] + 18 [−1 2 −3 2] = 9 [2 2 3 −1]

µ

µ

¶¶ µ ¶ 1 1 √ √ (e2 ) = projW e2 = e2 · [1 2 1 0] [1 2 1 0] 6 6 µ ¶¶ µ ¶ µ 1 1 √ [−1 2 −3 2] √ [−1 2 −3 2] + e2 · 3 2 3 2 1 1 1 = 3 [1 2 1 0] + 9 [−1 2 −3 2] = 9 [2 8 0 2] c 2016 Elsevier Ltd. All rights reserved.

Copyright °

204

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.2

µ µ ¶¶ µ ¶ 1 1 √ [1 2 1 0] (e3 ) = projW e3 = e3 · √ [1 2 1 0] 6 6 µ ¶¶ µ ¶ µ 1 1 √ √ [−1 2 −3 2] [−1 2 −3 2] + e3 · 3 2 3 2 =

(−1) 1 1 6 [1 2 1 0] + 6 [−1 2 −3 2] = 3 [1 0 2 −1]

and µ µ ¶¶ µ ¶ 1 1 √ [1 2 1 0] e4 · √ [1 2 1 0] 6 6 µ ¶¶ µ ¶ µ 1 1 √ [−1 2 −3 2] √ [−1 2 −3 2] + e4 · 3 2 3 2 1 1 = (0)[1 2 1 0] + 9 [−1 2 −3 2] = 9 [−1 2 −3 2] ⎡ ⎤ 2 2 3 −1 ⎢ 2 8 0 2 ⎥ ⎥.

Therefore, the matrix for  with respect to the standard basis is A = 19 ⎢ ⎣ 3 0 6 −3 ⎦ −1 2 −3 2 (e4 ) = projW e4 =

Method #2: The subspace W is a two-dimensional subspace of R4 , and  is the orthogonal projection onto that subspace. The subspace W is also the eigenspace 1 for , corresponding to the eigenvalue 1 = 1. The eigenspace 0 corresponding to the eigenvalue 2 = 0 is W ⊥ . We are given the basis  = {[1 2 1 0] [−1 0 −2 1]}} for W, but we also need a basis for W ⊥ . To find this basis for W ⊥ , we first expand the basis  for W to a basis  for R4 using the Enlarging Method, producing  = {[1 2 1 0] [−1 0 −2 1] [1 0 0 0] [0 1 0 0]}. Next, we apply the Gram-Schmidt Process to , obtaining the orthogonal basis  = {[1 2 1 0] [−1 2 −3 2] [7 −2 −3 1] [0 1 −2 −4]} for R4 . Now {[1 2 1 0] [−1 2 −3 2]} is a basis for W = 1 , and so Theorem 6.12 shows that {[7 −2 −3 1] [0 1 −2 −4]} is a basis for W ⊥ = 0 . Hence each vector in  is an eigenvector for . Thus, the matrix A for  with respect to the standard basis is given by A = PDP−1 , where ⎡ ⎤ ⎡ ⎤ 1 0 0 0 1 −1 7 0 ⎢ 0 1 0 0 ⎥ ⎢ 2 2 −2 1 ⎥ −1 ⎥ ⎢ ⎥ D = ⎢ ⎣ 0 0 0 0 ⎦ and P = ⎣ 1 −3 −3 −2 ⎦. Using row reduction to compute P 0 0 0 0 0 2 1 −4 ⎡ ⎤ 21 42 21 0 ⎢ ⎥ −7 14 −21 14 1 ⎢ ⎥. Performing the matrix product PDP−1 gives produces P−1 = 126 ⎣ 14 −4 −6 2 ⎦ 0 6 −12 −24 the matrix A computed above using Method #1.

(17) Let  be a spanning set for W, and let v ∈ W ⊥ . If u ∈ , then u ∈ W, and so by definition of W ⊥ , v · u = 0. Hence v is orthogonal to every vector in .

Conversely, suppose v · u = 0 for all u ∈ . Let w ∈ W. Then w = 1 u1 + · · · +  u for some u1      u ∈ . Hence v · w = 1 (v · u1 ) + · · · +  (v · u ) = 0 + · · · + 0 = 0. Thus v ∈ W ⊥ .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

205

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.2

(19) We first prove that  is a linear operator. Suppose {u1      u } is an orthonormal basis for W. Then (v1 + v2 ) = = = = =

projW (v1 + v2 ) ((v1 + v2 ) · u1 )u1 + · · · + ((v1 + v2 ) · u )u ((v1 · u1 ) + (v2 · u1 ))u1 + · · · + ((v1 · u ) + (v2 · u ))u (v1 · u1 )u1 + · · · + (v1 · u )u + (v2 · u1 )u1 + · · · + (v2 · u )u projW (v1 ) + projW (v2 ) = (v1 ) + (v2 )

Similarly, (v) = = = =

projW (v) = ((v) · u1 )u1 + · · · + ((v) · u )u (v · u1 )u1 + · · · + (v · u )u ((v · u1 )u1 + · · · + (v · u )u ) projW (v) = (v)

Hence,  is a linear operator.

Next, we prove that range() = W. First, note that if w ∈ W, then w = w + 0, where w ∈ W and 0 ∈ W ⊥ . Hence, by the uniqueness part of Corollary 6.16, (w) = projW w = w. Therefore, every w in W is in the range of . Similarly, Corollary 6.16 implies that projW v ∈ W for every v ∈ R .

Hence, range() = W. To show W ⊥ = ker(), we first prove that W ⊥ ⊆ ker(). If v ∈ W ⊥ , then since v · w = 0 for every w ∈ W, (v) = projW (v) = (v · u1 )u1 + · · · + (v · u )u = 0u1 + · · · + 0u = 0. Hence, v ∈ ker().

Finally, by the Dimension Theorem, we have dim(ker()) =  − dim(range()) =  − dim(W) = dim(W ⊥ ). Since W ⊥ ⊆ ker(), Theorem 4.13 implies that W ⊥ = ker().

(22) Suppose that  is any point in W and w is the vector from the origin to  . We need to show that kv − wk ≥ kv − projW vk; that is, the distance from  to  is at least as large as the distance from  to the terminal point of projW v. Let a = v − projW v and b = (projW v) − w First, we show that a ∈ W ⊥ . This follows from Corollary 6.16.

Next, we show that b ∈ W. Now, by Corollary 6.16, projW v ∈ W. Also, w ∈ W (since  is in W), and so b = (projW v) − w ∈ W. Finally, kv − wk2

= kv − projW v + (projW v) − wk2

= = = =

2

ka + bk = (a + b) · (a + b) a·a+a·b+b·a+b·b a·a+0+0+b·b kak2 + kbk2 ≥ kak2 = kv − projW vk2 

which completes the proof.

(25) (a) True. This is the definition of W ⊥ .

(b) True. This follows from Theorem 6.10. (c) False. By Theorem 6.11, W ∩ W ⊥ = {0}, for every subspace W of R .

(d) False. If the original basis {b1      b7 } is an orthogonal basis then this is true, but it is false in general. So, for example, if b = e for 1 ≤  ≤ 6, and b7 = e1 + e7 , then {b1      b7 } is a c 2016 Elsevier Ltd. All rights reserved.

Copyright °

206

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.3

basis for R7 . If W = span({b1      b4 }), then b7 ∈  W ⊥ because b7 · b1 = 1 6= 0. In this case, {b5  b6  b7 } is not a basis for W ⊥ . (e) True. This follows from Corollary 6.13. (f) False. The orthogonal complement of a subspace is not a setwise complement of the subspace.

For example, if W is the subspace {[ 0] |  ∈ R} of R2 , then W ⊥ = {[0 ] |  ∈ R}. However, the vector [1 1] of R2 is in neither of these subspaces. (g) True. This is Corollary 6.14. (h) True. The orthogonal complement of a line through the origin was discussed in Example 5 and illustrated in Figure 6.2 in Section 6.2 of the textbook. Since this was found to be a plane through the origin, this discussion and Corollary 6.14, which states that (W ⊥ )⊥ = W, completes the justification of the given statement. (i) True. This is part of Theorem 6.17. (j) False. The eigenvalues for an orthogonal projection are 0 and 1, not −1 and 1. Vectors perpendicular to the plane are sent to 0 by the projection. No vector is sent to its opposite. The matrix for the projection diagonalizes to a matrix with 0, 1, and 1 on the main diagonal. See Example 8 in Section 6.2 of the textbook. (Keep in mind that orthogonal reflections have matrices that diagonalize to a diagonal matrix with −1, 1, 1 on the main diagonal.) (k) True. This follows from Theorem 6.18 and Corollary 6.16. (l) True. This follows from Corollary 6.16.

Section 6.3 (1) (a) The matrix for  with respect to the standard basis is A = is a symmetric operator by Theorem 6.19.

∙

3 2 2 5

¸ . Since A is symmetric, 

[] (d) Let v1 = k[]k and let {v2  v3 } be an orthonormal basis for for the plane  +  +  = 0.

Then  = {v1  v2  v3 } is an orthonormal basis for R3 , and the matrix for  with respect to  is the diagonal matrix with 0, 1, 1 on its main diagonal. Hence,  is an orthogonally diagonalizable linear operator. Therefore,  is symmetric by Theorem 6.22. (e) Rotations in R3 about an axis through an angle of 3 have  = 1 as their only eigenvalue. Also, the eigenspace 1 is 1-dimensional, containing vectors parallel to the axis of rotation. Hence, the sum of the geometric multiplicities of the eigenvalues of  is 1. Since this is less than 3,  is not diagonalizable. Hence  is not orthogonally diagonalizable. Therefore, Theorem 6.22 implies that  is not symmetric. (g) Direct computation shows that (e1 ) = [4 0 3 0], (e2 ) = [0 4 0 3], (e3 ) = [3 0 9 0], and (e4 ) = [0 3 0 9]. The matrix for  with respect to the standard basis is the 4 × 4 matrix whose columns are these four vectors in the order listed. Since this matrix is symmetric,  is a symmetric operator by Theorem 6.19. # "

(2) (a) Let P =

3 5 4 5

4 5 − 35

, the matrix whose columns are the basis vectors for the eigenspaces 1

and −1 . Then the desired matrix A equals PDP−1 , where D is the diagonal matrix with the eigenvalues 1 and −1 on its main diagonal. Theorem 2.14 shows that P−1 = P, and so " #∙ # ¸" 3 ∙ ¸ 4 4 3 1 0 −7 24 5 5 5 5 1 , which is symmetric.

A= 4 = 25 4 0 −1 24 7 −3 −3 5

5

5

5

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

207

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.3

⎡

⎤ 12 12 −3 −2 ⎢ 3 −1 12 24 ⎥ ⎥, the matrix whose columns are the basis vectors for the (d) Let P = ⎢ ⎣ 4 7 0 −12 ⎦ 0 12 4 11

eigenspaces −1 and 1 . Then the desired matrix A equals PDP−1 , where D is the diagonal matrix with the eigenvalues −1, −1, 1 and 1 on its main diagonal. Using row reduction to find ⎡ ⎤ 12 7 1 −12 ⎢ 0 −4 3 12 ⎥ 1 ⎢ ⎥.

P−1 shows that P−1 = 169 ⎣ −11 12 24 −2 ⎦ 4 0 −12 3 ⎡ ⎤ −119 −72 −96 0 ⎢ 119 0 96 ⎥ 1 ⎢ −72 ⎥  which is symmetric.

Hence, A = PDP−1 = 169 ⎣ −96 0 119 −72 ⎦ 0 96 −72 −119

(3) In each part, we follow Steps 2 and 3 of the Orthogonal Diagonalization Method of Section 6.3. ¯ ¯ ¯  − 144 60 ¯¯ ¯ (a) Step 2(a): A () = ¯ = 2 − 169 = ( − 169), giving us eigenvalues 1 = 0 60  − 25 ¯

and 2 = 169. Using row reduction to find particular solutions to the systems [0I2 − A|0] and [169I2 − A|0] as in the Diagonalization Method of Section 3.4 produces the basis {[5 12]} for 0 and {[−12 5]} for 169 .

Step 2(b): Since 0 and 169 are 1-dimensional, the given bases are already orthogonal. To create 1 1 [5 12]} for 0 and { 13 [−12 5]} orthonormal bases, we need to normalize the vectors, yielding { 13 for 169 . ¡1 ¢ 1 Step 2(c):  = 13 [5 12] 13 [−12 5] . ¡1 ¢ 1 [5 12] 13 [−12 5] . Now P is the matrix Step 3: Since  is the standard basis for R2 ,  =  = 13 whose columns are the vectors in , and D is the diagonal matrix whose main diagonal entries ∙ ¸ ∙ ¸ 5 −12 0 0 1 , and D = . are the eigenvalues 0 and 169. Hence, P = 13 12 5 0 169

Now P−1 = P , and so D = P AP, which is easily verified by performing the matrix product. ¯ ¯ 4 ¯ ¯  − 17 − 89 9 9 ¯ ¯ ¯ ¯ 8 17 4 − 9 (c) Step 2(a): A () = ¯ − 9 ¯ = 3 − 52 + 7 − 3 = ( − 1)2 ( − 3), giving us 9 ¯ ¯ 4 4 ¯ ¯  − 11 9 9 9

eigenvalues 1 = 1 and 2 = 3. Using row reduction to find particular solutions to the systems [1I3 − A | 0] and [3I3 − A | 0] as in the Diagonalization Method of Section 3.4 produces the basis {[−1 1 0] [1 0 2]} for 1 and {[−2 −2 1]} for 3 . orthogonal.

To make it an Step 2(b): Since 3 is 1-dimensional, the given basis is already ª © orthonormal basis, we need to normalize the vector, yielding 13 [−2 −2 1] for 3 . Now 1 is 2-dimensional, and so we must use the Gram-Schmidt Process to turn the basis we have into an orthogonal basis. We let v1 = [−1 1 0]. Then ¶ µ [1 0 2] · [−1 1 0] v2 = [1 0 2] − [−1 1 0] [−1 1 0] · [−1 1 0]

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

208

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

=

Section 6.3

£1 1 ¤ the orthogonal basiso 2  2  2 . Multiplying by 2 to simplify produces v2 = [1 1 4]. Thus we have n

1 [1 1 4] . {v1  v2 } for 1 . Normalizing these vectors gives us the orthonormal basis √12 [−1 1 0] 3√ 2 ³ ´ 1 Step 2(c):  = √12 [−1 1 0] 3√ [1 1 4] 13 [−2 −2 1] .

2 ³ ´ 1 Step 3: Since  is the standard basis for R3 ,  =  = √12 [−1 1 0] 3√ [1 1 4] 13 [−2 −2 1]  2 Now P is the matrix whose columns are the vectors in , and D is the diagonal matrix whose main diagonal entries are the eigenvalues 1, 1 and 3. Hence, ⎤ ⎡ 1 ⎡ ⎤ − 23 − √12 3√ 2 1 0 0 ⎥ ⎢ 1 √1 √ − 23 ⎥ and D = ⎣ 0 1 0 ⎦ P=⎢ 2 3 2 ⎦ ⎣ 0 0 3 1 4 √ 0 3 2

3

Now P−1 = P , and so D = P AP, which is verified by performing the matrix product.

(e) Step 2(a): The hint in the textbook indicates that A () = (−2)2 (+3)(−5), giving eigenvalues 1 = 2, 2 = −3, and 3 = 5. Using row reduction to find particular solutions to the systems [2I4 − A | 0], [−3I4 − A | 0] and [5I4 − A | 0] as in the Diagonalization Method of Section 3.4 produces the basis {[3 2 1 0] [−2 3 0 1]} for 2 , {[1 0 −3 2]} for −3 and {[0 −1 2 3]} for 5 .

Step 2(b): Since −3 and 5 are 1-dimensional, the givennbases are already o orthogonal. To make 1 these bases orthonormal, we need to normalize, yielding √14 [1 0 −3 2] for 3 and n o √1 [0 −1 2 3] for 5 . Now 2 is 2-dimensional, and so typically, we would use the Gram14 Schmidt Process to turn the basis for 2 into an orthogonal basis. However, quick inspection shows us that the basis we have is already orthogonal! (This is not the typical situation.)

Thus, o n 1 1 we only need to normalize to get the orthonormal basis √14 [3 2 1 0] √14 [−2 3 0 1] for 2 . ³ ´ Step 2(c):  = √114 [3 2 1 0] √114 [−2 3 0 1] √114 [1 0 −3 2] √114 [0 −1 2 3] .

Step 3: Since  is the standard basis for R3 , ¶ µ 1 1 1 1  =  = √ [3 2 1 0] √ [−2 3 0 1] √ [1 0 −3 2] √ [0 −1 2 3]  14 14 14 14

Now P is the matrix whose columns are the vectors in , and D is the diagonal matrix whose main diagonal entries are the eigenvalues 2, 2, −3, and 5. Hence, ⎡ ⎤ ⎡ ⎤ 3 −2 1 0 2 0 0 0 ⎢ 0 2 1 ⎢ 2 3 0 −1 ⎥ 0 0 ⎥ ⎥ ⎥ P= √ ⎢ and D=⎢ ⎣ ⎦ ⎣ 0 −3 2 0 0 −3 0 ⎦ 14 1 0 1 2 3 0 0 0 5

Now P−1 = P , and so D = P AP, which is easily verified by performing the matrix product. ¯ ¯ ¯  − 11 −2 10 ¯¯ ¯  − 14 −5 ¯¯ = 3 − 152 − 225 + 3375 = ( − 15)2 ( + 15), (g) Step 2(a): A () = ¯¯ −2 ¯ 10 −5  + 10 ¯

giving us eigenvalues 1 = 15 and 2 = −15. Using row reduction to find particular solutions to the systems [15I3 − A | 0] and [−15I3 − A | 0] as in the Diagonalization Method of Section 3.4 produces the basis {[1 2 0] [−5 0 2]} for 15 and {[2 −1 5]} for −15 .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

209

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.3

Step 2(b): Since −15 is 1-dimensional, the given orthogonal. To make the basis n basis is already o 1 orthonormal, we need to normalize, yielding √30 [2 −1 5] for 3 . Now 15 is 2-dimensional, and so we must use the Gram-Schmidt Process to turn the basis for 15 into an orthogonal basis.

We let v1 = [1 2 0]. Then ¶ µ [−5 0 2] · [1 2 0] v2 = [−5 0 2] − [1 2 0] [1 2 0] · [1 2 0] = [−4 2 2]. Thus we have n the orthogonal basis o {v1  v2 } for 1 . Normalizing these vectors gives the orthonormal basis √15 [1 2 0] √16 [−2 1 1] . ³ ´ Step 2(c):  = √15 [1 2 0] √16 [−2 1 1] √130 [2 −1 5] . ³ ´ Step 3: Since  is the standard basis for R3 ,  =  = √15 [1 2 0] √16 [−2 1 1] √130 [2 −1 5] .

Now P is the matrix whose columns are the vectors in , and D is the diagonal matrix whose main diagonal entries are the eigenvalues 15, 15, and −15. Hence, ⎤ ⎡ 1 √ √2 − √26 ⎡ ⎤ 5 30 15 0 0 ⎥ ⎢ ⎢ √1 − √130 ⎥ 0 ⎦ and D = ⎣ 0 15 P = ⎢ √25 ⎥ 6 ⎦ ⎣ 0 0 −15 √1 √5 0 6 30 Now P−1 = P , and so D = P AP, which is easily verified by performing the matrix product.

(4) (a) We use the Orthogonal Diagonalization Method from Section 6.3: Step 1: The plane V is 2-dimensional, and we are given the images under  of two vectors in V. Since the set {[−10 15 6] [15 6 10]} containing these two vectors is linearly independent, it forms a basis for V. A quick inspection shows this basis is already orthogonal. Thus, to find an orthonormal basis for V, we only ¢ need to normalize. Doing so produces the orthonormal basis ¡1 1 [−10 15 6] 19 [15 6 10] for V.  = 19 Next, we need to compute the matrix A for  with respect ¡ 1 to . Since¢V is 12-dimensional, A [−10 15 6] = 19  ([−10 15 6]) = will be a 2×2 matrix.¡Applying  ¢to each vector in  yields  19 1 1 1 1 [50 −18 8] and  [15 6 10] =  ([15 6 10]) = [−5 36 22]. We need to express each of 19 19 19 19 these results in -coordinates. Hence, we row reduce ⎡ 10 15 ¯ 50 ¯ ⎡ ⎤ 5 ⎤ − 19 19 ¯¯ 19 − 19 1 0 ¯¯ −2 2 ⎢ 15 6 ¯ 18 36 ⎥ ⎣ 0 1 ¯ 2 1 ⎦ to obtain ⎣ 19 19 ¯ − 19 19 ⎦ ¯ ¯ 6 8 0 0 ¯ 0 0 10 ¯ 22 19

19

19

19

(Note: In performing the above row reduction by hand, or if entering the matrix into a computer or calculator, it is easier to begin by first performing three Type (I) operations to mentally multiply each row by 19.) Crossing out the last row of all zeroes result and using the ∙ in the above ¸ −2 2 matrix to the right of the bar gives the desired matrix A = .

2 1 ¯ ¯ ¯  + 2 −2 ¯ ¯ = 2 +  − 6 = ( − 2)( + 3), giving eigenvalues 1 = 2 Step 2(a): Now A () = ¯¯ −2  − 1 ¯ and 2 = −3. Using row reduction to find particular solutions to the systems [2I2 − A | 0] and [(−3)I2 − A | 0] as in the Diagonalization Method of Section 3.4 produces the basis {[1 2]} for 2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

210

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.3

and {[−2 1]} for −3 .

To make Step 2(b): Since 2 and −3 are 1-dimensional, the given nbases areoalready orthogonal. n o 1 1 them orthonormal bases, we need to normalize, yielding √5 [1 2] for 2 and √5 [−2 1] for −3 . ³ ´ Step 2(c):  = √15 [1 2] √15 [−2 1] . ¡1 ¢ 1 Step 3: Since  = 19 [−10 15 6] 19 [15 6 10] , we need to reverse the coordinatization isomorphism to calculate the vectors in a basis  for V. First, µ µ ¶ µ ¶¶ 1 1 1 1 1 √ [1 2] 7→ √ 1 [−10 15 6] + 2 [15 6 10] = √ [20 27 26] 19 19 5 5 19 5 Similarly, 1 1 √ [−2 1] 7→ √ 5 5

µ µ ¶ µ ¶¶ 1 1 1 (−2) [−10 15 6] + 1 [15 6 10] = √ [35 −24 −2] 19 19 19 5

Hence, an orthonormal basis for V consisting of eigenvectors for  is ¶ µ 1 1 √ [20 27 26] √ [35 −24 −2]  = 19 5 19 5 Now P is the matrix whose columns are the vectors in , and D, the matrix for  with respect to , is the diagonal matrix whose main diagonal entries are the eigenvalues 2 and −3. Hence, ∙ ¸ ∙ ¸ 1 1 −2 2 0 and D=  P= √ 1 0 −3 5 2 Now P−1 = P , and so D = P AP, which is easily verified by performing the matrix product.

(5) In each part, it is important to orthogonally diagonalize the matrix involved, not just diagonalize it.

Otherwise,by Corollary 6.23, the matrix A obtained might not be symmetric, as desired. ∙ ¸ 119 −108 1 and let  be the standard basis for R2 . First, we must orthogonally (a) Let B = 25 −108 56

diagonalize B. We use Steps 2 and 3 of the Orthogonal Diagonalization Method of Section 6.3. ¯ ¯ 108 ¯ ¯  − 119 ¯ ¯ 25 25 Step 2(a): Now B () = ¯ = 2 − 7 − 8 = ( − 8)( + 1), giving eigenvalues 108 56 ¯¯ ¯  − 25 25

1 = 8 and 2 = −1. Using row reduction to find particular solutions to the systems [8I2 − B | 0] and [(−1)I2 − B | 0] as in the Diagonalization Method of Section 3.4 produces the basis {[−4 3]} for 8 and {[3 4]} for −1 .

Step 2(b): Since 8 and −1 are 1-dimensional, the given© bases are ª already orthogonal. © ª To make them orthonormal bases, we need to normalize, yielding 15 [−4 3] for 8 and 15 [3 4] for −1 . ¢ ¡ Step 2(c):  = 15 [−4 3] 15 [3 4] . ¡ ¢ Step 3: Since  is the standard basis for R2 ,  =  = 15 [−4 3] 15 [3 4] . Now P is the matrix whose columns are the vectors in , and D is the diagonal matrix whose main diagonal entries ∙ ¸ ∙ ¸ −4 3 8 0 1 , and D = . are the eigenvalues 8 and −1. Hence, P = 5 3 4 0 −1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

211

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

−1

Now P





= P , and so B = PDP . Letting D2 =

cube roots of the eigenvalues 8 and −1 gives

∙

2 0 0 −1

Section 6.3

¸ , whose diagonal entries are the

¢3 ¡ B = PD32 P = PD2 (P P)D2 (P P)D2 P = PD2 P  ∙ ¸ 23 −36 1 Therefore, if we let A = PD2 P = 25 , then A3 = B, as desired. −36 2 ⎡ ⎤ 17 16 −16 41 −32 ⎦ and let  be the standard basis for R3 . First, we must orthogonally (c) Let B = ⎣ 16 −16 −32 41

diagonalize B. We use Steps 2 and 3 of the Orthogonal Diagonalization Method of Section 6.3. ¯ ¯ ¯  − 17 −16 16 ¯¯ ¯ 32 ¯¯ = 3 −992 +1539−6561 = (−81)(−9)2 , Step 2(a): Now B () = ¯¯ −16  − 41 ¯ 16 32  − 41 ¯

giving eigenvalues 1 = 81 and 2 = 9. Using row reduction to find particular solutions to the systems [81I3 − B | 0] and [9I3 − B | 0] as in the Diagonalization Method of Section 3.4 produces the basis {[−1 −2 2]} for 81 and {[−2 1 0] [2 0 1]} for 9 .

orthogonal. To make the basis Step 2(b): Since 81 is 1-dimensional, the given ª © basis is already orthonormal, we need to normalize, yielding 13 [−1 −2 2] for 81 . Now 9 is 2-dimensional, and so we must use the Gram-Schmidt Process to turn the basis for 9 into an orthogonal basis.

We let v1 = [−2 1 0]. Then ¶ µ [2 0 1] · [−2 1 0] v2 = [2 0 1] − [−2 1 0] [−2 1 0] · [−2 1 0] £ ¤ = 25  45  1 . Multiplying by 5 to eliminate fractions produces v2 = [2 4 5]. Thus we have an n o 1 [2 4 5] orthogonal basis {v1  v2 } for 9 . Normalizing gives the orthonormal basis √15 [−2 1 0] 3√ 5 for 9 . ³ ´ 1 Step 2(c):  = 13 [−1 −2 2] √15 [−2 1 0] 3√ [2 4 5] .

5 ³ ´ 1 Step 3: Since  is the standard basis for R3 ,  =  = 13 [−1 −2 2] √15 [−2 1 0] 3√ [2 4 5] .

5 Now P is the matrix whose columns are the vectors in , and D is the diagonal matrix whose main diagonal entries are the eigenvalues 81, 9 and 9. Hence, ⎤ ⎡ 1 2 ⎡ ⎤ − 3 − √25 3√ 5 81 0 0 ⎥ ⎢ 2 4 ⎥ √1 √ and D = ⎣ 0 9 0 ⎦ P=⎢ 5 3 5 ⎦ ⎣ −3 0 0 9 5 2 √ 0 3

3 5

⎡

⎤ 9 0 0 Now P−1 = P , and so B = PDP . Letting D2 = ⎣ 0 3 0 ⎦, whose diagonal entries are the 0 0 3 square roots of the eigenvalues 81, 9 and 9 gives ¢2 ¡ B = PD22 P = PD2 (P P)D2 P = PD2 P 

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

212

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.3

⎡

⎤ 11 4 −4 Therefore, if we let A = PD2 P = 13 ⎣ 4 17 −8 ⎦, then A2 = B, as desired. −4 −8 17

(6) For example, the matrix A in Example 7 of Section 5.6 is diagonalizable but not symmetric and hence not orthogonally diagonalizable. ¯ ¯ ∙ ¸ ¯  −  − ¯   ¯ ¯ = ( − )( − ) − 2 = 2 − ( + ) + ( − 2 ). Using (7) If A = , then A () = ¯   − − ¯ ³ ´ p the quadratic formula and simplifying yields the two eigenvalues 1 = 12  +  + ( − )2 + 42 ³ ´ p and 2 = 12  +  − ( − )2 + 42 . Now, D is the diagonal matrix with these eigenvalues on its " # p  +  + ( − )2 + 42 0 1 p main diagonal. Hence, D = 2 .

0  +  − ( − )2 + 42

(8) (b) Since  is symmetric, it is orthogonally diagonalizable by Theorem 6.22. But if 0 is the only eigenvalue of , the geometric multiplicity of  = 0 must be dim(V). Therefore, the eigenspace for 0 must be all of V, and so (v) = 0v = 0 for all v ∈ V. Therefore  must be the zero linear operator.

(13) (e) In Exercise 7 of Section 6.1, we saw that [−1 3 3] is an eigenvector for the matrix A corresponding to the eigenvalue  = 1. Therefore, the axis of rotation is in the direction of the vector [−1 3 3].

First, we enlarge the set {[−1 3 3]} to an orthogonal basis for R3 . To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the given vector, and use the Independence Test Method. Hence, we row reduce ⎡ ⎡ ⎤ 1 ⎤ 1 0 0 3 −1 1 0 0 ⎢ 1 ⎥ ⎣ 3 0 1 0 ⎦ to ⎣ 0 1 0 3 ⎦ 3 0 0 1 0 0 1 −1 giving the basis {[−1 3 3] [1 0 0] [0 1 0]}. We perform the Gram-Schmidt Process on this set.

Let v1 = [−1 3 3]. Then ¶ µ [1 0 0] · [−1 3 3] [−1 3 3] v2 = [1 0 0] − [−1 3 3] · [−1 3 3] £ 18 3 3 ¤ = [1 0 0] − (−1) 19 [−1 3 3] = 19  19  19 

Multiplying by 19 3 to eliminate fractions yields v2 = [6 1 1]. Finally, ¶ µ ¶ µ [0 1 0] · [6 1 1] [0 1 0] · [−1 3 3] v3 = [0 1 0] − [−1 3 3] − [6 1 1] [−1 3 3] · [−1 3 3] [6 1 1] · [6 1 1] ¤ £ 3 1 [−1 3 3] − 38 [6 1 1] = 0 12  − 12  = [0 1 0] − 19

Multiplying by 2 to simplify the form of the vector gives us v3 = [0 1 −1]. Hence, an orthogonal basis for R3 containing [−1 3 3] is {[−1 3 3] [6 1 1] [0 1 −1]}.

Next, we normalize the vectors in this orthogonal basis to get the orthonormal basis ½ ¾ 1 1 1 √ [−1 3 3] √ [6 1 1] √ [0 1 −1]  19 38 2 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

213

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 6.3

We create the orthogonal matrix ⎡

√6 38 √1 38 √1 38

⎢ Q=⎢ ⎣

0 − √119

√1 2 1 √ − 2

√3 19 √3 19

⎤

⎥ ⎥ ⎦

whose columns are the vectors in this orthonormal basis, using the first vector as the last column.

Computing Q AQ yields (approximately) ⎡ ⎤ ⎡ ⎤ 01363636 09906589 0 cos  − sin  0 ⎣ −09906589 01363636 0 ⎦  which is of the form ⎣ sin  cos  0 ⎦  0 0 1 0 0 1

Since cos  is positive and sin  is negative, the angle  is in the fourth quadrant. Hence, cos  = 01363636 implies that  ≈ 278◦ . This is a counterclockwise rotation about the vector [−1 3 3] as you look down from the point (−1 3 3) toward the plane through the origin spanned by [6 1 1] and [0 1 −1].

(14) (a) True. The given condition on  is the definition of a symmetric operator. The statement then follows from Lemma 6.20.

(b) False. A symmetric operator is only guaranteed to have a symmetric matrix with respect to an ordered orthonormal basis, not every ordered basis. For example, the symmetric operator on R2 µ∙ ¸¶ ∙ ¸∙ ¸ ∙ ¸  1 0  1 −1 given by  = has matrix A = with respect to the basis  0 2  0 2 {[1 0] [1 1]}. But A is not symmetric.

(c) True. This is part of Theorem 6.19.

(d) True. This follows from the definition of orthogonally diagonalizable and Theorem 6.22. (e) True. The Orthogonal Diagonalization Method states that D = P−1 AP, and that the matrix P is orthogonal. Thus, P−1 = P . Hence, D = P−1 AP

=⇒ PDP−1 = P(P−1 AP)P−1 =⇒ PDP−1 = I AI =⇒ PDP = A

(f) True. In the Orthogonal Diagonalization Method, P is the transition matrix from an ordered orthonormal basis of eigenvectors for A to standard coordinates. (We can assume we are using standard coordinates here because the problem refers only to a matrix rather than a linear transformation.)

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

214

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 6 Review

Chapter 6 Review Exercises (1) (a) Let v1 = [1 3 −2] v2 = [−1 1 1] v3 = [5 1 4] Then v1 · v2 = 0 v1 · v3 = 0 and v2 · v3 = 0 so  is orthogonal. Then, ¸ ∙ (v · v1 ) (v · v2 ) (v · v3 )   [v] = ||v1 ||2 ||v2 ||2 ||v3 ||2 £ −14 12 84 ¤ = 14  3  42 = [−1 4 2]

(2) (a) First, let v1 = [1 −1 −1 1]. Next, v2

= [5 1 1 5] −

µ

[5 1 1 5] · [1 −1 −1 1] [1 −1 −1 1] · [1 −1 −1 1]

= [5 1 1 5] − 84 [1 −1 −1 1] = [3 3 3 3] 

¶

[1 −1 −1 1]

We divide v2 by 3 to simplify its form, yielding v2 = [1 1 1 1]. Hence, an orthogonal basis for the subspace is {[1 −1 −1 1] [1 1 1 1]}.

(3) First, we must find an ordinary basis for R3 containing [6 3 −6] and [3 6 6]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the given set of vectors, and use the Independence Test Method. Hence, we row reduce ⎤ ⎡ ⎡ ⎤ 1 − 19 1 0 0 9 6 3 1 0 0 1 1 ⎥ ⎣ 3 6 0 1 0 ⎦ to ⎢ ⎣ 0 1 0 9 18 ⎦  1 −6 6 0 0 1 0 0 1 −1 2

giving the basis {[6 3 −6] [3 6 6] [1 0 0]}. We perform the Gram-Schmidt Process on this set.

Let v1 = [6 3 −6]. Since the given set of vectors is orthogonal, we let v2 = [3 6 6]. (If the two given vectors were not already orthogonal, we would need to apply Gram-Schmidt to obtain v2 .) Finally, ¶ µ ¶ µ [1 0 0] · [3 6 6] [1 0 0] · [6 3 −6] v3 = [1 0 0] − [6 3 −6] − [3 6 6] [6 3 −6] · [6 3 −6] [3 6 6] · [3 6 6] ¤ £ 6 3 [6 3 −6] − 81 [3 6 6] = 49  − 49  29  = [1 0 0] − 81 Multiplying by 92 to simplify the form of the vector gives v3 = [2 −2 1]. Hence, an orthogonal basis for R3 containing the two given vectors is {[6 3 −6] [3 6 6] [2 −2 1]}.

(6) (a) First we use the Gram-Schmidt Process on the basis {[8 1 −4] [16 11 −26]} for W to find the following orthogonal basis for W: {[8 this orthogonal basis for ª © 1 −4] [−8 8 −14]}. Normalizing W produces the orthonormal basis 19 [8 1 −4] 19 [−4 4 −7] . Hence, by definition, ¡ ¡ ¢¢ ¡ 1 ¢ w1 = projW v = [2 7 26] · 19 [8 1 −4] 9 [8 1 −4] ¢¢ ¡ 1 ¢ ¡ ¡ + [2 7 26] · 19 [−4 4 −7] 9 [−4 4 −7] 1 1 = 81 (−81)[8 1 −4] + 81 (−162)[−4 4 −7] = [0 −9 18]  Finally, w2 = projW ⊥ v = v − projW v =[2 7 26] − [0 −9 18] = [2 16 8] c 2016 Elsevier Ltd. All rights reserved.

Copyright °

215

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 6 Review

(7) (a) Let v = [1 29 −29 −2]. By Theorem 6.18, the minimum distance from  to W is kv − projW vk.

Thus, we need to compute projW v. To do this, we need an orthonormal basis for W. We begin with the given basis {[2 9 −6 0] [2 5 −12 12]} for W and apply the Gram-Schmidt Process to obtain {[2 −6 12]}. Normalizing the vectors in this basis produces the orthonormal ª © 19 −6 0] [0 −4 [2 9 −6 0] 17 [0 −2 −3 6] for W. Thus, by definition, basis 11 ¡ ¡1 ¢¢ ¡ 1 ¢ projW v = [1 29 −29 −2] · 11 [2 9 −6 0] 11 [2 9 −6 0] ¢¢ ¡ 1 ¢ ¡ ¡ + [1 29 −29 −2] · 17 [0 −2 −3 6] 7 [0 −2 −3 6] 1 1 = 121 (437)[2 9 −6 0] + 49 (17)[0 −2 −3 6] ≈ [7223140 31810255 −22710238 2081633]  Hence, the minimum distance from  to W is kv − projW vk ≈ k[1 29 −29 −2] − [7223140 31810255 −22710238 2081633]k ≈ k[−6223140 −2810255 −6289762 −4081633]k ≈ 10141294 (10) In this problem, we utilize the ideas presented in Example 8 of Section 6.2 in the textbook.

Let W = {[  ] | 2−3 + = 0} = {[  ] | [  ]·[2 −3 1] = 0}. Then, if Y = span({[2 −3 1]}), W = Y ⊥ , and, by Corollary 6.14, W ⊥ = Y. So, {[2 −3 1]} is a basis for W ⊥ . Also, W can be expressed as {[  −2+3]} = {[1 0 −2]+[0 1 3]}, and so {[1 0 −2] [0 1 3]} is a basis for W. Now W = 1 , the eigenspace corresponding to the eigenvalue 1 = 1, for , since the plane W is left fixed by .

Also, W ⊥ = −1 , the eigenspace corresponding to the eigenvalue 2 = −1 for , since all vectors perpendicular to W are sent to their negatives by  because they are reflected through W. So, we know by the Generalized Diagonalization Method of Section 5.6 that if A is the matrix for  with respect to the standard basis for R3 , then D = P−1 AP, where P is a matrix whose columns are a basis for R3 consisting of eigenvectors for , and D is a diagonal matrix with the eigenvalues for  on ⎡ ⎤ −1 0 0 its main diagonal. Note that D = P−1 AP implies that A = PDP−1 . Setting D = ⎣ 0 1 0 ⎦ 0 0 1 ⎡ ⎤ 2 1 0 0 1 ⎦ . Row reduction and, using the bases for W ⊥ = −1 and W = 1 from above, P = ⎣ −3 1 −2 3 ⎡ ⎤ ⎡ ⎤ 10 −15 5 3 6 −2 1 ⎣ 14 0 −28 ⎦. Hence, A = PDP−1 = 17 ⎣ 6 −2 3 ⎦. yields P−1 = 70 6 5 3 −2 3 6

(11) (b) The following argument works for any orthogonal projection onto a line through the origin in R3 : An orthogonal projection onto a line in R3 has two eigenvalues, 1 = 1 and 2 = 0 The eigenspace 1 is the given line and thus is 1-dimensional. Hence, the geometric multiplicity of 1 is 1. The eigenspace 0 is the orthogonal complement of the line, since the vectors perpendicular to the line are sent to 0 by . By Corollary 6.13, dim(0 ) = 3 − dim(1 ) = 2. Hence, the geometric multiplicity of 2 is 2. Since the geometric multiplicities of the eigenvalues add up to 3,  is diagonalizable, and each geometric multiplicity must equal its corresponding algebraic multiplicity. Therefore, the factor (−1) appears in  () raised to the first power, and the factor of  appears in  () raised to the second power. Hence,  () = 2 ( − 1) = 3 − 2 .

(12) (a) ([1 0 0 0]) = [0 3 0 0], ([0 1 0 0]) = [0 0 2 0], ([0 0 1 0]) = [0 0 0 1], and ([0 0 0 1]) =

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

216

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 6 Review ⎡

⎤ 0 0 0 0 ⎢ 3 0 0 0 ⎥ ⎥ [0 0 0 0]. Thus, the matrix for  with respect to the standard basis is ⎢ ⎣ 0 2 0 0 ⎦  which 0 0 1 0 is not symmetric, and so  is not a symmetric operator. ¯ ¯ ¯  + 17 − 13 − 16 ¯¯ 30 15 ¯ ¯ ¯  − 11 − 13 ¯ = 3 − 22 −  + 2 = ( − 2)( − 1)( + 1), giving (13) (a) Step 2(a): A () = ¯ − 13 15 15 ¯ ¯ ¯ −1 −1  − 11 ¯ 6

3

6

us eigenvalues 1 = 1, 2 = 2 and 3 = −1. Using row reduction to find particular solutions to the systems [1I3 − A | 0], [2I3 − A | 0] and [(−1)I3 − A | 0] as in the Diagonalization Method of Section 3.4 produces the bases {[−1 −2 1]}, {[1 2 5]}, and {[−2 1 0]} for 1 , 2 and −1 , respectively.

Step 2(b): Since all three bases are 1-dimensional, they are already orthogonal. o n To make each an orthonormal basis, we need to normalize each vector, yielding the bases √16 [−1 −2 1] , n o n o √1 [1 2 5] , and √1 [−2 1 0] for 1 , 2 and −1 , respectively.

30 5 ³ ´ 1 Step 2(c):  = √6 [−1 −2 1] √130 [1 2 5] √15 [−2 1 0] . ³ ´ Step 3: Since  is the standard basis for R3 ,  =  = √16 [−1 −2 1] √130 [1 2 5] √15 [−2 1 0] .

Now P is the matrix whose columns are the vectors in , and D is the diagonal matrix whose main diagonal entries are the eigenvalues 1, 2 and −1. Hence, ⎤ ⎡ ⎡ ⎤ − √16 √130 − √25 1 0 0 ⎥ ⎢ √2 √2 √1 ⎥ 0 ⎦ and D=⎣ 0 2 P=⎢ 30 5 ⎦ ⎣ − 6 0 0 −1 5 1 √ √ 0 6

30

Now P−1 = P , and so D = P AP, which is verified by performing the matrix product.

(16) (a) True. This is Theorem 6.1, and was proven as Result 7 in Section 1.3. (b) True. This is the step “Let v1 = w1 ” at the beginning of the formal Gram-Schmidt Process described in Section 6.1. You may end up with a scalar multiple of w1 if you decide to multiply by a scalar to put the vector in a simpler form. (c) True. Note that w ∈ span({w1      w−1 }) = span({v1      v−1 }). Thus, Theorem 6.3 shows that ¶ ¶ ¶ µ µ µ w · v1 w · v2 w · v−1 v1 + v2 + · · · + v−1  w = v1 · v1 v2 · v2 v−1 · v−1 and so

µ

w · v1 v1 · v1

− 45

# "

v = w − is the zero vector. "

(d) True. Note that onal matrix.

3 5 4 5

3 5

¶

v1 −

3 5

4 5

− 45

3 5

µ

w · v2 v2 · v2

#

∙

=

1 0 0 1

(e) False. A must also be square. The matrix A = c 2016 Elsevier Ltd. All rights reserved.

Copyright °

217

¶

∙

v2 − · · · − ¸

µ

w · v−1 v−1 · v−1

¶

v−1

as required by the definition of an orthog-

1 0 0 0 1 0

¸

is a counterexample.

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 6 Review

(f) False. The columns of an orthogonal matrix must be unit vectors. The matrix A = a counterexample.

∙

2 0 0 3

¸

is

(g) True. By definition, an orthogonal matrix has its transpose as its inverse. (h) True. v⊥w =⇒ v · w = 0. Theorem 6.9 then implies that 0 = v · w = Av · Aw = (v) · (w), and so (v) ⊥ (w). (i) True. Theorem 6.11 assures us that W ⊥ is a subspace of R for any subspace W of R .

(j) True. Theorem 6.17 implies this since the statement is just one of the two properties of a linear transformation for the function (v) = projW v.

(k) True. First, the uniqueness assertion of Corollary 6.16 guarantees that if w ∈ W, then projW w = w. Thus, W ⊆ range(). But the definition of the projection shows that projW v ∈ W for all v ∈ R , and so range() ⊆ W. Hence range() = W. (l) True. If w ∈ W, then projW ⊥ w = w − projW w = w − w = 0, and so w ∈ ker(). This shows that W ⊆ ker(). Next, if w ∈ ker(), then 0 = projW ⊥ w = w − projW w. Therefore, w = projW w. But projW w ∈ W, and so w ∈ W. This shows that ker() ⊆ W. Hence, ker() = W.

(m) False. The kernel of the projection operator is W ⊥ (see part (l), switching the roles of W and W ⊥ ). So, whenever W ⊥ is nontrivial, the projection operator has a nontrivial kernel, and so its matrix with respect to any basis is singular. Thus, the matrix can not be orthogonal (see part (g)). For a specific counterexample, consider∙ the subspace W = span({i}) in R2 . The matrix ¸ 1 0 for the orthogonal projection onto W is A = , which is not an orthogonal matrix, since 0 0 AA = A 6= I2 .

(n) True. Note that range() = W, and if w ∈ W, then projW w = w (see part (k)). Thus, for every v ∈ R , (v) ∈ W, and ((v)) = (v). Hence,  ◦  = .

(o) True. The eigenvalues are ±1. See Example 10 in Section 6.2 and boxed statement that appears just after it.

(p) True. If v1  v2 ∈ R , then (v1 ) · v2

= = = = = =

(projW v1 ) · (projW v2 + projW ⊥ v2 ) (projW v1 ) · (projW v2 ) + (projW v1 ) · (projW ⊥ v2 ) (projW v1 ) · (projW v2 ) + 0 (projW v1 ) · (projW v2 ) + (projW ⊥ v1 ) · (projW v2 ) (projW v1 + projW ⊥ v1 ) · (projW v2 ) v1 · (v2 )

Hence  is a symmetric operator by definition. As an illustration, note that the matrix for the orthogonal projection operator in Example 8 of Section 6.2 with respect to the standard basis is symmetric.

Note: We could look at this problem from a different perspective. If  = {u1      u } is an orthonormal basis for R , with {u1      u } being an orthonormal basis for W, then the matrix for  with respect to  is the diagonal matrix with  1’s and ( − ) 0’s on its main diagonal.

Therefore  is a symmetric operator by Theorem 6.19.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

218

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 6 Review

(q) False. By Exercise 9 of Section 6.3, the composition of symmetric operators is symmetric if and only if the operators commute. For a specific counterexample, consider the operators 1 and 2 on ∙ ¸ ∙ ¸ 1 −1 3 2 2 and .

R whose respective matrices with respect to the standard basis are −1 2 2 −1 The matrix for 2 ◦ 1 with respect to the standard basis is ∙ ¸ ∙ ¸ ∙ ¸ 3 2 1 −1 1 1 =  2 −1 −1 2 3 −4 which is not symmetric. Therefore, 2 ◦ 1 is not a symmetric operator.

(r) False. The basis  must be an orthonormal basis of eigenvectors, not just any orthonormal basis. ∙ ¸ 1 2 2 For a specific counterexample, consider the symmetric operator on R with matrix A = 2 3 with respect to the standard basis. Of course, the standard basis is an orthonormal basis for R2 .

The matrix for  with respect to this basis is just A, which is not diagonal.

(s) True. By Lemma 6.20, the matrix has at least one eigenvalue. Since the matrix is orthogonally diagonalizable, every eigenvalue must have its geometric and algebraic multiplicities equal to each other. (t) True. This follows from Corollary 6.23. (u) False. A symmetric operator must be orthogonally diagonalizable. So, any orthogonal matrix that is not diagonalizable can not be the matrix for a symmetric operator. For a specific counterexample consider a counterclockwise rotation about the origin in R2 through an angle , where  is not a multiple of 180◦ . Its matrix with respect to the standard basis is the orthogonal matrix ∙ ¸ cos  − sin   sin  cos  which is not diagonalizable, and is not a symmetric matrix. Such rotations are not symmetric operators. (v) True. This is the matrix version of Exercise 11 in Section 6.3.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

219

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.1

Chapter 7 Section 7.1 (1) For extra help with adding and multiplying complex numbers, see Appendix C. (a) [2 +  3 −] + [−1 + 3 −2 +  6] = [2 +  − 1 + 3 3 − 2 +  − + 6] = [1 + 4 1 +  6 − ]

(b) (−8 + 3)[4 2 − 3 −7 + ] = [(−8 + 3)(4) (−8 + 3)(2 − 3) (−8 + 3)(−7 + )] = [−12 − 32 −7 + 30 53 − 29] £ ¤ (d) (−4)[6 − 3 7 − 2 −8] = [−24 + 12 −28 + 8 32] = −24 + 12 −28 + 8 32 = [−24 − 12 −28 − 8 −32] (e) [−2 +  5 − 2¡3 + 4] ¢ · [1 +  4 − ¡ 3 −6] ¢ ¡ ¢ = (−2 + ) 1 +  + (5 − 2) 4 − 3 + (3 + 4) −6 = (−2 + )(1 − ) + (5 − 2)(4 + 3) + (3 + 4)(6) = (−1 + 3) + (26 + 7) + (−24 + 18) = 1 + 28

(3) For extra help with adding and multiplying complex numbers, see Appendix C. ∙ ¸ ∙ ¸ ∙ ¸ 2 + 5 −4 +  9− −3 (2 + 5) + (9 − ) (−4 + ) + (−3) (a) + = −3 − 6 8 − 3 5 + 2 4 + 3 (−3 − 6) + (5 + 2) (8 − 3) + (4 + 3) ∙ ¸ 11 + 4 −4 − 2 = 2 − 4 12 ⎤ ⎡ ⎡ ⎤ 1 +  −2 6 + 4 1− 2 6 − 4 ¡ ¢  ⎦ =⎣ 0 ⎦ 3− 5 3+ 5 (c) C∗ = C = ⎣ 0 10 0 7 + 3 −10 0 7 − 3 ⎡ ⎤ 1− 0 10 ⎦ 3− 0 = ⎣ 2 6 − 4 5 7 + 3 ∙ ¸ ∙ ¸ (−3)(5 − ) (−3)(−) (−3)(−3) −3 − 15 −3 9 (d) (−3)D = = (−3)(2 + 3) (−3)(0) (−3)(−4 + ) 9 − 6 0 3 + 12 ∙ ¸∙ ¸ 2 + 5 −4 +  9− −3 (f) −3 − 6 8 − 3 5 + 2 4 + 3 ∙ ¸ (2 + 5)(9 − ) + (−4 + )(5 + 2) (2 + 5)(−3) + (−4 + )(4 + 3) = (−3 − 6)(9 − ) + (8 − 3)(5 + 2) (−3 − 6)(−3) + (8 − 3)(4 + 3) ¸ ∙ ¸ ∙ 1 + 40 −4 − 14 (23 + 43) + (−22 − 3) (15 − 6) + (−19 − 8) = = 13 − 50 23 + 21 (−33 − 51) + (46 + ) (−18 + 9) + (41 + 12) ⎤ ⎡ ¸ ∙ 5 −  2 + 3 ¡ ¢  5 −  − −3 ⎦ (i) C D∗ = C D = C = C ⎣ − 0 2 + 3 0 −4 +  −3 −4 +  ⎡ ⎤⎡ ⎤ 1+ 0 −10 5 +  2 − 3 ⎦⎣  ⎦ 0 0 = ⎣ −2 3 +  6 + 4 5 7 − 3 −3 −4 − 

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

220

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.1

⎡

⎤ (1 + )(5 + ) + 0() + (−10)(−3) (1 + )(2 − 3) + 0(0) + (−10)(−4 − ) (−2)(2 − 3) + (3 + )(0) + 0(−4 − ) ⎦ = ⎣ (−2)(5 + ) + (3 + )() + 0(−3) (6 + 4)(5 + ) + 5() + (7 − 3)(−3) (6 + 4)(2 − 3) + 5(0) + (7 − 3)(−4 − ) ⎡ ⎤ ⎡ ⎤ (4 + 6) + 0 + 30 (5 − ) + 0 + (−10 + 40) 4 + 36 −5 + 39 ⎦ = ⎣ 1 − 7 −6 − 4 ⎦ (2 − 10) + (−1 + 3) + 0 (−6 − 4) + 0 + 0 =⎣ (26 + 26) + (5) + (−21 + 9) (24 − 10) + 0 + (−31 + 5) 5 + 40 −7 − 5

(4) (b) Let Z be an  ×  complex matrix and let W be an  ×  matrix. Note that (ZW)∗ and W∗ Z∗ are both  ×  matrices. We present two methods of proof that (ZW)∗ = W∗ Z∗ 

First method: Notice that the rule (AB) = B A holds for complex matrices, since matrix multiplication for complex matrices is defined in exactly the same manner as for real matrices, and hence the proof of Theorem 1.18 is valid for complex matrices as well. Thus we have: (ZW)∗ = (ZW) = W Z = (W )(Z ) (by part (4) of Theorem 7.2) = W∗ Z∗ 

Second method: We begin by computing the ( ) entry of (ZW)∗  Now, the ( ) entry of ZW = 1 1 + · · · +   . Hence, ( ) entry of (ZW)∗

= (( ) entry of ZW) = (1 1 + · · · +   ) = 1 1 + · · · +   

Next, we compute the ( ) entry of W∗ Z∗ to show that it equals the ( ) entry of (ZW)∗ . Let A = Z∗ , an  ×  matrix, and B = W∗ , a  ×  matrix. Then  =  and  =  . Hence, ( ) entry of W∗ Z∗

= ( ) entry of BA = 1 1 + · · · +   = 1 1 + · · · +   = 1 1 + · · · +   

Therefore, (ZW)∗ = W∗ Z∗ .

(5) In each part, let A represent the given matrix. (a) The matrix A is skew-Hermitian because all of its main diagonal entries are pure imaginary (their real parts all equal zero) and, for  6= ,  = − .

(b) The matrix A is not Hermitian because 11 is not a real number. Also, A is not skew-Hermitian because 11 is not a pure imaginary number. (c) The matrix A is Hermitian because all of its main diagonal entries are real, and, since all other entries are zero (and hence, for  6= ,  =  ).

(d) The matrix A is skew-Hermitian because all of its main diagonal entries are pure imaginary, and, since all other entries are zero (and hence, for  6= ,  = − ).

(e) The matrix A is Hermitian because all of its main diagonal entries are real, and, for  6= ,  = − .

(9) Let Z be Hermitian. Then ZZ∗ = ZZ = Z∗ Z. Similarly, if Z is skew-Hermitian, Z∗ = −Z, so ZZ∗ = Z(−Z) = −(ZZ) = (−Z)Z = Z∗ Z ¡¢ ¡ ¢ (11) (a) False. Consider that [ −1] · [1 ] =  1 + (−1)  = (1) + (−1)(−) = 2. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

221

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.2

(b) False. When computing the complex dot product, we find the complex conjugates of the entries in the second vector before multiplying corresponding coordinates. However, this is not done when computing each entry of the matrix product of complex matrices. For a specific example, consider the matrices Z and W in Example 2 in Section 7.1. In that example, the (1 1) entry of ZW is −5 − 8. However, the complex dot product of the 1st row of Z with the 1st column of W is (1 − ) (−2) + (2) (−1 + 3) + (−2 + ) (−2 + ) = (1 − ) (2) + (2) (−1 − 3) + (−2 + ) (−2 − ) = (2 + 2) + (6 − 2) + 5 = 13 (c) True. Both are equal to the conjugate transpose of Z. This is stated in Section 7.1 just before Example 3. (d) False. By part (5) of Theorem 7.1, v1 · (v2 ) = (v1 · v2 ), which does not necessarily equal (v1 · v2 ) when  is complex. For example, [1 1] · ([1 1]) = [1 1] · [ ] = 1(  ) + 1(  ) = −2, but ([1 1] · [1 1]) = (1( 1 ) + 1( 1 )) = (2) = 2. (e) False. The matrix H in Example 4 in Section 7.1 is Hermitian but not symmetric. In fact, a Hermitian matrix is symmetric if and only if all of its entries are real. (f) True. In fact, the transpose of a skew-Hermitian matrix K is skew-Hermitian, since ´ ¢ ¡  ¢∗ ³ ¡   K = (K ) = (K∗ ) = (−K) = − K . Hence, K is normal by Theorem 7.4.

Section 7.2

(1) Remember: if  is a complex number, 1 = || 2 . For further help with complex arithmetic, review Appendix C. ¯ ∙ ¸ 3 +  5 + 5 ¯¯ 29 + 33 (a) Start with the augmented matrix .

1 +  6 − 2 ¯ 30 − 12 In the first column, we perform the following operations: ¡3 ¢ 1 1 h1i [or, (I): h1i ← 10 − 10  h1i] (I): h1i ← 3+ (II): h2i ← −(1 + ) h1i + h2i

Next, we pivot in the second column: 1 h2i [or, (I): h2i ← (I): h2i ← 5 − 5

µ

¶ 1 1 +  h2i ] 10 10

3 Hence,  = 28 5 − 5 .

∙

¯ ¸ 1 2 +  ¯¯ 12 + 7  0 5 − 5 ¯ 25 − 31

∙

¯ ¸ 1 2 +  ¯¯ 12 + 7  3 0 1 ¯ 28 5 − 5

¢ 1 13 ¡ 3 Back substitution yields  = (12 + 7) − (2 + ) 28 5 − 5  = 5 + 5 . ¯ ⎡ ⎤ 3 −6 + 3 12 + 18 ¯¯ −51 + 9 25 − 2 ¯¯ −13 + 56 ⎦ , (c) Beginning with the augmented matrix ⎣ 3 + 2 1 + 7 1+ 2 9 +  ¯ −7 + 17 ⎡ ⎤ 1 0 4 − 3 2 + 5 − 5 + 2 ⎦. Thus, the solution set is we row reduce to obtain ⎣ 0 1 0 0 0 0 {( (2 + 5) − (4 − 3) (5 + 2) +  ) |  ∈ C}

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

222

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.2

(e) After applying the row reduction method to the first column, the matrix changes from ¯ ∙ ¸ ∙ ¸ 3 − 2 12 + 5 ¯¯ 3 + 11 1 2 + 3 −1 + 3 to  5 + 4 −2 + 23 ¯ −14 + 15 0 0 3 + 4 The second row shows that the system has no solutions.

(2) (b) Using a³cofactor expansion on the row yields ´ first ³

´ ³ ´ |A| =  (1 − )(2 − ) − ()(−2) − (2) (1 + )(2 − ) − ()(4) + (5) (1 + )(−2) − (1 − )(4)

= (1 − ) − (2)(3 − 3) + (5)(−6 + 2) = −15 − 23. Hence, |A| = −15 + 23. ¯ ¯ ¯ − 1 −  4 ¯¯ ¯ 1 +  −2 ¯¯.

Next, |A∗ | = ¯¯ 2 ¯ −5 − 2 +  ¯

Again, ³ using a cofactor expansion ´ on the first ³ row, this determinant ´ equals ³ ´ (−) (1 + )(2 + ) − (−2)(−) − (1 − ) (2)(2 + ) − (−2)(−5) + 4 (2)(−) − (1 + )(−5) = (−)(1 + ) − (1 − )(4 − 8) + 4(−5 + 3) = −15 + 23.

(3) In each part, let A represent the given matrix. Your computations may produce complex scalar multiples of the vectors given here, although they might not be immediately recognized as such. ¯ ¯ ¯  − (4 + 3) ¯ 1 + 3 ¯ (a) A () = ¯¯ −8 + 2  + (5 + 2) ¯ = ( − (4 + 3)) ( + (5 + 2)) − (−8 + 2)(1 + 3) = 2 + (1 − ) −  = ( − )( + 1).

Hence, the eigenvalues are 1 =  and 2 = −1. Now we solve for corresponding fundamental eigenvectors. ¯ " # ¯ ∙ ¸ 1 − 12 − 12  ¯¯ 0 −4 − 2 1 + 3 ¯¯ 0 For 1 = , we row reduce [I2 − A|0] = to . ¯ −8 + 2 5 + 3 ¯ 0 ¯ 0 0 0

Setting the independent variable 2 in the associated system equal to 2 to eliminate fractions yields the fundamental eigenvector [1 +  2]. Hence,  = {[1 +  2] |  ∈ C}. ¯ # ¯ ∙ ¸ " 7 6 − 17  ¯¯ 0 1 − 17 −5 − 3 1 + 3 ¯¯ 0 For 2 = −1, we row reduce [(−1)I2 −A|0] = to . ¯ −8 + 2 4 + 2 ¯ 0 ¯ 0 0 0

Setting the independent variable 2 in the associated system equal to 17 to eliminate fractions yields the fundamental eigenvector [7 + 6 17]. Hence, −1 = {[7 + 6 17] |  ∈ C}.

(c) Using cofactor expansion along the first row yields ¯ ¯ ¯  − (4 + 3) 4 + 2 −4 − 7 ¯¯ ¯  − (−2 + 5) −7 + 4 ¯¯ A () = ¯¯ −2 + 4 ¯ 4 + 2 −4 − 2  + (4 + 6) ¯

= ( − (4 + 3)) (( − (−2 + 5))( + (4 + 6)) − (−7 + 4)(−4 − 2)) − (4 + 2) ((−2 + 4)( + (4 + 6)) − (−7 + 4)(4 + 2)) + (−4 − 7) ((−2 + 4)(−4 − 2) − ( − (−2 + 5))(4 + 2))

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

223

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.2

= ( − (4 + 3))(2 + (6 + ) + (2 − 6)) − (4 + 2)((−2 + 4) + (4 + 2)) + (−4 − 7)((−4 − 2) + (−2 + 4)) ¡ 3 ¢ =  + (2 − 2)2 + (−19 − 28) + (−26 + 18) − ((−16 + 12) + (12 + 16)) + ((2 + 36) + (36 − 2)) = 3 + (2 − 2)2 + (−1 − 4) − 2 = ( − )2 ( + 2) (A computer algebra system would help here.) Hence, the eigenvalues are 1 =  and 2 = −2.

Now we solve for corresponding fundamental eigenvectors.

For 1 = , we row reduce ¯ ⎤ −4 − 2 4 + 2 −4 − 7 ¯¯ 0 2 − 4 −7 + 4 ¯¯ 0 ⎦ [I3 − A | 0] = ⎣ −2 + 4 4 + 2 −4 − 2 4 + 7 ¯ 0 ⎡

to

⎡

1 −1

⎣ 0

0 0

0

3 2 +

0 0

¯ ⎤ ¯ 0 ¯ ¯ 0 ⎦ ¯ ¯ 0

Setting the independent variables 2 and 3 in the associated system equal to 1 and 0, respectively, yields the fundamental eigenvector [1 1 0]. Then setting the independent variables 2 and 3 in the associated system equal to 0 and 2 (to eliminate fractions), respectively, gives the fundamental eigenvector [−3 − 2 0 2].

For 2 = −2, we row reduce ¯ ⎡ ⎤ −6 − 3 4 + 2 −4 − 7 ¯¯ 0 −5 −7 + 4 ¯¯ 0 ⎦ [(−2)I3 − A | 0] = ⎣ −2 + 4 4 + 2 −4 − 2 2 + 6 ¯ 0

to

¯ ⎤ 1 0 1 ¯¯ 0 ⎣ 0 1 − ¯ 0 ⎦  ¯ 0 0 0 ¯ 0 ⎡

Setting the independent variable 3 in the associated system equal to 1 yields the fundamental eigenvector [−1  1]. Hence,  = {[(−3 − 2) 0 2] + [1 1 0] |   ∈ C} and −2 = {[−1  1] |  ∈ C}

(4) (a) We use the solution to Exercise 3(a), above. Let A be the matrix given in Exercise 3(a). A is diagonalizable since two fundamental eigenvectors were found. The matrix P is the matrix whose columns are the fundamental eigenvectors we computed. The matrix D is the diagonal matrix whose main diagonal entries are the eigenvalues  and −1. Hence, ∙ ¸ ∙ ¸ ∙ ¸ 1 17 −7 − 6 1 +  7 + 6  0  P= and D =  Now, P−1 = 1+ 2 17 0 −1 3 + 5 −2 1 1 by using Theorem 2.14. Using 3+5 = 34 (3 − 5), we can verify that D = P−1 AP.

(6) (a) True. Performing the row operation (II): h2i ← − h1i + h2i on A produces a matrix B whose second row is zero. Hence, |B| = 0 (perform a cofactor expansion along its second row). But since B was obtained from A by performing a single Type (II) row operation, |A| = |B|. Therefore, |A| = 0. (b) False. Example 2 in Section 7.2 of the textbook illustrates a 3 × 3 matrix A having eigenvalue 3 having algebraic multiplicity 1. What is true, however, is that the sum of all the algebraic multiplicities of all the eigenvalues of a complex  ×  matrix must equal .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

224

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.3

(c) False. If the number of fundamental eigenvectors for some eigenvalue does not equal its algebraic multiplicity, the matrix is still not diagonalizable when thought of as a complex matrix. For example, the real matrix given in Example 7 of Section 3.4 has −2 as an eigenvalue with algebraic multiplicity 2 but only one fundamental eigenvector is produced for this eigenvalue. Hence, this matrix is not diagonalizable, even when thought of as a complex matrix. (d) False. The Fundamental Theorem of Algebra only guarantees that an th degree complex polynomial factors into  linear factors. However, if these factors are not different, the polynomial will not have  distinct roots. For example, 2 − 2 − 1 = ( − )2 has only one root, namely, .

Section 7.3 (2) In each part, let  represent the given set of vectors. (b) The set  is not linearly independent because [−3 + 6 3 9] = 3[2 +  − 3]. Thus, {[2 +  − 3]} is a basis for span(), and so dim(span()) = 1. ⎡ ⎤ 1 0  (d) The matrix formed using the vectors in  as columns row reduces to ⎣ 0 1 −2 ⎦. Therefore, 0 0 0 by the Independence Test Method, {[3 −  1 + 2 −] [1 +  −2 4 + ]} is a basis for span().

Hence, dim(span()) = 2.

(3) In each part, let  represent the given set of vectors. (b) Since [−3+6 3 9] is not a real scalar multiple of [2+ − 3],  is linearly independent. (Consider the 3rd coordinate: 9 6= 3 for any  ∈ R.) Hence,  is a basis for span(), and dim(span()) = 2.

(d) Separating the real and imaginary parts of each vector in , we can think of them as vectors in R6 . That is, rewrite each complex vector [ +   +   +  ] as [      ]. We then create a matrix using each 6-vector as a column. This gives ⎡ ⎤ ⎡ ⎤ 3 1 3 1 0 0 ⎢ −1 ⎢ 0 1 0 ⎥ 1 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 −2 −2 ⎥ ⎢ ⎥ ⎢ ⎥ which row reduces to ⎢ 0 0 1 ⎥  ⎢ 2 ⎥ ⎢ 0 0 0 ⎥ 0 5 ⎢ ⎥ ⎢ ⎥ ⎣ 0 ⎣ 0 0 0 ⎦ 4 3 ⎦ −1 1 −8 0 0 0

The Independence Test Method then shows that  is linearly independent. Therefore,  is a basis for span(), and dim(span()) = 3. ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 0 0 ¯¯  2 3 +  −3 + 5 ¯¯ 3 −  ¯ −5 − 5 ⎦ row reduces to ⎣ 0 1 0 ¯ 1 +  ⎦.

2 (4) (b) The matrix ⎣ −1 + 3 −2 ¯ ¯ 0 0 1 ¯ −1 4 1 −  −5 + 3 ¯ 7 +  Hence, [z] = [ 1 +  −1].

(5) With C2 considered as a real vector space, we use the ordered basis  = ([1 0] [ 0] [0 1] [0 ]). Now [([1 0])] [([0 1])]

= [0 1] = [0 0 1 0] [([ 0])] = [0 −] = [0 0 0 −1] = [1 0] = [1 0 0 0] and [([0 ])] = [− 0] = [0 −1 0 0]

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

225

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

0 0 ⎢ 0 0 Therefore, the matrix for  with respect to  is ⎢ ⎣ 1 0 0 −1

Section 7.3

⎤ 1 0 0 −1 ⎥ ⎥.

0 0 ⎦ 0 0

(8) We present two methods for solving this problem: Method 1: Let  be the standard basis for C2 ,  be the basis {[1 +  −1 + 3] [1 −  1 + 2]} for C2 , ⎡ ⎤ 3− 2+ 1 − 3 ⎦. and  be the standard basis for C3 . Then, by the given values for  on , A = ⎣ 5 − 3 Next, we compute the transition matrix P from  to . Since  is the standard basis, we can do this ∙ ¸ ∙ ¸ 1 + 2 −1 +  1+ 1− . by computing the inverse of . Theorem 2.14 yields P = −3+ 10 1 − 3 1 +  −1 + 3 1 + 2 ⎡ ⎤ −3 +  − 25 − 11 5  ⎢ ⎥ 1 3 ⎥. − Finally, by Theorem 5.6, A = A P = ⎢ 2 − 2 ⎣ ⎦ − 12 + 72 

− 85 − 45 

Method 2: We begin by expressing [1 0] and [0 1] as linear combinations of [1 +  −1 + 3] and [1 −  1 + 2] by row reducing ¯ # " ¯ ∙ ¸ 1 2 1 0 ¯¯ − 12 − 12  1 +  1 −  ¯¯ 1 0 5 − 5  to obtain ¯ −1 + 3 1 + 2 ¯ 0 1 0 1 ¯  − 25 − 15  Therefore,

¡ 1 1 ¢ and − −  [1 +  −1 + 3] + [1 −  1 + 2] ¡ 2 1 ¢ ¡ 1 2 2 2¢ [0 1] = 5 − 5  [1 +  −1 + 3] + − 5 − 5  [1 −  1 + 2] [1 0] =

Hence,

¡ 1 1 ¢ − −  [3 −  5 −] +  [2 +  1 − 3 3] ¤ £ 2 21 3 and = −3 +  2 − 2  − 12 + 72   ¡ 2 1 ¢ ¡1 2 ¢ −  [3 −  5 −] + − 5 − 5  [2 +  1 − 3 3] ([0 1]) = ¤ £ 5 2 5 11 = − 5 − 5  − − 85 − 45   ([1 0]) =

Using ([1 0]) and ([0 1]) as columns produces the same 3 × 2 matrix A as in Method 1.

(9) (a) True. This is the complex analog of Theorem 4.15. (b) False. If v = , then (v) = = (−1) = (−1) = −1. ¢ ¡ (()) However, (v) = () =   = (−) = 1.

(c) True. This follows from the complex analog of Theorem 4.16.

(d) False. Complex vector spaces (and their complex subspaces) must have even dimension as real vector spaces, not necessarily as complex vector spaces. For example, the subspace W = span({[1 0 0]}) of C3 has complex dimension 1.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

226

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.4

Section 7.4 (1) Remember to use the conjugates of the coordinates of the second vector when computing a complex dot product. (a) The given set is not orthogonal because [1 + 2 −3 − ] · [4 − 2 3 + ] = −10 + 10 6= 0. (c) The given set is orthogonal because [2 −1 ] · [1 − −1] = 0, [2 −1 ] · [0 1 ] = 0, and [1 − −1] · [0 1 ] = 0.

(3) (a) First, we must find an ordinary basis for C3 containing [1+  1]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for C3 to the vector we have and use the Independence Test Method. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1+ 1 0 0 1 0 0 1 ⎣  0 1 0 ⎦ to obtain ⎣ 0 1 0 −1 −  ⎦  1 0 0 1 0 0 1 − yielding the basis {[1 +   1] [1 0 0] [0 1 0]}.

We perform the Gram-Schmidt Process on this set. First, let v1 = [1 +   1]. Next, ¶ µ [1 0 0] · [1 +   1] [1 +   1] v2 = [1 0 0] − [1 +   1] · [1 +   1] £ ¤ 1− [1 +   1] = 12  − 14 − 14  − 14 + 14   = [1 0 0] − 4 Multiplying by 4 to eliminate fractions yields v2 = [2 −1 −  −1 + ].

Finally, ¶ µ [0 1 0] · [1 +   1] [1 +   1] v3 = [0 1 0] − [1 +   1] · [1 +   1] µ ¶ [0 1 0] · [2 −1 −  −1 + ] − [2 −1 −  −1 + ] [2 −1 −  −1 + ] · [2 −1 −  −1 + ] £ ¤ (−1 + ) (−) [1 +   1] − [2 −1 −  −1 + ] = 0 12  12   = [0 1 0] − 4 8

Multiplying by 2 to eliminate fractions produces v3 = [0 1 ]. Hence, the desired orthogonal set is {[1 +   1] [2 −1 −  −1 + ] [0 1 ]}. (b) According to part (1) of Theorem 7.7, a 3 × 3 matrix is unitary if and only if its rows form an orthonormal basis for C3 . Normalizing the orthogonal basis we computed in part (a) yields an orthonormal basis whose first vector is a scalar multiple of [1 +   1]. Using the vectors from this orthonormal basis as the rows of a matrix produces the unitary matrix ⎤ ⎡ 1+  1 ⎢ ⎢ ⎢ ⎣

2

2

2

√2 8

−1− √ 8

−1+ √ 8

0

√1 2

√ 2

⎥ ⎥ ⎥ ⎦

(4) For any  ×  matrix B, |B| = |B| by part (3) of Theorem 7.5. Therefore, ¯ ¯2 ¯ ¯ ¯ |A| ¯ = |A| |A| = |A| |A| = |A| |(A) | = |A| |A∗ | = |AA∗ | = |I | = 1 ¯ ¯

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

227

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.4

(7) (a) A is unitary ⇐⇒ A∗ = A−1 ⇐⇒ A = A−1 ⇐⇒ (A ) = (A−1 ) ⇐⇒ (A )∗ = (A−1 ) ⇐⇒ (A )∗ = (A )−1 ⇐⇒ A is unitary. ¯ ¯ ¯  − (1 − 6) 10 + 2 ¯ ¯ ¯ = ( − (1 − 6)) ( − 5) − (10 + 2)(−2 + 10) (10) (b) A () = ¯ −2 + 10 −5 ¯ = 2 + (−6 + 6) + (45 − 126).

The quadratic formula produces the eigenvalues 1 = 9 + 6 and 2 = −3 − 12. Now we solve for corresponding fundamental eigenvectors.

For 1 = 9 + 6, we row reduce " ¯ ∙ ¸ 1 8 + 12 10 + 2 ¯¯ 0 [(9 + 6)I2 − A | 0] = to −2 + 10 4 + 6 ¯ 0 0

1 1 2 − 2

0

¯ # ¯ 0 ¯  ¯ ¯ 0

Setting the independent variable 2 in the associated system equal to 2 to eliminate fractions yields the fundamental eigenvector [−1 +  2].

For 2 = −3 − 12, we row reduce ¯ ¯ ∙ ¸ ∙ ¸ −4 − 6 10 + 2 ¯¯ 0 1 −1 +  ¯¯ 0 [(−3 − 12)I2 − A | 0] = to ¯ 0  −2 + 10 −8 − 12 ¯ 0 0 0

Setting the independent variable 2 in the associated system equal to 1 yields the fundamental eigenvector [1 −  1].

Note that the two fundamental eigenvectors we have computed form an orthogonal basis for and using them as columns for a matrix produces the C2 . Normalizing each " of these vectors #

unitary matrix P =

−1+ √ 6 √2 6

1− √ 3 √1 3

−1

P

. Direct computation shows that ∗

AP = P AP =

∙

9 + 6 0 0 −3 − 12

¸



a diagonal matrix having the eigenvalues of A on its main diagonal. ¯ ¯ ¯ −1 −2 −  −1 + 2 ¯¯ ¯ ¯. By basketweaving, +3  (13) Let A represent the given matrix. Then, A () = ¯¯ −2 +  ¯ ¯ −1 − 2 − −2 ¯ this determinant equals ( − 1)( + 3)( − 2) + (−2 − )()(−1 − 2) + (−1 + 2)(−2 + )(−) − ( − 1)()(−) − (−2 − )(−2 + )( − 2) − (−1 + 2)( + 3)(−1 − 2) 3 = ( − 7 + 6) + (−5) + (−5) − ( − 1) − (5 − 10) − (5 + 15) = 3 − 18 − 8

Testing the divisors of (−8), the constant term, shows that −4 is a root of A (). Then, dividing A () 2 2 by ( + 4) yields √ A () = ( + √4)( − 4 − 2). Finally, using the quadratic √ formula on √  − 4 − 2 gives the roots 2 + 6 and 2 − 6. Hence, the eigenvalues of A are −4, 2 + 6, and 2 − 6, which are all real. (Using a computer algebra system makes these computations much easier.)

(15) (a) False. Instead, every Hermitian matrix is unitarily diagonalizable, by Corollary 7.10. But, for example, the Hermitian matrix given in Exercise 13 of Section 7.4 is not unitary, since its rows do not form an orthonormal basis for C3 . c 2016 Elsevier Ltd. All rights reserved.

Copyright °

228

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.5

(b) True. Since the dot product of two vectors in R is the same as the complex dot product of those vectors (because a real number is its own complex conjugate), then any orthonormal set of  vectors in R is also an orthonormal set of  vectors in C . (c) True. The definition of unitarily diagonalizable is that A is unitarily diagonalizable if and only if there is a unitary matrix P such that P−1 AP is diagonal. But, a matrix U is unitary if and only if U−1 = U∗ . Hence, A is unitarily diagonalizable if and only if there is a unitary matrix P such that P∗ AP is diagonal. Now notice that P is unitary if and only if P is unitary (by Exercise 6(a)) ¡ ¢ if and only if P is unitary (by Exercise 7(a)) if and only if P∗ is unitary. Therefore, letting Q = P∗ then shows that A is unitarily diagonalizable if and only if there is a unitary matrix Q such that QAQ∗ is diagonal, which is what we needed to show. (d) True. If the columns of A form an orthonormal basis for C , then part (2) of Theorem 7.7 implies that A is unitary. Hence, part (1) of Theorem 7.7 shows that the rows of A form an orthonormal basis for C . (e) False. In general, A∗ , not A , is the matrix for the adjoint of  with respect to the standard basis. This is explained in Section 7.4 of the textbook, just after Corollary 7.10.

Section 7.5 (1) (b) hx yi = (Ax) · (Ay) ⎛⎡ ⎤⎡ ⎤⎞ ⎛⎡ ⎤⎡ ⎤⎞ 5 4 2 3 5 4 2 −2 3 1 ⎦ ⎣ −2 ⎦⎠ · ⎝⎣ −2 3 1 ⎦ ⎣ 1 ⎦⎠ = ⎝⎣ −2 1 −1 0 4 1 −1 0 −1

= [15 −8 5] · [−8 6 −3] = −183. p Also, kxk = hx xi. But hx xi = (Ax) · (Ax) ⎛⎡ ⎤⎡ ⎤⎞ ⎛⎡ ⎤⎡ ⎤⎞ 5 4 2 3 5 4 2 3 3 1 ⎦ ⎣ −2 ⎦⎠ · ⎝⎣ −2 3 1 ⎦ ⎣ −2 ⎦⎠ = ⎝⎣ −2 1 −1 0 4 1 −1 0 4

= [15 −8 5] · [15 −8 5] = 314. √ Hence, kxk = 314. ¯ R R  ¯ (3) (b) hf  gi = 0  sin   = (− cos ) ¯ − 0 (− cos )  0 ¯ ¯ R  ¯ ¯ = (− cos ) ¯ + ( sin ) ¯ − 0  sin   0 0 ¯ ¯ ¯ ¯ = (− cos ) ¯ + ( sin ) ¯ − hf  gi.

0

(using integration by parts) (using integration by parts again)

0

Adding hf  gi to both ends of the equation produces ¢¯ ¡ ¢¯ ¡  2 hf  gi = − cos  ¯0 +  sin  ¯0 ¡ ¢ ¡ ¢ = (− cos ) − (−0 cos 0) + ( sin ) − (0 sin 0) =  + 1 Dividing by 2 yields hf  gi = 12 ( + 1). ¯ p ¯ ¡ ¢ R   R  2 1 2 ¯ Next, kf k = hf  f i. But, hf  f i = 0    = 0   = 2  ¯ = 12 2 − 1 .

0 q 1 2 Hence, kf k = 2 ( − 1).

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

229

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.5

p p hx xi =  hx xi (by property (5) of an inner product space) q =  hx xi (by part (3) of Theorem 7.12) p p = ||2 hx xi = || hx xi = || kxk. p (9) (a) The distance between f and g is defined to be kf − gk = h(f − g)  (f − g)i. Now, Z  h(f − g)  (f − g)i = ( − sin ) ( − sin )  Z0  ¡2 ¢  − 2 sin  + sin2   =

(6) kxk =

0

Z 

=

0

We compute each integral individually.

First, Z



0

2

  − 2

Z 

 sin   +

0

Z 

sin2  

0

¯ 3 2  = 13 3 ¯0 = 3 

Next, using integration by parts, Z  Z   sin   = (− cos )|0 − (− cos )  = (− cos )|0 + sin |0 = ( − 0) − (0 − 0) =  0

0

2

Finally, using the identity sin  = 12 (1 − cos 2), Z  Z  ¯ ¡ 1 ¢ 2 1 1 1 1 ¯ sin   = 2 (1 − cos 2)  = 2 ( − 2 sin 2) 0 = 2  − 0 − (0 − 0) = 2  0

0

Combining these produces

3

Hence, kf − gk =

q

3

h(f − g)  (f − g)i = 3 − 2 + 12  = 3 − 3 2  3 3 3 − 2 .

(10) (b) The angle  between x and y is defined to be arccos Now,

µ

¶ hx yi . First we compute hx yi. kxk kyk

hx yi = (Ax) · (Ay) ⎛⎡ ⎤⎡ ⎤⎞ ⎛⎡ ⎤⎡ ⎤⎞ −2 0 1 2 −2 0 1 5 2 ⎦ ⎣ −1 ⎦⎠ · ⎝⎣ 1 −1 2 ⎦ ⎣ −2 ⎦⎠ = ⎝⎣ 1 −1 3 −1 −1 3 3 −1 −1 2 = [−1 9 4] · [−8 11 15] = 167

Next, kxk2

= hx xi = (Ax) · (Ax) ⎛⎡ ⎤⎡ ⎤⎞ ⎛⎡ ⎤⎡ ⎤⎞ −2 0 1 2 −2 0 1 2 2 ⎦ ⎣ −1 ⎦⎠ · ⎝⎣ 1 −1 2 ⎦ ⎣ −1 ⎦⎠ = ⎝⎣ 1 −1 3 −1 −1 3 3 −1 −1 3

= [−1 9 4] · [−1 9 4] = 98 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

230

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.5

Similarly, kyk2

= hy yi = (Ay) · (Ay) ⎛⎡ ⎤⎡ ⎤⎞ ⎛⎡ ⎤⎡ ⎤⎞ −2 0 1 5 −2 0 1 5 2 ⎦ ⎣ −2 ⎦⎠ · ⎝⎣ 1 −1 2 ⎦ ⎣ −2 ⎦⎠ = ⎝⎣ 1 −1 3 −1 −1 2 3 −1 −1 2

= [−8 11 15] · [−8 11 15] = 410

³ ´ √ √ √ Therefore, kxk = 98 and kyk = 410, and so  = arccos √98167 ≈ arccos(083313) ≈ 0586 410 radians, or, 336◦ .

(14) To prove that a set of vectors {v1      v } is orthogonal, we must show that hv  v i = 0 for  6= . ® ­ ® ­ (a) Note that 2   + 1 = (1)(0) + (0)(1) + (0)(1) = 0, 2   − 1 = (1)(0) + (0)(1) + (0)(−1) = 0, and h + 1  − 1i = (0)(0) + (1)(1) + (1)(−1) = 0. Hence, the given set is orthogonal.

(c) Note that h[5 −2] [3 4]i = (5)(3) − (5)(4) − (−2)(3) + (2)(−2)(4) = −15 6= 0. Therefore, the given set is not orthogonal.

(19) We begin with the basis {w1  w2  w3 } for P2 , with w1 = 2 −  + 1, w2 = 1, and w3 = , and apply the Gram-Schmidt Process. (Note: The Enlarging Method is not needed here since it can be seen by inspection that {w1  w2  w3 } = {2 −  + 1 1 } is linearly independent, and since dim(P2 ) = 3, {w1  w2  w3 } is a basis for P2 .)

First, let v1 = w1 = 2 −  + 1. Next, v2 = w2 − But hw2  v1 i = and hv1  v1 i = = Hence,

Z 1

−1

Z 1

−1

hw2  v1 i v1  hv1  v1 i

¯1 (1)(2 −  + 1)  = ( 13 3 − 12 2 + )¯−1 = 83 

(2 −  + 1)(2 −  + 1)  =

Z 1

−1

¯1 ( 15 5 − 12 4 + 3 − 2 + )¯−1 = 22 5  v2 = (1) −

(4 − 23 + 32 − 2 + 1) 

(83) 2 ( −  + 1) (225)

Multiplying by 33 to eliminate fractions yields v2 = −202 + 20 + 13. Finally, v3 = w3 −

hw3  v1 i hw3  v2 i v1 − v2  hv1  v1 i hv2  v2 i

However, hw3  v1 i =

Z 1

−1

()(2 −  + 1)  =

Z 1

−1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

¯1 (3 − 2 + )  = ( 14 4 − 13 3 + 12 2 )¯−1 = − 23  231

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

hw3  v2 i = =

Z 1

2

()(−20 + 20 + 13)  =

−1

¯ 3 13 2 ¯1 40 (−54 + 20 3  + 2  ) −1 = 3 

hv2  v2 i = =

Z 1

Hence, v3 = () −

(−203 + 202 + 13) 

−1

and

(−202 + 20 + 13)(−202 + 20 + 13) 

−1

Z 1

−1

=

Z 1

Section 7.5

(4004 − 8003 − 1202 + 520 + 169) 

¡ 5 ¢¯1 80 − 2004 − 403 + 2602 + 169 ¯−1 = 418

(−23) 2 (403) ( −  + 1) − (−202 + 20 + 13) (225) (418)

Multiplying by 19 to eliminate fractions gives v3 = 152 + 4 − 5. Thus, we get the orthogonal basis {v1  v2  v3 }, with v1 = 2 −  + 1, v2 = −202 + 20 + 13, and v3 = 152 + 4 − 5.

(22) (b) First, we prove part (4) of Theorem 7.19 as follows, using a proof similar to the proof of Theorem 6.12: Let {v1      v } be an orthogonal basis for V, with W = span({v1      v }). Let X = span({v+1      v }). Since {v+1      v } is linearly independent, it is a basis for W ⊥ if X = W ⊥  We will show that X ⊆ W ⊥ and W ⊥ ⊆ X  To show X ⊆ W ⊥ , we must prove that any vector x of the form +1 v+1 + · · · +  v (for some scalars +1       ) is orthogonal to every vector w ∈ W. Now if w ∈ W, then w = 1 v1 + · · · +  v  for some scalars 1        Hence, x · w = (+1 v+1 + · · · +  v ) · (1 v1 + · · · +  v ) which equals zero when expanded because each vector in {v+1      v } is orthogonal to every vector in {v1      v }. Hence, x ∈ W ⊥  and so X ⊆ W ⊥ .

To show W ⊥ ⊆ X , we must show that any vector x in W ⊥ is also in span({v+1      v }).

Let x ∈ W ⊥  Since {v1      v } is an orthogonal basis for V, Theorem 7.16 tells us that x=

(x · v ) (x · v+1 ) (x · v ) (x · v1 ) v1 + · · · + v + v+1 + · · · + v  (v1 · v1 ) (v · v ) (v+1 · v+1 ) (v · v )

However, since each of v1      v is in W, we know that x · v1 = · · · = x · v = 0 Hence, x=

(x · v ) (x · v+1 ) v+1 + · · · + v  (v+1 · v+1 ) (v · v )

and so x ∈ span({v+1      v }). Thus, W ⊥ ⊆ X .

Next, we prove part (5) of Theorem 7.19 as follows: Let W be a subspace of V of dimension . By Theorem 7.17, W has an orthogonal basis {v1      v }. Expand this basis to an orthogonal basis for all of V. (That is, first expand to any basis for V by Theorem 4.15, then use the Gram-Schmidt Process. Since the first  vectors are already orthogonal, this expands {v1      v } to an orthogonal basis {v1      v } for V.) Then, by part (4) of Theorem 7.19, {v+1      v } is a basis for W ⊥ , and so dim(W ⊥ ) =  − . Hence dim(W) + dim(W ⊥ ) = . c 2016 Elsevier Ltd. All rights reserved.

Copyright °

232

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.5

(d) By part (3) of Theorem 7.19, W ⊆ (W ⊥ )⊥ . Next, by part (5) of Theorem 7.19, dim(W) =  − dim(W ⊥ ) =  − ( − dim((W ⊥ )⊥ )) = dim((W ⊥ )⊥ ). Thus, by Theorem 4.13, or its complex analog, W = (W ⊥ )⊥ .

(23) We begin with the basis  = {3 + 2   − 1 1 2 } for P3 containing the polynomials 3 + 2 and  − 1. (Notice that the Enlarging Method is not needed here since it can be seen by inspection that  = {3 + 2   − 1 1 2 } is linearly independent (since each element is a polynomial of a different degree), and since dim(P3 ) = 4,  is a basis for P3 .)

We use the Gram-Schmidt Process on . Let v1 = 3 + 2 . Next, ­ ®  − 1 3 + 2 (3 + 2 ) = ( − 1) − 02 (3 + 2 ) =  − 1 v2 = ( − 1) − 3 h + 2  3 + 2 i Also, ­ 3 ® 1  + 2 h1  − 1i (3 + 2 ) − ( − 1) = (1) − 3 h + 2  3 + 2 i h − 1  − 1i

v3

1 1 = (1) − 02 (3 + 2 ) − (−1) 2 ( − 1) = 2  + 2 

Multiplying by 2 to eliminate fractions yields v3 =  + 1. Finally, ­2 ­2 3 ­2 ® ® ®   + 1    + 2  − 1 2 3 2 ( +  ) − ( − 1) − ( + 1) v4 = ( ) − 3 h + 2  3 + 2 i h − 1  − 1i h + 1  + 1i = (2 ) − 12 (3 + 2 ) − 02 ( − 1) − 02 ( + 1) = − 12 3 + 12 2 

Multiplying by −2 to simplify the form of the result produces v4 = 3 − 2 . Hence, by part (4) of Theorem 7.19, W ⊥ = span({ + 1 3 − 2 }).

(25) As in the hint, let {v1      v } be an orthonormal basis for W. Now if v ∈ V, let w1 = hv v1 i v1 + · · · + hv v i v

and w2 = v − w1 

Then, w1 ∈ W because w1 is a linear combination of basis vectors for W. We claim that w2 ∈ W ⊥ .

To see this, let u ∈ W. Then u = 1 v1 + · · · +  v for some 1       . Then hu w2 i = hu v − w1 i = h1 v1 + · · · +  v  v − (hv v1 i v1 + · · · + hv v i v )i = h1 v1 + · · · +  v  vi − h1 v1 + · · · +  v  hv v1 i v1 + · · · + hv v i v i =

 X =1

 hv  vi −

  X X =1 =1

 hv v i hv  v i 

But hv  v i = 0 when  6=  and hv  v i = 1 when  = , since {v1      v } is an orthonormal set.

Hence,     X X X X  hv  vi −  hv v i =  hv  vi −  hv  vi = 0 hu w2 i = =1

=1

=1

=1

⊥

Since this is true for every u ∈ W, we conclude that w2 ∈ W .

Finally, we want to show uniqueness of decomposition. Suppose that v = w1 +w2 and v = w10 +w20  where w1  w10 ∈ W and w2 , w20 ∈ W ⊥ . We want to show that w1 = w10 and w2 = w20  Now, w1 − w10 = w20 − w2 . Also, w1 − w10 ∈ W, but w20 − w2 ∈ W ⊥  Thus, w1 − w10 = w20 − w2 ∈ W ∩ W ⊥  By part (2) of Theorem 7.19, w1 − w10 = w20 − w2 = 0 Hence, w1 = w10 and w2 = w20 . c 2016 Elsevier Ltd. All rights reserved.

Copyright °

233

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

(26) From Example 8 in Section 7.5 of the textbook we know that basis for W = span({cos  sin }).

(We quickly verify this: hcos  sin i =

Z 

(cos  sin )  =

−

µ

n

Section 7.5 o

√1 cos  √1 sin   

is an orthonormal

¶¯ ¯ 1 2 sin  ¯¯ = (0 − 0) = 0 2 −

Hence, {cos  sin } is an orthogonal basis for W. We normalize these vectors to get an orthonormal basis. Now, µ ¶¯ Z  Z  ¯   1 1 (1 + cos 2)  =  + sin 2 ¯¯ = ( + 0) − (− + 0) =  cos2   = hcos  cos i = 2 2 2 2 − − − and

¶¯ ¯ 1    − sin 2 ¯¯ = ( − 0) − (− − 0) =  2 2 2 − n o √ √ Therefore, kcos k =  and ksin k = , giving the orthonormal basis √1 cos  √1 sin  for W.) Z 

Z 

1 (1 − cos 2)  = sin   = hsin  sin i = − − 2 2

µ

Using f = 1  ,

Àµ ¶ ¿ Àµ ¶ ¿ 1 1 1 1 √ cos  + f  √ sin  √ sin   w1 = projW f = f  √ cos      so we have two more inner products to compute. First, ¿ À µ ¶¯ Z  Z  ¯ ¡  ¢ 1 1 1 1 √ √  sin  ¯¯ − √ f  √ cos   sin   =  cos   =    −     − − =

µ

(using integration by parts), ¶¯ µ ¶¯ Z  ¯ ¯ −1  1 1  √  sin  ¯¯ − √  cos  ¯¯ + √ (− cos )        − − −

(using integration by parts again).

E D R  1  cos   to both ends of this equation produces Adding f  √1 cos  = √  −

À µ ¶ ¶¯ µ ¶¯ µ  ¿ ¯ ¯ 1 − 1 −1   1 √  sin  ¯¯ − √  cos  ¯¯ = (0 − 0) − √ − √ = −√  2 f  √ cos  =           − − D E since  =  − − . Dividing by 2 yields f  √1 cos  = − 2√1  . Similarly, ¿ À 1 f  √ sin  = 

=

1 √   µ

Z 

 sin   =

−

µ

¶¯ Z  ¯ −1  1 √  cos  ¯¯ − √ (− cos )      − −

(using integration by parts), ¶ ¯ µ ¶ ¯ Z  ¯ ¯ 1 1 −1  √  cos  ¯¯ + √  sin  ¯¯ − √ ( sin )        − − − (using integration by parts again).

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

234

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 7.5

D E R  1 Adding f  √1 sin  = √  sin   to both ends of this equation produces  −

¿ À µ ¶ ¶¯ µ ¶¯ µ  ¯ ¯ 1 −1  − 1 1   ¯ ¯ √  cos  ¯ + √  sin  ¯ = √ − √ 2 f  √ sin  = + (0 − 0) = √            − −

D E Dividing by 2 yields f  √1 sin  = 2√1  . Therefore, w1 = projW f =

¶µ ¶ µ ¶µ ¶ µ 1 1 1 1 1 √ √ cos  + √ sin  = (sin  − cos ) − √ 2 2   2  

1 1 Also, w2 = f − w1 = 1  − 2 sin  + 2 cos .

Finally, we must verify that hw1  w2 i = 0. Now, ¶µ ¶ Z  µ 1 1 1  1 hw1  w2 i = (sin  − cos )  − sin  + cos   2  2 2 − Z  Z  Z  1 1 1 = ( sin )  − 2 sin2   − ( cos )  2 − 4 − 2 − Z  Z  1 1 − 2 cos2   + 2 sin  cos   4 − 2 −

We consider each integral separately. First, ¿ À µ ¶ Z  1 1 1 1 1 1 √ f  √ sin  = √ (from our earlier computations) =  ( sin )  = √ 2 − 4 2   2  2  Next, 1 42

Z 

sin2   =

−

Similarly 1 2

1 1 1  hsin  sin i = 2 () (from our earlier computations) = 2 4 4 4

Z 

1 ( cos )  = √ 2  −

and

1 42

Z 

¿ À µ ¶ 1 1 −1 1 √ f  √ cos  = √ =− 4  2  2 

cos2   =

−

1 1  () = 2 4 4

Finally, 1 2 2

Z 

sin  cos   =

−

1 1 hsin  cos i = 2 hcos  sin i = 0 (from our earlier computations).

22 2

Hence, hw1  w2 i =

1 1 1 1 − − (− ) − = 0 4 4 4 4

completing the problem.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

235

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 7 Review

(29) (b) First, we establish that range() = W. We start by showing that W ⊆ range(). Let v ∈ W. Then we claim that (v) = v. Note that v ∈ W implies that the unique decomposition of v into v = w1 + w2 with w1 ∈ W and w2 ∈ W ⊥ is w1 = v and w2 = 0. (This is because for these vectors w1 and w2 , we have w1 + w2 = v, and because by Theorem 7.20 this decomposition is unique.) Hence, since (v) = projW v = w1 , we get (v) = v. From this we can conclude that W ⊆ range(), since every vector in W is the image of itself. Also, since the codomain of  is W, range() ⊆ W. Hence, range() = W. Next, we establish that ker() = W ⊥ . First we show that W ⊥ ⊆ ker(). Let v ∈ W ⊥ .

We claim that (v) = 0. Note that v ∈ W ⊥ implies that the unique decomposition of v into v = w1 + w2 with w1 ∈ W and w2 ∈ W ⊥ is w1 = 0 and w2 = v. (This is because for these vectors w1 and w2 , we have w1 + w2 = v, and because by Theorem 7.20 this decomposition is unique.) Hence, since (v) = projW v = w1 , we get (v) = 0. From this we can conclude that W ⊥ ⊆ ker(). We still need to prove that ker() ⊆ W ⊥ . So, let x ∈ ker(). Then x = u1 + u2 with u1 ∈ W and u2 ∈ W ⊥ . Also, (x) = u1 . But, x ∈ ker() implies u1 = (x) = 0. Therefore, x = u1 + u2 = 0 + u2 = u2 ∈ W ⊥ . This completes the proof.

(30) (a) False. The correct equation is kxk = || kxk. For example, in C2 with the usual p complex dot [ 0] · [ 0] = product, consider x = [1 0] and  = .

Then kxk = k[1 0]k = k[ 0]k = q p √ √ ()(  ) + (0)(0) = 1 = 1. Also, kxk = [1 0] · [1 0] = 1 = 1. Hence, || kxk = ||(1) = (1)(1) = 1 = kxk, but  kxk = (1) = (−)(1) = − 6= kxk. p (b) False. The distance between two vectors x and y is defined to be kx − yk = hx − y x − yi, which is always real and positive (since x and y are distinct) by parts (1) and (2) of the definition of an inner product. Hence, the distance can not be pure imaginary. (c) False. For a set  of vectors to be orthonormal,  must be an orthogonal set. This implies that  is linearly independent, but the converse is not true. There arenmany sets independent io h √ of√linearly 2 2 in R2 using the unit vectors that are not orthogonal sets. For example,  = [1 0] 2  2 usual dot product is such a set. (d) True. For example, R2 has the usual dot product as one inner product, and the inner product of Example 2 in Section 7.5 in the textbook as a different inner product. (e) False. The uniqueness assertion is that, using a single, fixed inner product, and a single, fixed subspace W, the decomposition of a particular vector described in the Projection Theorem is unique. However, different inner products could result in different decompositions. For example, consider the subspace W = span({[1 0]}) in R2 . Let v = [0 1]. Using the dot product as the inner product on R2 , the decomposition of v is v = [0 0] + [0 1], with [0 0] ∈ W and [0 1] ∈ W ⊥ (since [0 1] · [1 0] = 0). However, consider the inner product from Example 2 in Section 7.5 in the textbook. It is easily verified that {[1 0]} is an orthonormal basis for W with this inner product, and so, projW v = h[0 1] [1 0]i [1 0] = (−1)[1 0] = [−1 0]. Therefore, the decomposition of v using this inner product is v = [−1 0] + [1 1].

Chapter 7 Review Exercises (1) (a) v · w = [ 3 −  2 + 3] · [−4 − 4 1 + 2 3 − ]

= (−4 − 4) + (3 − )(1 + 2) + (2 + 3)(3 − ) = (−4 + 4) + (3 − )(1 − 2) + (2 + 3)(3 + ) = (−4 − 4) + (1 − 7) + (3 + 11) = 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

236

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 7 Review

(b) (1 + 2)(v · z) = (1 + 2)([ 3 −  2 + 3] · [2 + 5 2 − 5 − ])

= (1 + 2)((2 + 5) + (3 − )(2 − 5) + (2 + 3)(−)) = (1 + 2)((2 − 5) + (3 − )(2 + 5) + (2 + 3)()) = (1 + 2)((5 + 2) + (11 + 13) + (−3 + 2)) = (1 + 2)(13 + 17) = −21 + 43

((1 + 2)v) · z = ((1 + 2)[ 3 −  2 + 3]) · [2 + 5 2 − 5 − ] = [−2 +  5 + 5 − 4 + 7] · [2 + 5 2 − 5 − ]

= (−2 + )(2 + 5) + (5 + 5)(2 − 5) + (−4 + 7)(−) = (−2 + )(2 − 5) + (5 + 5)(2 + 5) + (−4 + 7)() = (1 + 12) + (−15 + 35) + (−7 − 4) = −21 + 43

v · ((1 + 2)z) = [ 3 −  2 + 3] · ((1 + 2)[2 + 5 2 − 5 − ]) = [ 3 −  2 + 3] · [−8 + 9 12 −  2 − ] = ()(−8 + 9) + (3 − )(12 − ) + (2 + 3)(2 − ) = ()(−8 − 9) + (3 − )(12 + ) + (2 + 3)(2 + ) = (9 − 8) + (37 − 9) + (1 + 8) = 47 − 9

(4) (a) Using the row operations h1i ← (−) h1i h2i ← (−1 − ) h1i + h2i h3i ← (−2 + ) h1i + h3i

3 2 − 13 ) h2i h2i ← ( 13 h1i ← (−1 + ) h2i + h1i, and h3i ← (−1 + 2) h2i + h3i, ¯ ¯ ⎡ ⎤ ⎡ ⎤  1 +  ¯¯ −1 + 2 1 0 ¯¯ 4 + 3 the matrix ⎣ 1 +  5 + 2 ¯¯ 5 − 3 ⎦ row reduces to ⎣ 0 1 ¯¯ −2 ⎦. Thus, the system 2 −  2 − 5 ¯ 1 − 2 0 0 ¯ 0

has the unique solution  = 4 + 3,  = −2.

(d) Using the row operations

h1i ← ( 12 − 12 ) h1i  h2i ← (−4 − 3) h1i + h2i

h2i ← ( 12 − 12 ) h2i, and h1i ← (−1 + 2) h2i + h1i, ¯ ¯ ∙ ¸ ∙ ¸ 1+ 3− 5 + 5 ¯¯ 23 −  1 0 3 −  ¯¯ 2 +  the matrix row reduces to .

4 + 3 11 − 4 19 + 16 ¯ 86 − 7 0 1  ¯ 7+ Hence,  is an independent variable. Setting the value of  equal to  yields the solution set

{[(2 + ) − (3 − ) (7 + ) −  ] |  ∈ C} = {[(2 − 3) + (1 + ) 7 + (1 − ) ] |  ∈ C}

¯ ¯ ¯  + 3 −5 10 ¯¯ ¯ (6) (a) Step 1: A () = |I3 − A| = ¯¯ −2  + 3 −8 ¯¯. Using basketweaving gives ¯ −2 3 −7 ¯

A () = 3 − 2 +  − 1 = (2 + 1)( − 1) = ( − )( + )( − 1)

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

237

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 7 Review

Step 2: The eigenvalues of A are the roots of A (). Hence, there are three eigenvalues, 1 = , 2 = −, and 3 = 1.

Step 3: We solve for fundamental eigenvectors.

For 1 = , we solve the homogeneous system whose augmented matrix is ¯ ¯ ⎡ ⎤ ⎡ ⎤ ¯ 0 3 +  −5 10 1 0 2 +  ¯¯ 0 ¯ −8 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 −1 +  ¯¯ 0 ⎦  [(I3 − A) | 0] = ⎣ −2 3 +  ¯ 0 −2 3 −7 +  ¯ 0 0 0 0

This yields the fundamental eigenvector [−2 −  1 −  1].

For 2 = −, we solve the homogeneous system whose augmented matrix is ¯ ¯ ⎡ ⎤ ⎡ ⎤ ¯ 0 3 −  −5 10 1 0 2 −  ¯¯ 0 ¯ −8 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 −1 −  ¯¯ 0 ⎦  [((−)I3 −A) | 0] = ⎣ −2 3 −  ¯ 0 −2 3 −7 −  ¯ 0 0 0 0

This yields the fundamental eigenvector [−2 +  1 +  1].

For 3 = 1, we solve the homogeneous system whose augmented matrix is ¯ ¯ ⎡ ⎤ ⎡ ⎤ 4 −5 10 ¯¯ 0 1 0 0 ¯¯ 0 4 −8 ¯¯ 0 ⎦  which reduces to ⎣ 0 1 −2 ¯¯ 0 ⎦  [(I3 − A) | 0] = ⎣ −2 −2 3 −6 ¯ 0 0 0 0 ¯ 0 This yields the fundamental eigenvector [0 2 1].

Step 4: Since  = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable.

Step 5: We form the 3 × 3 matrix whose columns are the 3 vectors we have found: ⎡ ⎤ −2 −  −2 +  0 1 +  2 ⎦ P=⎣ 1− 1 1 1

Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the same order) is ⎡ ⎤ ⎤ ⎡ −1 −  1 + 2 −2 − 4  0 0 D = ⎣ 0 − 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = 12 ⎣ −1 +  1 − 2 −2 + 4 ⎦.

2 −2 6 0 0 1

It is easy to verify for these matrices that D = P−1 AP.

(7) (a) One possibility: Consider : C → C given by (z) = z. Note that (v +w) = (v + w) = v +w = (v) + (w). But  is not a linear operator on C because () = −, but (1) = (1) = , so the rule “(v) = (v)” is not satisfied. (b) The example given in part (a) is a linear operator on C, thought of as a real vector space. In that case we may use only real scalars, and so, if v =  + , then (v) = ( + ) =  −  = ( − ) = (v).

Note: for any function  from any vector space to itself (real or complex), the rule “(v+w) = (v) + (w)” implies that (v) = (v) for any rational scalar . Thus, any example for real vector spaces for which (v + w) = (v) + (w) is satisfied, but (v) = (v) is not satisfied for some , must involve an irrational value of . Examples of such functions do exist, but the ideas behind their construction are slightly beyond the level of topics covered in this text. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

238

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 7 Review

(8) (a) First, we must find an ordinary basis for C4 containing [1  1 −] and [1 +  2 −  0 0]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for C4 to the given vector and use the Independence Test Method. Hence, we row reduce ⎤ ⎡ ⎡ ⎤ 1 0 0 0 0  1 1+ 1 0 0 0 ⎥ ⎢ 2 1 2 1 ⎥ ⎢  2− 0 1 0 0 ⎥ ⎢ 0 1 1 0 5 + 5 5 + 5 ⎥ ⎢ ⎥ to ⎢ ⎢ 0 0 1 − 1 − 3 0 −1 − 8 ⎥  ⎣ 1 0 0 0 1 0 ⎦ ⎣ 5 5 5 5 ⎦ − 0 0 0 0 1 0 0 0 0 1 − giving the basis {[1  1 −] [1 +  2 −  0 0] [1 0 0 0] [0 0 1 0]}.

We perform the Gram-Schmidt Process on this set. Let v1 = [1  1 −]. Then, ¶ µ [1 +  2 −  0 0] · [1  1 −] [1  1 −] v2 = [1 +  2 −  0 0] − [1  1 −] · [1  1 −] = [1 +  2 −  0 0] + 14 [1  1 −] = 14 [4 + 5 7 − 4  1] 

Multiplying by 4 to eliminate fractions yields v2 = [4 + 5 7 − 4  1]. Next, ¶ µ [1 0 0 0] · [1  1 −] [1  1 −] v3 = [1 0 0 0] − [1  1 −] · [1  1 −] µ ¶ [1 0 0 0] · [4 + 5 7 − 4  1] − [4 + 5 7 − 4  1] [4 + 5 7 − 4  1] · [4 + 5 7 − 4  1] (4 − 5) [4 + 5 7 − 4  1] = [0 1 0 0] − 14 [1  1 −] − 108 1 = 27 [10 −2 + 6 −8 −  −1 + 8]  Multiplying by 27 to eliminate fractions gives us v3 = [10 −2 + 6 −8 −  −1 + 8]. Finally, ¶ µ [0 0 1 0] · [1  1 −] [1  1 −] v4 = [0 0 1 0] − [1  1 −] · [1  1 −] µ ¶ [0 0 1 0] · [4 + 5 7 − 4  1]] − [4 + 5 7 − 4  1] [4 + 5 7 − 4  1] · [4 + 5 7 − 4  1] µ ¶ [0 0 1 0] · [10 −2 + 6 −8 −  −1 + 8] − [10 −2 + 6 −8 −  −1 + 8] [10 −2 + 6 −8 −  −1 + 8] · [10 −2 + 6 −8 −  −1 + 8] (−8 + ) (−) [4 + 5 7 − 4  1] − [10 −2 + 6 −8 −  −1 + 8] = [0 0 1 0] − 14 [1  1 −] − 108 270 £ ¤ = 0 0 12  12   Multiplying by 2 to eliminate fractions produces v4 = [0 0 1 ]. Hence, an orthogonal basis for C4 in which the first two vectors spans the same subspace as {[1  1 −] [1 +  2 −  0 0]} is {[1  1 −] [4 + 5 7 − 4  1]  [10 −2 + 6 −8 −  −1 + 8]  [0 0 1 ]} (9) (a) Note that A is Hermitian, so A is unitarily diagonalizable. Now, ¯ ¯ ¯  − 3 −1 −  ¯ ¯ ¯ A () = |I2 − A| = ¯ −1 +  −2 ¯

= ( − 3)( − 2) − (−1 − )(−1 + ) = 2 − 5 + 4 = ( − 1)( − 4)

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

239

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 7 Review

The eigenvalues of A are the roots of A (). Hence, there are two eigenvalues, 1 = 1, and 2 = 4.

Next we solve for fundamental eigenvectors.

For 1 = 1, we solve the homogeneous system whose augmented matrix is [(I2 − A) | 0]. Now, ¯ # " ¯ ∙ ¸ 1 12 + 12  ¯¯ 0 −2 −1 −  ¯¯ 0  [(I2 − A) | 0] =  which reduces to ¯ −1 +  −1 ¯ 0 ¯ 0 0 0

Eliminating fractions, this yields the fundamental eigenvector v1 = [−1 −  2].

For 2 = 4, we solve the homogeneous system whose augmented matrix is [(4I2 − A) | 0]. Now, ¯ ¯ ∙ ¸ ∙ ¸ 1 −1 −  ¯¯ 0 1 −1 −  ¯¯ 0 [(4I2 − A) | 0] = ¯ 0  which reduces to ¯ 0  −1 +  2 0 0 This yields the fundamental eigenvector v2 = [1 +  1].

Now we normalize the two fundamental eigenvectors we have found (which are orthogonal) to get √ √ an orthonormal basis  for C2 . Since kv1 k = 6 and kv2 k = 3, we get ½ ¾ 1 1  = √ [−1 −  2] √ [1 +  1]  6 3 We form the 2 × 2 matrix whose columns are the 2 vectors in : ⎤ ⎡ √1 (−1 − ) √1 (1 + ) 6 3 ⎦ P=⎣ √2 6

√1 3

The matrix whose diagonal entries are the eigenvalues of A (in the same order) is D =

∙

1 0 0 4

¸ .

Also, it is easy to verify for these matrices that D = P∗ AP. ⎡ ⎤ 80 105 + 45 60 185 60 − 60 ⎦, and so A is normal. Hence, by (10) Both AA∗ and A∗ A equal ⎣ 105 − 45 60 60 + 60 95 Theorem 7.9, A is unitarily diagonalizable. ° p ° (13) The distance between  and 3 equals ° − 3 ° = h( − 3 ) ( − 3 )i. But ® ­ ( − 3 ) ( − 3 ) =

=

= Thus the distance is

q

Z 1 0

¡

 − 3

¢2

 =

Z 1 0

(2 − 24 + 6 ) 

¡ 1 3 2 5 1 7 ¢¯1 ¯ = (1 − 2 + 1) − 0 3 − 5 + 7 3 5 7 0

35 42 15 8 105 − 105 + 105 = 105 

8 105 ≈ 0276.

(14) We are given w1 = [1 0 0], w2 = [0 1 0], and w3 = [0 0 1]. Following the steps in the Generalized

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

240

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 7 Review

Gram-Schmidt Process, we get v1 = w1 = [1 0 0]. Next, v2

h[0 1 0] [1 0 0]i [1 0 0] h[1 0 0] [1 0 0]i [−1 1 −3] · [1 −1 2] [1 0 0] = [0 1 0] − [1 −1 2] · [1 −1 2] = [0 1 0] −

4 = [0 1 0] − (−8) 6 [1 0 0] = [ 3  1 0]

We multiply by 3 to eliminate fractions to produce v2 = [4 3 0]. Finally, v3

h[0 0 1] [4 3 0]i h[0 0 1] [1 0 0]i [1 0 0] − [4 3 0] h[1 0 0] [1 0 0]i h[4 3 0] [4 3 0]i [2 3 1] · [1 −1 −1] [2 3 1] · [1 −1 2] [1 0 0] − [4 3 0] = [0 0 1] − [1 −1 2] · [1 −1 2] [1 −1 −1] · [1 −1 −1] = [0 0 1] −

5 = [0 0 1] − 16 [1 0 0] − (−2) 3 [4 3 0] = [ 2  2 1]

We multiply by 2 to eliminate fractions, producing v3 = [5 4 2]. Hence, the desired orthogonal basis is {[1 0 0] [4 3 0] [5 4 2]}.

(16) (a) True. Every real number is also a complex number, so an -tuple of real numbers is also an -tuple of complex numbers. (b) False. The angle between vectors in C is not defined. Since v · w can be a complex number, v·w could be complex, making the inverse of the cosine of this value a complex the expression kvkkwk number. (c) True. (A∗ Aw) · z = (A∗ (Aw)) · z = Aw · (A∗ )∗ z (by Theorem 7.3) = Aw · Az.

(d) False. Example 5 in Section 7.1 provides a counterexample.

(e) False. If A is skew-Hermitian, A∗ = −A, and so  = − for each diagonal entry  of A. If  =  + , then  = − implies  −  = − − . This can only be true if  = 0.

But,  can be any real number. Hence, the main diagonal entries of skew-Hermitian matrices are pure ∙imaginary¸numbers but not necessarily zeroes. A counterexample to the given statement is  1 A=  −1  (f) True. The Fundamental Theorem of Algebra shows that the characteristic polynomial for an  ×  matrix must factor into  linear factors. Each linear factor contributes to the algebraic multiplicity of one of the eigenvalues of the matrix.

(g) False. The system has a unique solution if and only if |A| 6= 0. However, systems with |A| = 0 can have infinitely many solutions. For example, the system Az = 0, with A being any square matrix with determinant zero (such as a 2 × 2 matrix with every entry equal to 1) has the trivial solution as well as other nontrivial solutions. (h) True. The vector [−1 − 1 +  1 + ] equals [ 1 +  1 − ]. Any nonzero scalar multiple of an eigenvector is also an eigenvector corresponding to the same eigenvalue because eigenspaces are closed under scalar multiplication. (i) False. Otherwise every∙ square ¸complex matrix would be diagonalizable. For a specific counterex 1 ample, consider A = . This matrix has  =  as its only eigenvalue, having algebraic 0  multiplicity 2. However,  is one-dimensional, spanned by [1 0]. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

241

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Chap 7 Review

¯ ¯2 ¯ ¯ (j) True. By parts (1) and (3) of Theorem 7.5, |AA∗ | = |A| |A∗ | = |A| (|A|) = ¯|A|¯ . (The last step uses the fact that, for any complex number ,  = ||2 . See Appendix C.)

(k) True. If the vector space satisfies the 10 properties of a vector space using complex scalars, it will also satisfy them using real scalars, since the real numbers are a subset of the complex numbers.

(l) True. This is stated in the text just after the definition of an orthogonal set of vectors in C . The proof for the complex case is almost identical to the proof for the real case. (m) False. Orthogonality in C is determined using the complex dot product, which involves using the complex conjugate. Thus, for the rows of a complex matrix to form an orthonormal basis for C " 3 " # # 7 − 25 − 45  − 24 5 25  ∗  we need A A = I . For example, if A = , then A A = , while 3 7 − 45  − 24 − 25 5 25  A∗ A = I2 . Notice that the rows of A are orthogonal unit vectors in C2 .

(n) True. This works because if all of the entries of A are real, A = A∗ . (o) True. If A is Hermitian, then A is unitarily diagonalizable, by Corollary 7.10. But, by Theorem 7.11, all of the eigenvalues of A are real. Hence, A is unitarily similar to a diagonal matrix with its real eigenvalues on its main diagonal; that is, a matrix with all real entries. (p) True. If A is skew-Hermitian, then A is unitarily diagonalizable by Corollary 7.10. Thus, the geometric multiplicity of each eigenvalue must equal its algebraic multiplicity, just as in the case of a diagonalizable real matrix. (q) True. This warning is given in Section 7.4, just after Example 1. (r) True. This is discussed in Example 1 in Section 7.5. (s) True. Let {u1      u } be an orthonormal basis for W. Expand this to an orthonormal basis {u1      u } for V. This will be a basis of fundamental eigenvectors for  because {u1      u } is a basis for 1 , and {u+1      u } is a basis for 0 . Hence,  is unitarily diagonalizable. (t) True. This is part (6) of Theorem 7.19.

(u) True. As in Example 11 of Section 7.5, start with the basis {1  2       } for PC . Performing the Gram-Schmidt Process on this basis and then normalizing will produce an ordered orthonormal basis in which the vectors have the correct degrees. p (v) True. The distance between two vectors v and w is kv − wk = h(v − w) (v − w)i.

(w) False. For example, 1 is not orthogonal to cos() because Z 1

−1

1

1 cos()  = sin()|−1 = sin(1) − sin(−1) = 2 sin(1) 6= 0

However, if instead we had considered this set inside the inner product space R  of continuous realvalued functions defined on [− ], with inner product given by hf  gi = − f ()g() , then the set would be an orthogonal set. (See the remarks after Example 8 in Section 7.5.)

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

242

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.1

Chapter 8 Section 8.1 (1) In parts (a) through (d), use the fact the adjacency matrix for a graph is defined to be the matrix whose ( ) entry is the number of edges between the vertices  and  . In parts (e) through (h), use the fact the adjacency matrix for a digraph is defined to be the matrix whose ( ) entry is the number of directed edges from the vertex  to the vertex  . Inspection shows that the matrices in parts (a), (b), (c), and (d) are symmetric, while those in parts (e), (f), (g), and (h) are not. ⎡ ⎡ ⎡ ⎤ ⎤ ⎤ 0 1 1 1 0 0 0 1 2 0 0 1 ⎢ 2 0 0 1 0 ⎥ ⎢ 1 0 1 1 ⎥ (c) 3 : ⎣ 0 0 0 ⎦ ⎢ ⎥ ⎥ (a) 1 : ⎢ ⎢ ⎥ ⎣ 1 1 0 1 ⎦ 0 0 0 (b) 2 : ⎢ 0 0 1 0 1 ⎥ ⎣ 0 1 0 0 1 ⎦ 1 1 1 0 1 0 1 1 1 ⎡ ⎡ ⎡ ⎤ ⎤ ⎤ 0 1 0 0 0 2 1 0 0 1 0 1 0 0 ⎢ 0 0 1 0 ⎥ ⎢ 0 1 1 1 ⎥ ⎢ 1 0 2 1 0 0 ⎥ ⎢ ⎢ ⎥ ⎥ ⎥ (f) 2 : ⎢ ⎢ 0 2 0 0 1 1 ⎥ (e) 1 : ⎣ 1 0 0 1 ⎦ ⎣ 0 1 0 1 ⎦ ⎥ (d) 4 : ⎢ ⎢ 1 1 0 0 2 0 ⎥ 1 1 0 0 0 0 0 1 ⎢ ⎥ ⎣ 0 0 1 2 0 1 ⎦ 0 0 1 0 1 0 ⎡ ⎡ ⎤ ⎤ 0 2 2 0 0 0 2 0 0 0 0 0 0 ⎢ 1 0 0 0 1 ⎥ ⎢ 0 0 1 0 0 0 0 0 ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ 0 0 0 1 0 0 0 0 ⎥ ⎥ 1 0 0 2 0 (g) 3 : ⎢ ⎢ ⎢ ⎥ ⎥ ⎢ 0 0 0 0 1 0 0 0 ⎥ ⎣ 0 0 1 0 1 ⎦ ⎢ ⎥ (h) 4 : ⎢ ⎥ 0 2 0 2 0 ⎢ 0 0 0 0 0 2 0 0 ⎥ ⎢ 0 0 0 0 0 0 1 0 ⎥ ⎢ ⎥ ⎣ 0 0 0 0 0 0 0 1 ⎦ 1 0 0 0 0 0 0 0 (2) For a matrix to be the adjacency matrix for a simple graph or digraph, it must be a square matrix, all of whose entries are nonnegative integers. In addition, the adjacency matrix for a graph must be symmetric.

A can not be the adjacency matrix for either a graph or digraph because it is not square.

B can not be the adjacency matrix for either a graph or digraph because its (2 2) entry equals −1, which is negative.

C can be the adjacency matrix for a digraph only since C is not symmetric (see Figure 5 on the next page).

D can not be the adjacency matrix for either a graph or digraph because it has non-integer entries.

E can not be the adjacency matrix for either a graph or digraph because its (3 2) entry equals −2, which is negative.

F can be the adjacency matrix for either a graph or digraph (see Figure 6 on the next page).

G can be the adjacency matrix for a digraph only, since G is not symmetric (see Figure 7 on the next page).

H can be the adjacency matrix for a digraph only, since H is not symmetric (see Figure 8 on the next page).

I can be the adjacency matrix for a graph or digraph (see Figure 9 on the next page). c 2016 Elsevier Ltd. All rights reserved.

Copyright °

243

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.1

J can be the adjacency matrix for a graph or digraph (see Figure 10 on the previous page).

K can be the adjacency matrix for a graph or digraph (see Figure 11 on the previous page).

(4) The adjacency matrix for the graph in Figure 8.6 in Section 8.1 of the textbook is ⎡ ⎤ 0 1 0 2 2 ⎢ 1 0 2 1 0 ⎥ ⎢ ⎥ ⎥ A=⎢ ⎢ 0 2 1 0 1 ⎥ ⎣ 2 1 0 0 1 ⎦ 2 0 1 1 0

whose ( ) entry is the number of edges between vertices  and  . Direct computation shows that ⎡ ⎡ ⎤ ⎤ 9 2 4 3 2 12 20 10 22 25 ⎢ 2 6 2 2 5 ⎥ ⎢ 20 8 19 15 8 ⎥ ⎢ ⎢ ⎥ ⎥ 2 3 ⎢ ⎥ ⎥ A = ⎢ 4 2 6 3 1 ⎥ A = ⎢ ⎢ 10 19 11 11 17 ⎥  ⎣ 3 2 3 6 4 ⎦ ⎣ 22 15 11 12 15 ⎦ 2 5 1 4 6 25 8 17 15 9 ⎡ ⎤ 114 54 75 69 56 ⎢ 54 73 43 56 74 ⎥ ⎢ ⎥ ⎥ and A4 = ⎢ ⎢ 75 43 66 56 42 ⎥  ⎣ 69 56 56 74 67 ⎦ 56 74 42 67 82

(a) The number of paths of length 3 from 2 to 4 is the (2 4) entry of A3 . Hence, there are 15 such paths. (c) The number of paths of lengths 1, 2, 3, and 4 from 4 to 2 are the (4 2) entries of A, A2 , A3 , and A4 , respectively. Adding these four matrix entries yields 1 + 2 + 15 + 56 = 74. Hence, the number of paths of length ≤ 4 from 4 to 2 is 74. (e) The number of paths of length  from 4 to 3 is the (4 3) entry of A . Thus, to find the shortest path from 4 to 3 , we look for the smallest positive integer power  for which the (4 3) entry of A is nonzero, indicating that a path of length  exists. Inspecting the matrix A, above, we see that there is no path of length 1 from 4 to 3 , since A has 0 as its (4 3) entry. However, the (4 3) entry of A2 is 3. Therefore, there are three paths of length 2 from 4 to 3 , and so the shortest path from 4 to 3 has length 2.

(5) The adjacency matrix for the digraph in Figure 8.7 in Section 8.1 of the textbook is ⎡ ⎤ 0 1 1 0 0 ⎢ 0 0 1 1 0 ⎥ ⎢ ⎥ ⎥ A=⎢ ⎢ 2 1 0 1 0 ⎥ ⎣ 0 0 1 1 0 ⎦ 1 1 0 0 0

whose ( ) entry is the number of directed edges from the vertex  to the vertex  . Direct compu⎡ ⎡ ⎤ ⎤ 2 1 1 2 0 2 3 5 4 0 ⎢ 2 1 1 2 0 ⎥ ⎢ 2 3 5 4 0 ⎥ ⎢ ⎢ ⎥ ⎥ ⎥ and A3 = ⎢ 8 4 4 8 0 ⎥.

0 2 4 2 0 tation shows that A2 = ⎢ ⎢ ⎢ ⎥ ⎥ ⎣ 2 1 1 2 0 ⎦ ⎣ 2 3 5 4 0 ⎦ 0 1 2 1 0 4 2 2 4 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

245

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.1

(a) The number of paths of length 3 from 2 to 4 is the (2 4) entry of A3 . Hence, there are 4 such paths. (c) The number of paths of lengths 1, 2, and 3 from 3 to 2 are the (3 2) entries of A, A2 , and A3 , respectively. Adding these three matrix entries yields 1 + 2 + 4 = 7. Hence, the number of paths of length ≤ 3 from 3 to 2 is 7.

(e) An inspection of the digraph shows that no directed edges are going to 5 . Hence, there are no paths from any vertex of any length going to 5 . This can also be seen in the adjacency matrix.

Since the 5th column of A has all zeroes, every power A of A (for any integer   0) will also have all zeroes in its 5th column (because the 5th column of A equals A−1 multiplied by the 5th column of A).

(6) In each part, we use the adjacency matrix for the graph in Figure 8.6 and their powers as computed above for Exercise 4. (a) The number of cycles of length 3 connecting 2 to itself is the (2 2) entry of A3 . Using the matrix A3 from Exercise 4 yields 8 cycles of length 3 connecting 2 to itself in Figure 8.6. (c) The number of cycles of length ≤ 4 connecting 4 to itself is the sum of the (4 4) entries of A, A2 , A3 , and A4 . Using the (4 4) entries from the matrices from Exercise 4 yields 0 + 6 + 12 + 74 = 92 cycles of length ≤ 4 connecting 4 to itself in Figure 8.6.

(7) In each part, we use the adjacency matrix for the graph in Figure 8.7 and their powers as computed above for Exercise 5. (a) The number of cycles of length 3 connecting 2 to itself is the (2 2) entry of A3 . Using the matrix A3 from Exercise 5 yields 3 cycles of length 3 connecting 2 to itself in Figure 8.7. (c) The number of cycles of length ≤ 4 connecting 4 to itself is the sum of the (4 4) entries of A, A2 , A3 , and A4 . Since we did not compute A4 in Exercise 5, we still need the (4 4) entry of A4 .

This is the dot product of the 4th row of A3 with the 4th column of A, resulting in 12. Using this and the (4 4) entries from the matrices from Exercise 5 yields 1 + 2 + 4 + 12 = 19 cycles of length ≤ 4 connecting 4 to itself in Figure 8.7.

(8) (a) If the vertex is the th vertex, then the th row and th column entries of the adjacency matrix all equal zero, except possibly for the ( ) entry, which would equal the number of loops at the th vertex.

(9) (a) Since the ( ) entry indicates the number of loops at the th vertex, the trace, which is the sum of the main diagonal entries, equals the total number of loops in the graph or digraph.

(10) In each part, we use Theorem 8.3. Compute A + A2 + · · · + A−1 , where  is the number of vertices in the corresponding graph. If this sum has no zeros off the main diagonal, then the graph is connected.

If some entry off the main diagonal equals zero, the graph is disconnected. ⎡ ⎤ 26 0 20 28 ⎢ 0 3 0 0 ⎥ ⎥ (a) A + A2 + A3 = ⎢ ⎣ 20 0 14 24 ⎦. Because this matrix has zero entries off the main diagonal, 28 0 24 20 the corresponding graph is disconnected.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

246

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

6 ⎢ 2 (b) A + A2 + A3 = ⎢ ⎣ 5 1

2 2 4 3

5 4 3 1

Section 8.1

⎤ 1 3 ⎥ ⎥. Because this matrix has no zero entries off the main diagonal, 1 ⎦ 1

the corresponding graph is connected.

(11) (a) Since 2 has the same edges as 1 , along with one new loop at each vertex, 2 has the same number of edges connecting any two distinct vertices as 1 . Thus, the entries off the main diagonal of the adjacency matrices for the two graphs are the same. But 2 has one more loop at each vertex than 1 has. Hence, the entries on the main diagonal of the adjacency matrix for 2 are all 1 larger than the entries of A, so the adjacency matrix for 2 is found by adding I to A. This does not change any entries off the main diagonal, but adds 1 to every entry on the main diagonal. (b) Connectivity only involves the existence of a path between distinct vertices. If there is a path from  to  in 1 , for  6= , the same path connects these two vertices in 2 . If there is a path from  to  in 2 , then a similar path can be found from  to  in 1 by merely deleting any loops that might appear in the path from  to  in 2 . Thus,  is connected by a path to  in 1 if and only if  is connected by a path to  in 2 . Hence, 1 is connected if and only if 2 is connected. (c) Suppose there is a path of length  from  to  in 1 , where  ≤ . Then, we can find a path of length  from  to  in 2 by first following the known path of length  from  to  that exists in 1 (and hence in 2 ), and then going ( − ) times around the newly added loop at  .

(13) (a) The digraph 2 in Figure 8.2 is strongly connected since there is a path from any vertex to any other vertex. (This can be verified by inspection, or by using part (b) of this exercise — that is, letting A represent the adjacency matrix for this digraph and noting that A + A2 + A3 has all nonzero entries off the main diagonal.) The digraph in Figure 8.7 is not strongly connected, since there is no path directed to 5 from any other vertex (see the answer to Exercise 5(e), above). (d) For the first adjacency matrix, let C = (A + A )Then ⎡ ⎡ ⎤ ⎤ 0 0 0 2 0 56 0 84 28 0 ⎢ 0 2 0 0 1 ⎥ ⎢ 0 62 0 0 58 ⎥ ⎢ ⎢ ⎥ ⎥ 2 3 4 ⎢ ⎢ ⎥ C = ⎢ 0 0 0 3 0 ⎥ , and, C + C + C + C = ⎢ 84 0 126 42 0 ⎥ ⎥ ⎣ 2 0 3 0 0 ⎦ ⎣ 28 0 42 182 0 ⎦ 0 1 0 0 2 0 58 0 0 62

However, some of the entries off the main diagonal of this sum are zero, so the digraph for A is not weakly connected.

For the second adjacency matrix, let D = (B + B ) Then ⎡ ⎡ ⎤ ⎤ 0 0 2 2 0 80 20 24 20 32 ⎢ 0 0 0 1 0 ⎥ ⎢ 20 6 4 6 4 ⎥ ⎢ ⎢ ⎥ ⎥ ⎥ , and, D + D2 + D3 + D4 = ⎢ 24 4 52 44 32 ⎥  2 0 0 0 1 D=⎢ ⎢ ⎢ ⎥ ⎥ ⎣ 2 1 0 0 0 ⎦ ⎣ 20 6 44 46 12 ⎦ 0 1 1 0 2 32 4 32 12 52

Since this sum has no zero entries off the main diagonal, the digraph for B is weakly connected.

(14) (b) Yes, it is a dominance digraph, because no tie games are possible and because each team plays every other team. Thus if  and  are two given teams, either  defeats  or vice versa. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

247

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.1

(15) We prove Theorem 8.1 by induction on . For the base step,  = 1. This case is true by the definition of the adjacency matrix. For the inductive step, we assume the theorem is true for , and prove that it is true for  + 1. Let B = A . Then A+1 = BA. Note that the ( ) entry of A+1 = (th row of B) · (th column of A) = =

=

 X =1  X

 X

 

=1

(number of paths of length  from  to  ) · (number of paths of length 1 from  to  ) number of paths of length  + 1 from  to  that pass through  as the next-to-last vertex

=1

= (number of paths of length  + 1 from  to  ) The last equation is true because any path of length  + 1 from  to  must be composed of a path of length  from  to  , for some  followed by a path of length 1 from  to  .

(16) (a) True. In a graph, no distinction is made regarding the direction of any edge. Thus, there is an edge connecting  to  if and only if there is an edge connecting  to  . (The same edge is used in both cases.) Hence, the ( ) entry of the adjacency matrix must have the same value as the ( ) entry of the matrix. (b) False. By definition, the adjacency matrix of a digraph can contain only nonnegative numbers as entries. But, if the digraph has a directed edge going from  to  , with  6= , then the ( ) entry of its adjacency matrix would be positive. Hence, in order to be skew-symmetric, the ( ) entry of the matrix would have to be negative. Also, any loops in the digraph would make a main diagonal entry nonzero, which would make the matrix fail to be skew-symmetric. (c) True. The (1 2) entry of A gives the number of paths of length  from 1 to 2 . If this entry is zero for all  ≥ 1, then there are no paths of any length from 1 to 2 .

(d) False. In a graph, both the ( ) and the ( ) entries of the adjacency matrix give the number of edges connecting  and  for  6= . Hence, summing entries in the adjacency matrix would count all such edges twice. For example, the graph 1 in Example 1 of Section 8.1 of the textbook has 5 edges, but the sum of the entries in its adjacency matrix is 10. Note that 2 in that same example has 8 edges, and the sum of the entries in its adjacency matrix is 15. (The loop connecting 2 to itself is only counted once in the adjacency matrix.) (e) True. In a digraph, each directed edge is counted once and only once in the associated adjacency matrix. That is, a directed edge from  to  contributes to the ( ) entry of the matrix, and to no other entry. (f) True. To get from 1 to 3 in  +  steps, first follow the path of length  from 1 to 2 , and then follow the path of length  from 2 to 3 . (g) True. The ( ) entry of the adjacency matrix gives the number of edges connecting  to  .

Adding these numbers for 1 ≤  ≤  gives the total number of edges connecting some vertex to  . (h) False. The ( ) entry of the adjacency matrix gives the number of directed edges going from  to  . Adding these numbers for 1 ≤  ≤  gives the total number of directed edges going toward  . Hence, this does not count all of the directed edges going away from  , with the exception of the loops. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

248

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.2

(i) False. If the (3 4) entry of A4 is 5, then there are 5 paths of length 4 from 3 to 4 , not necessarily 4 paths of length 5. (j) False. The (5 2) entry of A + A2 + A3 + A4 is the number of paths of length ≤ 4 from 5 to 2 .

The number of paths of length 4 from 5 to 2 is the (5 2) entry of A4 . (k) False. A graph with adjacency matrix A can be connected if the (2 3) entry of A4 is 0 (For example, consider the graph of 5 vertices (1 through 5 ) having only the following 4 edges: an edge between 1 and 2 , an edge between 2 and 3 , an edge between 3 and 4 , and an edge between 4 and 5 . This is easily seen to be a connected graph. But, if A is the adjacency matrix for this graph, then the (2 3) entry of A4 is 0) However, if the (2 3) entry of A + A2 + A3 + A4 is known to be 0 for an adjacency matrix A, then by Theorem 8.3, the corresponding graph could not be connected. (l) False. A graph with adjacency matrix A can be connected if the (2 3) entry of A + A2 + A3 is 0 (For example, consider the graph of 5 vertices (1 through 5 ) having only the following 4 edges: an edge between 2 and 1 , an edge between 1 and 4 , an edge between 4 and 5 , and an edge between 5 and 3 . This is easily seen to be a connected graph. But, if A is the adjacency matrix for this graph, then the (2 3) entry of A + A2 + A3 is 0) However, if the (2 3) entry of A + A2 + A3 + A4 is known to be 0 for an adjacency matrix A, then by Theorem 8.3, the corresponding graph could not be connected. (m) True. If the (2 3) entry of A+A2 +A3 +A4 +A5 is 0 then, since all entries of A5 are nonnegative, we see that the (2 3) entry of A + A2 + A3 + A4 must also be 0. (Since each entry of A , for any integer   0, counts the number of paths of length  connecting two vertices, each entry of A must be nonnegative.)

Section 8.2 (1) (a) The given circuit has two junctions and two loops. By Kirchhoff’s First Law, both junctions produce the same equation: 1 = 2 + 3 . The two loops starting and ending at the voltage source 36 are: (1) 1 → 2 → 1 (2) 1 → 3 → 1 .

Kirchhoff’s Second Law gives an Ohm’s Law equation for each of these two loops: 36 − 2 (4Ω) − 1 (2Ω) = 0 36 + 7 − 3 (9Ω) − 1 (2Ω) = 0

(loop 1) (loop 2)

These relationships lead to the following system of linear equations: ⎧ ⎨ 1 − 2 − 3 = 0 = 36  21 + 42 ⎩ + 93 = 43 21

Thus, we row reduce ⎡

¯ ⎤ 1 −1 −1 ¯¯ 0 ⎣ 2 4 0 ¯¯ 36 ⎦ 2 0 9 ¯ 43

to obtain

This gives the solution 1 = 8, 2 = 5, and 3 = 3. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

249

¯ ⎤ 1 0 0 ¯¯ 8 ⎣ 0 1 0 ¯ 5 ⎦ ¯ 0 0 1 ¯ 3 ⎡

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.2

(c) The given circuit has four junctions and four loops. By Kirchhoff’s First Law, the junctions produce the following equations: 1 6 6 1

= = = =

2 3 3 2

+ + + +

6 4 4 6

+ 5  + 5

The last two of these equations are redundant. Next, the four loops starting and ending at the voltage source 42 are: (1) 1 → 6 → 3 → 6 → 1 (2) 1 → 6 → 4 → 6 → 1 (3) 1 → 6 → 5 → 6 → 1 (4) 1 → 2 → 1 Kirchhoff’s Second Law gives an Ohm’s Law equation for each of these four loops: 42 − 3 (6Ω) − 1 (2Ω) = 0 42 − 4 (9Ω) − 1 (2Ω) = 0 42 − 5 (9Ω) − 1 (2Ω) = 0 42 + 7 − 2 (5Ω) − 1 (2Ω) = 0

(loop 1) (loop 2) (loop 3) (loop 4)

These relationships lead to the following system of linear equations: ⎧ 1 − 2 − 6 ⎪ ⎪ ⎪ ⎪ +  +   ⎪ 3 4 5 − 6 ⎪ ⎨ + 63 21 + 94 21 ⎪ ⎪ ⎪ ⎪ + 95 2 ⎪ 1 ⎪ ⎩ 21 + 52

Thus, we row reduce ⎡ 1 −1 0 ⎢ 0 0 1 ⎢ ⎢ 2 0 6 ⎢ ⎢ 2 0 0 ⎢ ⎣ 2 0 0 2 5 0

0 1 0 9 0 0

¯ ⎤ 0 −1 ¯¯ 0 1 −1 ¯¯ 0 ⎥ ⎥ 0 0 ¯¯ 42 ⎥ ⎥ 0 0 ¯¯ 42 ⎥ ⎥ 9 0 ¯¯ 42 ⎦ 0 0 ¯ 49

⎡ to obtain

1 ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎣ 0 0

0 1 0 0 0 0

0 0 1 0 0 0

= = = = = =

0 0 0 1 0 0

This gives the solution 1 = 12, 2 = 5, 3 = 3, 4 = 2, 5 = 2, and 6 = 7.

0 0 42 42 42 49

0 0 0 0 1 0

¯ ⎤ 0 ¯¯ 12 0 ¯¯ 5 ⎥ ⎥ 0 ¯¯ 3 ⎥ ⎥ 0 ¯¯ 2 ⎥ ⎥ 0 ¯¯ 2 ⎦ 1 ¯ 7

(2) (a) True. However, some of the equations produced by the junctions may be redundant, as in the solution to Exercise 1(c), above. (b) True. This is Ohm’s law:  = .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

250

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.3

Section 8.3 (1) We use A and B to represent the matrices determined by the given points as described in Theorem 8.4. (a) Using the given data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 3 −8 ⎢ 1 1 ⎥ ⎢ ⎥ ⎥ and B = ⎢ −5 ⎥  A=⎢ ⎣ 1 0 ⎦ ⎣ −4 ⎦ 1 2 −1 ∙ ¸ ∙ ¸ 4 6 −18 and A B = .

Direct computation yields A A = 6 14 −31 Row reducing the augmented matrix ¯ ∙ ¸ 4 6 ¯¯ −18 6 14 ¯ −31

gives

∙

¯ ¸ 1 0 ¯¯ −33  0 1 ¯ −08

and so the line of best fit is  = −08 − 33. Plugging in  = 5 produces  ≈ −73. (c) Using the given data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 −4 10 ⎢ 1 −3 ⎥ ⎢ 8 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ A = ⎢ 1 −2 ⎥ and B = ⎢ ⎢ 7 ⎥ ⎣ 1 −1 ⎦ ⎣ 5 ⎦ 1 0 4 ∙ ¸ ∙ ¸ 5 −10 34   and A B = .

Direct computation yields A A = −10 30 −83 Row reducing the augmented matrix ¯ ∙ ¸ 5 −10 ¯¯ 34 −10 30 ¯ −83

gives

∙

¯ ¸ 1 0 ¯¯ 38  0 1 ¯ −15

and so the line of best fit is  = −15 + 38. Plugging in  = 5 produces  ≈ −37.

(2) We use A and B to represent the matrices determined by the given points as described in Theorem 8.4. (a) Using the given data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 −4 16 8 ⎢ 1 −2 4 ⎥ ⎢ 5 ⎥ ⎢ ⎥ ⎢ A=⎣ and B = ⎣ ⎥  1 0 0 ⎦ 3 ⎦ 1 2 4 6 ⎡ ⎤ ⎡ ⎤ 4 −4 24 22 24 −64 ⎦ and A B = ⎣ −30 ⎦.

Direct computation yields A A = ⎣ −4 24 −64 288 172 Row reducing the augmented matrix ¯ ⎡ ⎤ 4 −4 24 ¯¯ 22 ⎣ −4 24 −64 ¯¯ −30 ⎦ 24 −64 288 ¯ 172

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

251

gives

¯ ⎤ 1 0 0 ¯¯ 3600 ⎣ 0 1 0 ¯ 0350 ⎦  ¯ 0 0 1 ¯ 0375 ⎡

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.3

and so the least squares quadratic polynomial is  = 03752 + 035 + 36. (c) Using the given data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 −4 16 −3 ⎢ 1 −3 9 ⎥ ⎢ −2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ A = ⎢ 1 −2 4 ⎥ and B = ⎢ ⎢ −1 ⎥  ⎣ 1 ⎦ ⎣ 0 0 0 ⎦ 1 1 1 1 ⎡ ⎤ ⎡ ⎤ 5 −8 30 −5 30 −98 ⎦ and A B = ⎣ 21 ⎦.

Direct computation yields A A = ⎣ −8 30 −98 354 −69 Row reducing the augmented matrix ¯ ⎡ ⎤ 5 −8 30 ¯¯ −5 ⎣ −8 30 −98 ¯¯ 21 ⎦ 30 −98 354 ¯ −69

gives

¯ ⎤ 1 0 0 ¯¯ 0266 ⎣ 0 1 0 ¯ 0633 ⎦  ¯ 0 0 1 ¯ −0042 ⎡

and so the least-squares quadratic polynomial is  = −00422 + 0633 + 0266.

(3) We use A and B to represent the matrices determined by the given points as described in Theorem 8.4. (a) Using the given data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 −3 9 −27 −3 ⎢ 1 −2 4 −8 ⎥ ⎢ −1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ and B = ⎢ 0 ⎥  1 −1 1 −1 A=⎢ ⎢ ⎥ ⎢ ⎥ ⎣ 1 ⎣ 1 ⎦ 0 0 0 ⎦ 1 1 1 1 4 ⎡ ⎡ ⎤ ⎤ 5 −5 15 −35 1 ⎢ −5 ⎢ ⎥ 15 −35 99 ⎥ ⎥ and A B = ⎢ 15 ⎥.

Direct computation yields A A = ⎢ ⎣ 15 −35 ⎣ ⎦ 99 −275 −27 ⎦ −35 99 −275 795 93 Row reducing the augmented matrix ⎡

5 −5 15 ⎢ −5 15 −35 ⎢ ⎣ 15 −35 99 −35 99 −275

¯ ⎤ −35 ¯¯ 1 99 ¯¯ 15 ⎥ ⎥ −275 ¯¯ −27 ⎦ 795 ¯ 93

gives

¯ 1 0 0 0 ¯¯ 37 35 ⎢ ¯ ⎢ 0 1 0 0 ¯ 25 ⎢ ¯ 14 ⎢ 0 0 1 0 ¯ 25 ¯ 28 ⎣ ¯ 0 0 0 1 ¯ 1 ⎡

25 37 2 and so the least-squares cubic polynomial is  = 14 3 + 25 28  + 14  + 35 .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

252

4

⎤

⎥ ⎥ ⎥ ⎥ ⎦

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.3

(4) We use A and B to represent the matrices determined by the computed points as described in Theorem 8.4. (a) First, we use the function  = 4 and the given -values to generate the following data points: (−2 16), (−1 1), (0 0), (1 1), (2 16). Using this data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 −2 4 16 ⎢ 1 −1 1 ⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ and B = ⎢ 0 ⎥  1 0 0 A=⎢ ⎢ ⎥ ⎢ ⎥ ⎣ 1 ⎣ 1 ⎦ 1 1 ⎦ 1 2 4 16 ⎡ ⎤ ⎡ ⎤ 5 0 10 34 Direct computation yields A A = ⎣ 0 10 0 ⎦ and A B = ⎣ 0 ⎦.

10 0 34 130 Row reducing the augmented matrix ¯ ⎡ ⎤ 5 0 10 ¯¯ 34 ⎣ 0 10 0 ¯ 0 ⎦ ¯ 10 0 34 ¯ 130

gives

¯ ⎤ 1 0 0 ¯¯ −20571 ⎣ 0 1 0 ¯ 00000 ⎦  ¯ 0 0 1 ¯ 44286 ⎡

and so the least-squares quadratic polynomial is  = 442862 − 20571.

(c) First, we use the function  = ln  and the given -values to generate the following data points: (1 0), (2 06931), (3 10986), (4 13863). Note that, while we only list 4 digits after the decimal point here, all computations were actually performed on a calculator displaying 12 significant digits of accuracy. Using this data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 1 1 00000 ⎢ 1 2 4 ⎥ ⎢ 06931 ⎥ ⎥ ⎢ ⎥ A=⎢ ⎣ 1 3 9 ⎦ and B = ⎣ 10986 ⎦  1 4 16 13863 ⎡ ⎡ ⎤ ⎤ 4 10 30 31781 Direct computation yields A A = ⎣ 10 30 100 ⎦ and A B = ⎣ 102273 ⎦.

30 100 354 348408 Row reducing the augmented matrix ¯ ⎡ ⎤ 4 10 30 ¯¯ 31781 ⎣ 10 30 100 ¯ 102273 ⎦ ¯ 30 100 354 ¯ 348408

gives

¯ ⎤ 1 0 0 ¯¯ −08534 ⎣ 0 1 0 ¯ 09633 ⎦  ¯ 0 0 1 ¯ −01014 ⎡

and so the least-squares quadratic polynomial is  = −010142 + 09633 − 08534.

(e) First, we use the function  = cos  and the given -values to generate the following data points: √

√

(− 2  0), (− 4  22 ), (0 1), ( 4  22 ), ( 2  0). Making decimal approximations yields (−15708 0), (−07854 07071), (0 1), (07854 07071), (15708 0). Note that, while we only list 4 digits after the decimal point here, all computations were actually performed on a calculator displaying 12

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

253

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.3

significant digits of accuracy. Using this data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 −15708 24674 −38758 0 ⎢ 1 −07854 06169 −04845 ⎥ ⎢ 07071 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 0 A=⎢ 1 ⎥ and B = ⎢ 1 ⎥ ⎣ 1 ⎦ ⎣ 07854 06169 04845 07071 ⎦ 1 15708 24674 38758 0

Direct computation yields ⎡ ⎤ 5 0 61685 0 ⎢ 0 61685 0 129371 ⎥ ⎥ A A = ⎢ ⎣ 61685 0 ⎦ 129371 0 0 129371 0 305128 Row reducing the augmented matrix ¯ ⎡ ⎤ ¯ 24142 5 0 61685 0 ¯ ⎢ 0 ⎥ 61685 0 129371 ¯¯ 0 ⎢ ⎥ ¯ ⎣ 61685 0 ⎦ 129371 0 ¯ 08724 ¯ 0 129371 0 305128 0

⎡

⎤ 24142 ⎢ 0 ⎥ ⎥ and A B = ⎢ ⎣ 08724 ⎦  0

gives

⎡

⎢ ⎢ I4 ⎣

and so the least-squares cubic polynomial is  = 03 − 039542 + 09706.

¯ ⎤ ¯ 09706 ¯ ¯ 0 ⎥ ¯ ⎥ ¯ −03954 ⎦  ¯ ¯ 0

(5) We use A and B to represent the matrices determined by the given points as described in Theorem 8.4. This problem uses the following set of data points: (1 3), (3 33), (5 37), (7 41), (9 46). (a) Using the given data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 1 30 ⎢ 1 3 ⎥ ⎢ 33 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ and B = ⎢ 37 ⎥  1 5 A=⎢ ⎢ ⎥ ⎢ ⎥ ⎣ 1 7 ⎦ ⎣ 41 ⎦ 1 9 46 ∙ ¸ ∙ ¸ 5 25 187 and A B = .

Direct computation yields A A = 25 165 1015 Row reducing the augmented matrix ¯ ∙ ¸ 5 25 ¯¯ 187 25 165 ¯ 1015

gives

∙

¯ ¸ 1 0 ¯¯ 274  0 1 ¯ 020

and so the line of best fit is  = 02 + 274. Substituting  = 8 into this equation yields  = (02)(8) + 274 = 434. Thus, the tower was leaning at 434◦ in the 8th month. (c) Using the given data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 1 1 30 ⎢ 1 3 9 ⎥ ⎢ 33 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ and B = ⎢ 37 ⎥  1 5 25 A=⎢ ⎢ ⎥ ⎢ ⎥ ⎣ 1 7 49 ⎦ ⎣ 41 ⎦ 1 9 81 46 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

254

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.3

⎡

⎤ ⎡ ⎤ 5 25 165 187 Direct computation yields A A = ⎣ 25 165 1225 ⎦ and A B = ⎣ 1015 ⎦.

165 1225 9669 6987 Row reducing the augmented matrix ¯ ⎡ ⎤ 5 25 165 ¯¯ 187 ⎣ 25 165 1225 ¯ 1015 ⎦ ¯ 165 1225 9669 ¯ 6987

¯ ⎤ 1 0 0 ¯¯ 28614 ⎣ 0 1 0 ¯ 01286 ⎦  ¯ 0 0 1 ¯ 0007143 ⎡

gives

and so the least-squares quadratic polynomial is  = 00071432 + 01286 + 28614. Substituting  = 12 into this equation yields  = (0007143)(12)2 + (01286)(12) + 28614 = 54332. Thus, the tower will be leaning at 543◦ in the 12th month. (The townspeople decided to put a clock on the tower. They figured that if it had the inclination, it should have the time.)

(7) We use A and B to represent the matrices determined by the given points as described in Theorem 8.4. Using the given data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 −2 4 6 0 0 ⎦ and B = ⎣ 2 ⎦  A=⎣ 1 1 3 9 8 ⎡ ⎤ ⎡ ⎤ 3 1 13 16 Direct computation yields A A = ⎣ 1 13 19 ⎦ and A B = ⎣ 12 ⎦.

13 19 97 96 Row reducing the augmented matrix

¯ ⎤ 3 1 13 ¯¯ 16 ⎣ 1 13 19 ¯ 12 ⎦ ¯ 13 19 97 ¯ 96 ⎡

gives

¯ ⎤ 1 0 0 ¯¯ 2 ⎥ ⎢ ¯ ⎣ 0 1 0 ¯ − 25 ⎦  ¯ 0 0 1 ¯ 4 ⎡

5

and so the least-squares quadratic polynomial is  = 45 2 − 25  + 2. Since  = 6 when  = −2,  = 2 when  = 0, and  = 8 when  = 3, this polynomial is the exact quadratic through the given three points.

(9) (a) The matrices A and B are ⎡

⎤ ⎡ ⎤ 4 −3 12 5 ⎦ and B = ⎣ 32 ⎦  A=⎣ 2 3 1 21 ∙ ¸ ∙ ¸ 29 1 175 and A B = .

Direct computation yields A A = 1 35 145

Row reducing the augmented matrix ∙

¯ ¸ 29 1 ¯¯ 175 1 35 ¯ 145

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

gives

255

"

¯ # 1 0 ¯¯ 230 39  ¯ 0 1 ¯ 155 39

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.4

155 giving the least-squares solutions 1 = 230 39 , 2 = 39 . Note that ⎧ 41 − 32 = 11 23  which is almost 12 ⎪ ⎪ ⎨ 21 + 52 = 31 23  which is almost 32 . ⎪ ⎪ ⎩ 31 + 2 = 21 23  which is close to 21

Hence, the least-squares solution comes close to satisfying the original inconsistent system.

(10) (a) True. Since the distance of each data point to the line is 0, the sum of the squares of these distances is also 0. That is the minimum possible value for the sum  , since all such sums are nonnegative. (b) False. The computed coefficient of 3 in the degree three least-squares polynomial might be zero. For example, if all of the data points lie on a straight line, that line will be the degree  least-squares polynomial for every  ≥ 1. (c) False. For example, it is easy to check that the least-squares line computed in Example 1 of Section 8.3 of the textbook does not pass through any of the data points used to compute the line.

(d) False. Theorem 8.4 states that A is an  × ( + 1) matrix. Hence, A is a ( + 1) ×  matrix.

Therefore, A A is a ( + 1) × ( + 1) matrix.

Section 8.4 (1) A is not stochastic, since A is not square; A is not regular, since A is not stochastic.

B is not stochastic, since the entries of column 2 do not sum to 1; B is not regular, since B is not stochastic.

C is stochastic; C is regular, since C is stochastic and has all nonzero entries.

D is stochastic; D is not regular, since every positive power of D is a matrix whose rows are the rows of D permuted in some order, and hence every such power contains zero entries.

E is not stochastic, since the entries of column 1 do not sum to 1; E is not regular, since E is not stochastic.

F is stochastic; F is not regular, since every positive power of F has all second row entries zero.

G is not stochastic, since G is not square; G is not regular, since G is not stochastic. ⎡ 1 1 1 ⎤ 2

⎢ 1 H is stochastic; H is regular, since H = ⎢ ⎣ 4 2

(2) (a) First, p1 = Mp =

"

Then p2 = Mp1 =

"

⎡

⎢ (c) First, p1 = Mp = ⎣

1 4 3 4

1 3 2 3

1 4 3 4 1 4 1 2 1 4

1 3 2 3 1 3 1 3 1 3

#"

#" 1 2 1 6 1 3

2 3 1 3

#

5 18 13 18

1 4

= #

"

=

⎤⎡ 1 ⎤ ⎥⎢ ⎦⎣

4 1 2 1 4

4

4

1 2

1 4

1 4

1 2

5 18 13 18

"

#

. #

67 216 149 216

≈

⎡ 17 ⎤

⎥ ⎢ ⎦=⎣

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

⎥ ⎥, which has no zero entries. ⎦

48 1 3 5 16

256

⎥ ⎦.

∙

03102 06898

¸ .

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡ 1

⎢ Then p2 = Mp1 = ⎣

4 1 2 1 4

1 3 1 3 1 3

1 2 1 6 1 3

⎤ ⎡ 17 ⎤ ⎥⎢ ⎦⎣

48 1 3 5 16

⎡ 205 ⎤

⎥ ⎢ ⎦=⎣

576 49 144 175 576

Section 8.4

⎡

⎤ 03559 ⎥ ⎣ ⎦ ≈ 03403 ⎦.

03038

(3) (a) We must solve the system (M − I2 )x = 0, under the additional condition that 1 + 2 = 1. Thus, we row reduce ¯ ¯ ⎤ ⎡ 1 ⎤ ⎡ 1 ¯ 1 0 ¯¯ 25 −2 3 ¯ 0 ⎥ ⎢ 1 ¯ ⎥ ⎢ ¯ ⎣ 2 − 13 ¯ 0 ⎦ to ⎣ 0 1 ¯ 35 ⎦  ¯ ¯ 1 1 ¯ 1 0 0 ¯ 0 ¤ £ giving the steady state vector p = 25  35 .

(4) (a) We will round numbers to four places after the decimal point. Then, ∙ ¸ ∙ ¸ ∙ 04167 03889 04028 03981 04005 2 3 4 M =  M =  M = 05833 06111 05972 06019 05995 ∙ ¸ ∙ ¸ ∙ 04001 03999 04000 04000 04000 5 6 7 M =  M =  M = 05999 06001 06000 06000 06000

03997 06003 04000 06000

¸ ¸





Since all entries of M6 and M7 agree to 4 places after the decimal point, we stop here. Hence, the steady state vector for this Markov chain is given by any of the columns of M7 , namely [04000 06000], rounded to 4 significant digits.

(5) (a) The initial probability vector p is [03 015 045 01]. (Note that 10% of the citizens did not vote in the last election.) Then ⎡ ⎤⎡ ⎤ ⎡ ⎤ 07 02 02 01 030 0340 ⎢ 01 06 01 01 ⎥ ⎢ 015 ⎥ ⎢ 0175 ⎥ ⎥⎢ ⎥ ⎢ ⎥ p1 = Mp = ⎢ ⎣ 01 02 06 01 ⎦ ⎣ 045 ⎦ = ⎣ 0340 ⎦  01 00 01 07 010 0145

Thus, in the next election, 34% will vote for Party A, 17.5% will vote for Party B, 34% will vote for Party C, and 14.5% will not vote at all.

For the election after that, ⎡ ⎤⎡ ⎤ ⎡ ⎤ 07 02 02 01 0340 03555 ⎢ 01 06 01 01 ⎥ ⎢ 0175 ⎥ ⎢ 01875 ⎥ ⎥⎢ ⎥ ⎢ ⎥ p2 = Mp1 = ⎢ ⎣ 01 02 06 01 ⎦ ⎣ 0340 ⎦ = ⎣ 02875 ⎦  01 00 01 07 0145 01695

Hence, in that election, 3555% will vote for Party A, 1875% will vote for Party B, 2875% will vote for Party C, and 1695% will not vote at all.

(b) A lengthy computation shows that M100 p1 ≈ [036 020 024 020]. Hence, in a century, the votes would be 36% for Party A and 24% for Party C. For an alternate approach, we assume that by the end of a century the system will have achieved its steady state. Hence, we can solve for the steady-state vector. Since the transition matrix M is regular (you can easily verify that M2 has no zero entries), part (5) of Theorem

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

257

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.4

8.7 shows that we can solve for p by solving the system (M − I4 )x = 0, under the additional condition that 1 + 2 + 3 + 4 = 1. Hence, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ −03 02 02 01 ¯¯ 0 1 0 0 0 ¯¯ 036 ⎢ 01 −04 ⎢ 0 1 0 0 ¯ 020 ⎥ 01 01 ¯¯ 0 ⎥ ⎢ ⎥ ⎢ ¯ ⎥ ⎢ 01 ¯ 0 ⎥ to ⎢ 0 0 1 0 ¯ 024 ⎥  02 −04 01 ⎢ ¯ ⎥ ⎢ ¯ ⎥ ⎣ 01 ⎣ 0 0 0 1 ¯ 020 ⎦ 00 01 −03 ¯¯ 0 ⎦ ¯ 10 10 10 10 ¯ 1 0 0 0 0 ¯ 000

Therefore, the steady-state vector is [036 020 024 020]. Again, we get that, in a century, the votes would be 36% for Party A and 24% for Party C. (6) (a) To compute the transition matrix M, we use the fact that the ( ) entry  of M is the probability that the rat moves from room  to room , where we use the labels A, B, C, D, and E and 1, 2, 3, 4, and 5 for the rooms interchangeably. So, for example,  = 12 for each  because the rat has a 50% chance of staying in the same room. To compute 21 , note that each of the 4 doorways leaving room A has the same probability of being used when a doorway is used, namely, 1 1 1 1 1 4 . But, since the rat only uses a doorway only 2 of the time, 21 = 2 × 4 = 8 . Similarly, to 1 compute 41 , again note that each door out of room A has probability 4 of being used any time a door is used. Now combining the facts that a doorway is ¡used only ¢ half the time, and that there are two doors from room A to room D implies that 41 = 14 × 2 × 12 = 14 . Similar computations produce all of the other entries of M giving ⎡ 1 1 1 1 ⎤ 0 2 6 6 5 ⎢ ⎥ ⎢ 1 1 0 0 1 ⎥ ⎢ 8 2 5 ⎥ ⎢ ⎥ ⎢ 1 1 ⎥ M = ⎢ 18 0 12 10  10 ⎥ ⎢ ⎥ ⎢ 1 ⎥ 1 ⎥ ⎢ 4 0 16 12 5 ⎦ ⎣ 1 0 13 61 15 2 (b) If M is the stochastic matrix in part (a), then ⎡ 41 1 ⎢ ⎢ ⎢ ⎢ ⎢ 2 M =⎢ ⎢ ⎢ ⎢ ⎣

120

6

1 5

13 60

9 100

1 8

27 80

13 240

13 200

1 5

3 20

13 240

73 240

29 200

3 25

13 48

13 120

29 120

107 300

13 60

9 80

1 3

1 5

13 60

28 75

which has all entries nonzero. Thus, M is regular.

⎤

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(c) If the rat starts in room C, its initial state vector is p = [0 0 1 0 0]. By Theorem 8.6, the state vector p2 is given by M2 p. Using the matrix M2 computed in part (b) and the fact that p is the £ 13 73 29 1 ¤  240  120  5 . Hence, vector e3 , we see that M2 p is just the third column of M2 , namely 15  240 29 the probability of the rat being in room D after two time intervals is 120 .

(d) Since the transition matrix M is regular, part (5) of Theorem 8.7 shows that we can solve for p by solving the system (M−I5 )x = 0, under the additional condition that 1 +2 +3 +4 +5 = 1. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

258

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.4

Hence, we row reduce ¯ ¯ ⎤ ⎡ 1 ⎤ ⎡ 1 1 1 0 ¯¯ 0 −2 1 0 0 0 0 ¯¯ 15 6 6 5 ⎢ 1 ⎥ ⎢ ¯ 3 ⎥ 1 ¯ 0 0 ⎥ ⎢ 8 − 12 ⎢ 0 1 0 0 0 ¯ 20 5 ¯¯ 0 ⎥ ⎥ ⎢ ⎥ ⎢ ¯ 3 ⎥ 1 1 1 ¯ ⎢ 1 ⎥ ⎢ ¯ 0 0 0 1 0 0 0 −2 ⎢ 8 ¯ ⎥ ⎢ ¯ 10 10 20 ⎥ ⎢ 1 ⎥ to ⎢ ¯ 1 ⎥ 1 1 1 ¯¯ ⎥ ⎢ 4 ⎥ ⎢ ¯ 0 −2 6 5 ¯ 0 ⎥ ⎢ ⎢ 0 0 0 1 0 ¯ 4 ⎥ ⎢ ¯ ⎥ ⎢ ¯ 1 ⎥ 1 1 1 − 12 ¯ 0 ⎦ ⎣ 0 ⎣ 0 0 0 0 1 ¯ 4 ⎦ 3 6 5 ¯ ¯ 1 1 1 1 1 ¯ 1 0 0 0 0 0 ¯ 0 £1 3 3 1 1¤ Therefore, the steady-state vector is 5  20  20  4  4 . Over time, the rat frequents rooms B and C the least and rooms D and E the most.

(9) We need to show that the product of any  stochastic matrices is stochastic. We prove this by induction on . The Base Step is  = 2. Let A and B be  ×  stochastic matrices. Then the entries of AB are nonnegative, since the entries of both A and B are nonnegative. Furthermore, the sum of the entries in the th column of AB =  X (AB)

=

=1

= =

 X =1  X

(th row of A) · (th column of B) (1 1 + 2 2 + · · · +   )

=1 Ã  X =1

!

1 1 +

à  X =1

2

!

2 + · · · +

à  X =1

!

 

= (1)1 + (1)2 + · · · + (1) (since A is stochastic, and each of the summations is a sum of the entries of a column of A) = 1 since B is stochastic. Hence AB is stochastic.

For the Inductive Step, we assume that the product of any  stochastic  ×  matrices is stochastic, and we must show that the product of any  + 1 stochastic  ×  matrices is stochastic. So, let A1      A+1 be stochastic  ×  matrices. Then A1 A2 · · · A A+1 = (A1 A2 · · · A )A+1 . Now A1 A2 · · · A is stochastic by the inductive hypothesis. Hence, (A1 A2 · · · A )A+1 is stochastic by the Base Step.

(10) We prove Theorem 8.6 by induction on . Suppose M is a  ×  matrix.

P Base Step ( = 1): th entry in Mp = (th row of M) · p = =1   =

 X

(probability of moving from state  to  )(probability of being in state  )

=1

= probability of being in state  after 1 step of the process = th entry of p1  Inductive Step: Assume p = M p is the probability vector after  steps of the process. Then, after an additional step of the process, the computation in the Base Step shows that the probability vector p+1 = Mp = M(M p) = M+1 p. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

259

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.5

(11) (a) False. While the sum of the entries in each column of a stochastic matrix must equal 1, the sum of the entries in each row are not required to equal 1. For example, this is true of the stochastic matrix M in Example 1 of Section 8.4 of the textbook. Now the columns of the transpose of a matrix equal the rows of the original matrix. Thus, the transpose of a stochastic matrix might not have the sum of its entries along each column equal to 1, and so might not be stochastic. (b) True. If A is upper triangular, then A is also upper triangular for all  ≥ 1. Hence, A will have zeroes below the main diagonal. (c) True. By part (5) of Theorem 8.7, the given statement is true for the vector p = p . (d) True. By part (3) of Theorem 8.7, if the Markov chain has a regular transition matrix, then there is a limit vector for every initial vector (and all of these limit vectors are equal). However, starting with the initial vector p, the given condition implies that M p = q when  is odd, and M p = p when  is even. Hence, lim M p can not exist, since M p oscillates between p and →∞

q. Therefore, M can not be regular. (e) False. Each entry of the transition matrix gives the probability of changing from one specific state to another.

Section 8.5 (1) (a) First, we must convert the letters in the message “PROOF BY INDUCTION” into numbers by listing each letter’s position in the alphabet: P

R

O

O

F

B

Y

I

N

D

U

C

T

I

O

N

16

18

15

15

6

2

25

9

14

4

21

3

20

9

15

14

Next, we take these numbers in pairs to form eight 2-vectors, and multiply the matrix ∙ ¸ 3 −4 A= times each of these 2-vectors in turn. This yields: 5 −7 ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ 16 −24 15 −15 6 10 25 39 A = , A = , A = , A = , 18 −46 15 −30 2 16 9 62 ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ 14 26 21 51 20 24 15 −11 A = , A = , A = , A = .

4 42 3 84 9 37 14 −23

Therefore, the encoded message is

−24 −46 −15 −30 10 16 39 62 26 42 51 84 24 37 −11 −23  ⎡

⎤ −2 −1 0 (2) (a) If A is the given key matrix, by row reducing [A|I3 ] we find A−1 to be A−1 = ⎣ −6 −4 1 ⎦.

9 4 1 Next, we take the numbers representing the message to be decoded in triples to form eight 3-vectors, and multiply the matrix A−1 times each of these 3-vectors in turn. This yields:

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

260

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

⎤ ⎡ ⎤ −62 8 A−1 ⎣ 116 ⎦ = ⎣ 15 ⎦ , 107 13 ⎡ ⎤ ⎡ ⎤ −160 19 A−1 ⎣ 301 ⎦ = ⎣ 7 ⎦, 251 15 ⎡ ⎤ ⎡ ⎤ −78 8 A−1 ⎣ 148 ⎦ = ⎣ 5 ⎦, 129 19

⎡

⎤ ⎡ ⎤ −32 5 A−1 ⎣ 59 ⎦ = ⎣ 23 ⎦, 67 15 ⎡ ⎤ ⎡ ⎤ −122 15 A−1 ⎣ 229 ⎦ = ⎣ 4 ⎦ , 188 6 ⎡ ⎤ ⎡ ⎤ −111 15 A−1 ⎣ 207 ⎦ = ⎣ 21 ⎦.

183 12

Section 8.6

⎡

⎤ ⎡ ⎤ −142 18 A−1 ⎣ 266 ⎦ = ⎣ 11 ⎦ , 223 9 ⎡ ⎤ ⎡ ⎤ −122 15 A−1 ⎣ 229 ⎦ = ⎣ 18 ⎦, 202 20

We take the results and associate with each number its corresponding letter in the alphabet: 8

15

13

5

23

15

18

11

9

19

7

15

H

O

M

E

W

O

R

K

I

S

G

O

15

4

6

15

18

20

8

5

19

15

21

12

O

D

F

O

R

T

H

E

S

O

U

L

Therefore, the message is “HOMEWORK IS GOOD FOR THE SOUL.” (3) (a) True. This is because the code used for a given letter is dependent on the letters that surround it in the message, not just upon the letter itself. Thus, the same letter may be replaced by several different numbers within the same message. (b) True. If the encoding matrix is singular, then there is no inverse matrix that can be used to decode the message. Decoding would involve solving systems having an infinite number of solutions. (c) False. The message is typically broken up into shorter strings of length , where  is less than the length of the entire message (see Example 1 in Section 8.5 of the textbook). Of course, the higher the value of , the more difficult it generally is to break the code. However, large values for  also make it more time consuming to invert the key matrix and to perform the necessary matrix products.

Section 8.6 √ (1) (c) For this parabola, the coefficients , , and  are  = 3,  = 1, and  = −2 3. Since,  6= , " √3 # ∙ ¸ 1 ³ ´ ¡ √ ¢ cos  − sin  2 2 1  1  =  = 2 arctan − = 2 arctan − 3 = − 6 . Hence, P = . √ sin  cos  3 − 12 2 √

√

Therefore,  = 23  + 12  and  = − 12  + 23 . Next, we substitute these expressions for  and  into the original equation for the conic section, producing !2 à !à Ã√ Ã√ √ !2 √ ! √ 1 1 1 1 3 3 3 3 +  + − +  −2 3 +  − +  3 2 2 2 2 2 2 2 2 ³

√ ´ − 1 + 12 3

Ã√

1 3 +  2 2

!

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

à √ ! ³ √ ´ 1 3 + 12 − 3  + 26 = 0 − + 2 2 261

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.6

(d) For this ellipse, ³ the´ coefficients ¡, ,¢ and  are  = 29,  = 36, and  = −24. Since,  6= , 1  ≈ 06435 radians (about 36◦ 520 ). We compute exact values  = 2 arctan − = 12 arctan 24 7 for sin  and cos . Now 2 = arctan( 24 7 ). So, using the right triangle in Figure 14,

25

24

2θ

7

Figure 14 7 , and so we see that cos 2 = 25

cos  =

Hence, P =

∙

r

r 7 p 1 + 25 4 3 16 = and sin  = 1 − cos2  = 1 − =  2 5 25 5 ¸ " 4 −3 # 5 5 − sin  = . Therefore,  = 45 − 35  and  = 35 + 45 . Next, we 3 4 cos 

1 + cos 2 = 2

cos  sin 

s

5

5

substitute these expressions for  and  into the original equation for the conic section, producing ¢2 ¡ ¢2 ¡ ¢¡ ¢ ¡ 29 45  − 35  ¡ + 36 35¢ + 45 ¡ − 24 ¢45  − 35  35  + 45  − 118 45  − 35  + 24 35  + 45  − 55 = 0 Expanding the terms on the left side of the equation yields

464 2 696 261 2 324 2 864 576 2 288 2 168 288 2 472 354 25  − 25  + 25  + 25  + 25  + 25  − 25  − 25  + 25  − 5  + 5  96 + 72 5  + 5  − 55 = 0

Combining like terms, rearranging terms, and dividing by 5 yields the following equation in coordinates: 42 − 16 + 9 2 + 18 = 11, which can be expressed, after completing the squares 2 2 in both  and , as (−2) + (+1) = 1. From this form, we can see that the conic section is an 9 4 ellipse and that its center in -coordinates is (2 −1). Its center in -coordinates is " # ∙ ¸ 11 ¢ ¡ 2 5 2 P = ; that is, the point 11 5  5 . Figures 15 and 16 on the next page show the 2 −1 5

graph of the ellipse in each of the two coordinate systems.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

263

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

From this form, we can see that the conic section is a hyperbola and that its center in ∙ ¸ ∙ ¸ 0 0 coordinates is (0 0). Its center in -coordinates is P = ; that is, the origin. Figures 0 0

17 and 18 show the graph of the hyperbola in each of the two coordinate systems.

(2) (a) True. This is because the coefficients of 2 and  2 are equal. (b) False. Since the entire plane is rotated about the origin by the change in coordinates, the center of a hyperbola is rotated to a new position, unless the center is at the origin, in which case the √ (− 2)2 2 − 2 =1 center stays fixed. For example. the hyperbola  −  −  = 0, which changes to 2 ¡√ ¢ ◦ after a 45 clockwise rotation, has its center move from (1 1) in -coordinates to 2 0 in -coordinates. (c) True. This is explained in Section 8.6 of the textbook, just after Example 1. (d) False. For example, the conic section 2 −  = 0 is a parabola with vertex at the origin whose graph is symmetric with respect to the -axis, but not with respect to the -axis. In general, if there is no  term, any axis of symmetry must be parallel to either the - or -axis, but it does not have to be one of these axes. For example, the ellipse ( − 1)2 + ( − 1)2 = 1 (which, when expanded, has no  term) has the lines  = 1 and  = 1 as horizontal and vertical axes of symmetry.

Section 8.7 (1) In each part we use the following list of vertices for the graphic in Figure 8.26(a) in Section 8.7 of the textbook: (5 3), (5 7), (8 7), (10 5) and (8 3). (a) A translation in the direction of the vector [4 −2] is computed by merely adding this vector to each of the vertices in the graphic. This produces the new vertices (9 1), (9 5), (12 5), (14 3) and (12 1). c 2016 Elsevier Ltd. All rights reserved.

Copyright °

265

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

(c) A reflection through a line of slope  through the origin is computed by multiplying the vector ∙ ¸ 1 − 2 2 1 . In this problem,  = 3, corresponding to each vertex by the matrix 1+2 2 2 − 1 ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ −08 06 5 −22 5 02 and so the matrix used is A = . Now, A = ,A = , 06 08 3 54 7 86 ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ 8 −22 10 −5 8 −46 A = ,A = and A = . Therefore, rounding each of 7 104 5 10 3 72 these results to the nearest whole number, the vertices for the reflected graphic are (−2 5), (0 9), (−2 10), (−5 10) and (−5 7).

(2) In each part we use the following list of vertices for the graphic in Figure 8.26(b) in Section 8.7 of the textbook: (6 6), (8 4), (10 6), (14 2) and (18 6). (b) A rotation about the origin through an angle  is computed by multiplying the vector correspond∙ ¸ cos  − sin  ing to each vertex by . In this problem,  = 120◦ , and so the matrix is sin  cos  " √ # ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ − 12 − 23 −05 −0866 6 −820 8 −746 √ A= ≈ . Now, A ≈ ,A ≈ , 3 0866 −05 6 220 4 493 − 12 2 ∙ ∙ ∙ ¸ ∙ ¸ ¸ ∙ ¸ ¸ ∙ ¸ 10 −1020 14 −873 18 −1420 A ≈ ,A ≈ and A ≈ . Therefore, round6 566 2 1112 6 1259

ing each of these results to the nearest whole number, the vertices for the rotated graphic are (−8 2), (−7 5), (−10 6), (−9 11) and (−14 13). The new graphic is illustrated in Figure 19 on the next page.

(d) A scaling about the origin with scale factors  in the -direction and  in the -direction is ∙ ¸  0 computed by multiplying the vector corresponding to each vertex by the matrix . In 0  " # ∙ ¸ ∙ ¸ 1 0 6 3 2 1 . Now, A = , this problem,  = 2 and  = 3, and so the matrix is A = 6 18 0 3 ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ 8 4 10 5 14 7 18 9 A = ,A = ,A = and A = . Therefore, 4 12 6 18 2 6 6 18 the vertices for the scaled graphic are (3 18), (4 12), (5 18), (7 6) and (9 18). The new graphic is illustrated in Figure 20.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

266

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Figure 19

Figure 20

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

267

Section 8.7

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

(3) In each part we use the following list of vertices for the graphic in Figure 8.26(c) in Section 8.7 of the textbook: (2 4), (2 10), (67), (8 10) and (10 4). In homogeneous coordinates, we express these points as the vectors [2 4 1], [2 10 1], [6 7 1], [8 10 1] and [10 4 1]. (a) When using homogeneous coordinates, a rotation about the origin through an angle  is computed ⎡ ⎤ cos  − sin  0 cos  0 ⎦ . In this by multiplying the vector corresponding to each vertex by A = ⎣ sin  0 0 1 √ ⎤ ⎡ √2 ⎡ ⎤ − 22 0 0707 −0707 0 √ √2 ⎥ ⎢ 2 0707 0 ⎦. Also, problem,  = 45◦ , and so the matrix is A = ⎣ 22 0 ⎦ ≈ ⎣ 0707 2 0 0 1 0 0 1

a reflection through a line of slope  through the origin is computed by multiplying the vector ⎡ ⎤ 1 − 2 2 0 1 ⎣ 2 ⎦ . In this 0 2 − 1 corresponding to each vertex by the matrix B = 1+ 2 2 0 0 1+ ⎡ ⎤ 06 08 0 problem,  = 12 , and so the matrix used is B = ⎣ 08 −06 0 ⎦. The composition of the 0 0 1

rotation followed by the reflection is therefore represented by the matrix ⎡ ⎤ 0990 0141 0 BA ≈ ⎣ 0141 −0990 0 ⎦  0 0 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 2545 2 3394 6 6930 Then, BA ⎣ 4 ⎦ ≈ ⎣ −3676 ⎦, BA ⎣ 10 ⎦ ≈ ⎣ −9615 ⎦ , BA ⎣ 7 ⎦ ≈ ⎣ −6080 ⎦, 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 8 9332 10 10464 BA ⎣ 10 ⎦ ≈ ⎣ −8767 ⎦ and BA ⎣ 4 ⎦ ≈ ⎣ −2545 ⎦. Therefore, since each of the product 1 1 1 1

vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number we get the following vertices for the moved graphic: (3 −4), (3 −10), (7 −6), (9 −9) and (10 −3).

(c) When using homogeneous coordinates, a scaling about the origin with scale factors  in the direction and  in the -direction is computed by multiplying the vector corresponding to each ⎡ ⎤ ⎤ ⎡ 3 0 0  0 0 ⎢ ⎥ vertex by A = ⎣ 0  0 ⎦ . In this problem,  = 3 and  = 12 , and so A = ⎣ 0 12 0 ⎦. Also, 0 0 1 0 0 1

a reflection through a line of slope  through the origin is computed by multiplying the vector ⎡ ⎤ 1 − 2 2 0 1 ⎣ 2 ⎦. In this 0 2 − 1 corresponding to each vertex by the matrix B = 1+ 2 0 0 1 + 2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

268

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

⎡

⎤ −06 08 0 problem,  = 2, and so B = ⎣ 08 06 0 ⎦. The composition of the scaling followed by the 0 0 1 reflection is therefore represented by the matrix ⎡ ⎤ −18 04 0 BA = ⎣ 24 03 0 ⎦  0 0 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 −2 2 04 6 −8 Then, BA ⎣ 4 ⎦ = ⎣ 6 ⎦, BA ⎣ 10 ⎦ = ⎣ 78 ⎦ , BA ⎣ 7 ⎦ = ⎣ 165 ⎦, 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 8 −104 10 −164 BA ⎣ 10 ⎦ = ⎣ 222 ⎦ and BA ⎣ 4 ⎦ = ⎣ 252 ⎦. Therefore, since each of the product 1 1 1 1

vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number we get the following vertices for the moved graphic: (−2 6), (0 8), (−8 17), (−10 22) and (−16 25).

(4) In each part we use the following list of vertices for the graphic in Figure 8.27(a) in Section 8.7 of the textbook: (5 4), (5 9), (8 6), (8 9), (8 11) and (11 4). In homogeneous coordinates, we express these points as the vectors [5 4 1], [5 9 1], [8 6 1], [8 9 1], [8 11 1] and [11 4 1]. (a) Since the given rotation is not about the origin, we use the Similarity Method. First, we must consider the translation along the vector [−8 −9] that moves the center of the rotation, (8 9), to the ⎡ ⎤ 1 0 −8 origin. The matrix for this translation in homogeneous coordinates is T1 = ⎣ 0 1 −9 ⎦. Next, 0 0 1

we need to compute the matrix for a rotation of 120◦ about the origin. When using homogeneous coordinates, a rotation about the origin through an angle  is computed by multiplying the vector ⎡ ⎤ cos  − sin  0 cos  0 ⎦ . In this problem,  = 120◦ , corresponding to each vertex by the matrix A = ⎣ sin  0 0 1 √ ⎡ 1 ⎤ ⎡ ⎤ − 2 − 23 0 −05 −0866 0 √ ⎢ 3 ⎥ 0 ⎦. Finally, we need and so the matrix used is A = ⎣ 2 − 12 0 ⎦ ≈ ⎣ 0866 −05 0 0 1 0 0 1 to translate along the vector [8 9] to bring the origin back to the point (8 9). The matrix for this ⎡ ⎤ 1 0 8 translation is T2 = ⎣ 0 1 9 ⎦ . Hence, the rotation of 120◦ about (8 9) is represented by the 0 0 1

matrix

⎡

⎤ −0500 −0866 19794 B = T2 AT1 ≈ ⎣ 0866 −0500 6572 ⎦  0 0 1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

269

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

⎡

⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 5 13830 5 9500 8 10598 Then, B ⎣ 4 ⎦ ≈ ⎣ 8902 ⎦ , B ⎣ 9 ⎦ ≈ ⎣ 6402 ⎦, B ⎣ 6 ⎦ ≈ ⎣ 10500 ⎦, 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 8 8 8 6268 11 10830 B ⎣ 9 ⎦ = ⎣ 9 ⎦ , B ⎣ 11 ⎦ ≈ ⎣ 8000 ⎦ and B ⎣ 4 ⎦ ≈ ⎣ 14098 ⎦. Therefore, since each 1 1 1 1 1 1 of the product vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding these results to the nearest whole number we get the following vertices for the moved graphic: (14 9), (10 6), (11 11), (8 9), (6 8) and (11 14). (c) Since the given scaling is not about the origin, we use the Similarity Method. First, we must consider the translation along the vector [−8 −4] that moves the center of the scaling, (8 4), to ⎡ ⎤ 1 0 −8 the origin. The matrix for this translation in homogeneous coordinates is T1 = ⎣ 0 1 −4 ⎦.

0 0 1

Next we need to find the matrix for a scaling about the origin with factors 2 in the -direction and 13 in the -direction. When using homogeneous coordinates, a scaling about the origin with scale factors  in the -direction and  in the -direction is computed by multiplying the vector ⎡ ⎤  0 0 corresponding to each vertex by the matrix A = ⎣ 0  0 ⎦. In this problem,  = 2 and  = 13 , 0 0 1 ⎤ ⎡ 2 0 0 ⎥ ⎢ and so the matrix is A = ⎣ 0 13 0 ⎦. Finally, we need to translate along the vector [8 4] to 0 0 1 ⎡ ⎤ 1 0 8 bring the origin back to the point (8 4). The matrix for this translation is T2 = ⎣ 0 1 4 ⎦.

0 0 1 Hence, the scaling with factor 2 in the -direction and 13 in the -direction about the point (8 4) is represented by the matrix ⎤ ⎡ 2 0 −8 ⎢ 8 ⎥ B = T2 AT1 = ⎣ 0 13 3 ⎦ 0

0

1

⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 2 ⎤ ⎡ ⎤ ⎡ 8 ⎤ ⎡ ⎤ ⎡ 8 ⎤ 5 2 5 8 8 ⎢ ⎥ ⎥ 14 ⎥ ⎣ 6 ⎦=⎢ ⎣ 9 ⎦=⎢ Then, B ⎣ 4 ⎦ = ⎣ 4 ⎦, B ⎣ 9 ⎦ = ⎣ 17 , B , B ⎦ ⎣ ⎦ ⎣ 17 3 3 3 ⎦, 1 1 1 1 1 1 1 1 ⎡

⎡ ⎤ ⎡ ⎤ ⎤ ⎡ 8 ⎤ 11 14 8 ⎥ ⎢ ⎣ 4 ⎦ = ⎣ 4 ⎦. Therefore, since each of the product vectors has B ⎣ 11 ⎦ = ⎣ 19 3 ⎦ and B 1 1 1 1 ⎡

a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

270

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

nearest whole number we get the following vertices for the moved graphic: (2 4), (2 6), (8 5), (8 6), (8 6) and (14 4).

(5) (b) We use the following list of vertices for the graphic in Figure 8.27(b) in Section 8.7 of the textbook: (7 3), (7 6), (10 8), (13 6) and (13 3). In homogeneous coordinates, we express these points as the vectors [7 3 1], [7 6 1], [10 8 1], [13 6 1] and [13 3 1].

Since the given reflection is not about a line through the origin, we use the Similarity Method.

First, we consider the translation along the vector [0 10] that moves the -intercept of the line of reflection, (0 −10), to the origin. The matrix for this translation in homogeneous coordinates ⎡ ⎤ 1 0 0 is T1 = ⎣ 0 1 10 ⎦. Next, we find the matrix for the reflection about the line  = 4 0 0 1

(since we have shifted the -intercept to the origin). A reflection through a line of slope  through the origin is computed by multiplying the vector corresponding to each vertex by A = ⎡ ⎤ ⎡ ⎤ 1 − 2 −15 8 0 2 0 1 ⎣ 2 ⎦. In this problem,  = 4, and so A = 1 ⎣ 8 15 0 ⎦.

0 2 − 1 1+2 17 2 0 0 17 0 0 1+

Finally, we need to translate along the vector [0 −10] to bring the origin back to the point ⎡ ⎤ 1 0 0 (0 −10). The matrix for this translation is T2 = ⎣ 0 1 −10 ⎦. Hence, the reflection about 0 0 1 the line  = 4 − 10 is represented by the matrix ⎡ ⎤ −15 8 80 1 ⎣ 8 15 −20 ⎦  B = T2 AT1 = 17 0 0 17 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 7 −0059 7 1353 10 −0353 Then, B ⎣ 3 ⎦ ≈ ⎣ 4765 ⎦, B ⎣ 6 ⎦ ≈ ⎣ 7412 ⎦, B ⎣ 8 ⎦ ≈ ⎣ 10588 ⎦ , 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 13 −3941 13 −5353 B ⎣ 6 ⎦ ≈ ⎣ 10235 ⎦ and B ⎣ 3 ⎦ ≈ ⎣ 7588 ⎦.

1 1 1 1

Therefore, since each of the product vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number we get the following vertices for the moved graphic: (0 5), (1 7), (0 11), (−4 10) and (−5 8).

(6) In each part we use the following list of vertices for the graphic in Figure 8.27(c) in Section 8.7 of the textbook: (5 3), (6 6), (8 9), (9 4), (9 7) and (12 8). In homogeneous coordinates, we express these points as the vectors [5 3 1], [6 6 1], [8 9 1], [9 4 1], [9 7 1] and [12 8 1]. (a) We begin with the rotation. Since the given rotation is not about the origin, we use the Similarity Method. First, we consider the translation along the vector [−12 −8] that moves the center of the rotation, (12 8), to the origin. The matrix for this translation in homogeneous coordinates is ⎡ ⎤ 1 0 −12 T1 = ⎣ 0 1 −8 ⎦. Next, we need to compute the matrix for a rotation of 60◦ about the origin.

0 0 1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

271

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

When using homogeneous coordinates, a rotation about the origin through an angle  is computed ⎡ ⎤ cos  − sin  0 cos  0 ⎦ . by multiplying the vector corresponding to each vertex by the matrix A = ⎣ sin  0 0 1 √ ⎤ ⎡ 1 ⎡ ⎤ − 23 0 2 05 −0866 0 √ ⎥ ⎢ 3 1 ⎣ 0866 05 0 ⎦. To finish In this problem,  = 60◦ , and so A = ⎢ 0 ⎥ ⎦≈ ⎣ 2 2 0 0 1 0 0 1 the rotation, we need to translate along the vector [12 8] to bring the origin back to the point ⎡ ⎤ 1 0 12 (12 8). The matrix for this is T2 = ⎣ 0 1 8 ⎦. Hence, the rotation of 60◦ about (12 8) is 0 0 1 represented by B = T2 AT1 .

Next, we compute the matrix for the reflection about  = 12  + 6. Since the given reflection is not about a line through the origin, we use the Similarity Method. First, we must consider the translation along the vector [0 −6] that moves the -intercept of the line of reflection, (0 6),⎤to ⎡ 1 0 0 the origin. The matrix for this translation in homogeneous coordinates is T3 = ⎣ 0 1 −6 ⎦.

0 0 1

Next, we need to find the matrix for the reflection about the line  = 12  (since we have shifted the -intercept to the origin). A reflection through a line of slope  through the origin is computed⎤ ⎡ 1 − 2 2 0 1 ⎣ 2 ⎦.

0 2 − 1 by multiplying the vector corresponding to each vertex by C = 1+ 2 2 0 0 1 +  ⎡ ⎤ 06 08 0 In this problem,  = 12 , and so C = ⎣ 08 −06 0 ⎦. Finally, we need to translate along 0 0 1 the vector [0 6] to bring the origin back to the point (0 6). The matrix for this translation is ⎡ ⎤ 1 0 0 T4 = ⎣ 0 1 6 ⎦. Hence, the reflection about the line  = 12  + 6 is represented by the matrix 0 0 1

D = T4 CT3 .

Finally, the composition of the rotation and the reflection is represented by the matrix ⎡ ⎤ 0993 −0120 −2157 DB = T4 CT3 T2 AT1 ≈ ⎣ −0120 −0993 23778 ⎦  0 0 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 5 2448 6 3082 8 4709 Then, DB ⎣ 3 ⎦ ≈ ⎣ 20201 ⎦, DB ⎣ 6 ⎦ ≈ ⎣ 17103 ⎦ , DB ⎣ 9 ⎦ ≈ ⎣ 13886 ⎦, 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 9 6300 9 5941 12 88 DB ⎣ 4 ⎦ ≈ ⎣ 18730 ⎦, DB ⎣ 7 ⎦ ≈ ⎣ 15752 ⎦ and DB ⎣ 8 ⎦ = ⎣ 144 ⎦. Therefore, 1 1 1 1 1 1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

272

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

since each of the product vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number we get the following vertices for the moved graphic: (2 20), (3 17), (5 14), (6 19), (6 16) and (9 14). The new graphic is illustrated in Figure 21.

Figure 21 (c) We begin with the scaling. Since the given scaling is not about the origin, we use the Similarity Method. First, we must consider the translation along the vector [−9 −4] that moves the center of the scaling, (9 4), to the origin. The matrix for this translation in homogeneous coordinates is ⎡ ⎤ 1 0 −9 T1 = ⎣ 0 1 −4 ⎦. Next we need to find the matrix for a scaling about the origin with factors 13 0 0 1 in the -direction and 2 in the -direction. When using homogeneous coordinates, a scaling about the origin with scale factors  in the -direction and  in the -direction is computed by multiplying the ⎡ ⎤  0 0 vector corresponding to each vertex by A = ⎣ 0  0 ⎦. In this problem,  = 13 and  = 2, and so 0 0 1 ⎡ 1 ⎤ 0 0 3 A = ⎣ 0 2 0 ⎦. Finally, we need to translate along the vector [9 4] to bring the origin back to 0 0 1 ⎡ ⎤ 1 0 9 the point (9 4). The matrix for this is T2 = ⎣ 0 1 4 ⎦. Hence, the scaling with factor 13 in the 0 0 1 -direction and 2 in the -direction about the point (9 4) is represented by the matrix B = T2 AT1 .

Next, we compute the matrix for the rotation. To do this, we first must consider the translation

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

273

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

along the vector [−2 −9] that moves the center of the rotation, (2 9), to the origin. The matrix ⎡ ⎤ 1 0 −2 for this translation in homogeneous coordinates is T3 = ⎣ 0 1 −9 ⎦. Next, we need to find the 0 0 1

matrix for a rotation of 150◦ about the origin. When using homogeneous coordinates, a rotation about the origin through an angle  is computed by multiplying the vector corresponding to each ⎡ ⎤ cos  − sin  0 cos  0 ⎦ . In this problem,  = 150◦ , and so the matrix is vertex by the matrix C = ⎣ sin  0 0 1 ⎤ ⎡ √3 ⎡ ⎤ − 2 − 12 0 −0866 −05 0 √ ⎥ ⎢ 1 ⎣ 05 −0866 0 ⎦. To finish the rotation, we need to translate C = ⎢ − 23 0 ⎥ ⎦ ≈ ⎣ 2 0 0 1 0 0 1

along the vector [2 9] to bring the origin back to the point (2 9). The matrix for this translation is ⎡ ⎤ 1 0 2 T4 = ⎣ 0 1 9 ⎦. Hence, the rotation of 150◦ about (2 9) is represented by the matrix D = T4 CT3 .

0 0 1

Finally, the composition of the rotation and the reflection is represented by the matrix ⎡ ⎤ −0289 −1 5036 DB = T4 CT3 T2 AT1 ≈ ⎣ 0167 −1732 22258 ⎦  0 0 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 5 0593 6 −2696 8 −6274 Then, DB ⎣ 3 ⎦ ≈ ⎣ 17896 ⎦, DB ⎣ 6 ⎦ ≈ ⎣ 12866 ⎦ , DB ⎣ 9 ⎦ ≈ ⎣ 8003 ⎦, 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 9 −1562 9 −4562 12 −6428 DB ⎣ 4 ⎦ ≈ ⎣ 16830 ⎦, DB ⎣ 7 ⎦ ≈ ⎣ 11634 ⎦ and DB ⎣ 8 ⎦ ≈ ⎣ 10402 ⎦. Therefore, 1 1 1 1 1 1

since each of the product vectors obtained has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by dropping that 1. Doing this and rounding each of these results to the nearest whole number yields the following vertices for the moved graphic: (1 18), (−3 13), (−6 8), (−2 17), (−5 12) and (−6 10). The new graphic is illustrated in Figure 22.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

274

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.7

Figure 22 (9) (b) We use the following list of vertices for the graphic in Figure 8.27(b) in Section 8.7 of the textbook: (7 3), (7 6), (10 8), (13 6) and (13 3). In homogeneous coordinates, we express these points as the vectors [7 3 1], [7 6 1], [10 8 1], [13 6 1] and [13 3 1].

Now according to part (b) of Exercise 7 in Section 8.7, a reflection about the line  =  +  ⎡ ⎤ 1 − 2 2 −2 1 ⎣ 2 ⎦ . In this problem, the line 2 2 − 1 is represented by the matrix B = 1+ 2 2 0 0 1+ is  = 4 − 10, and so  = 4 and  = −10. Hence, ⎡ ⎤ −15 8 80 1 ⎣ 8 15 −20 ⎦  B= 17 0 0 17 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 7 −0059 7 1353 10 −0353 Then, B ⎣ 3 ⎦ ≈ ⎣ 4765 ⎦, B ⎣ 6 ⎦ ≈ ⎣ 7412 ⎦, B ⎣ 8 ⎦ ≈ ⎣ 10588 ⎦, 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 13 −3941 13 −5353 B ⎣ 6 ⎦ ≈ ⎣ 10235 ⎦ and B ⎣ 3 ⎦ ≈ ⎣ 7588 ⎦. Therefore, since each of the product 1 1 1 1

vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number we get the following vertices for the moved graphic: (0 5), (1 7), (0 11), (−4 10) and (−5 8).

(11) (b) Consider the reflection about the -axis, whose matrix is given in Section 8.7 of the textbook as

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

275

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

−1 0 A=⎣ 0 1 0 0 ⎡ 0 −1 0 B=⎣ 1 0 0

Section 8.8

⎤ 0 0 ⎦ , and a counterclockwise rotation of 90◦ about the origin, whose matrix is 1 ⎤ ⎡ ⎤ ⎡ ⎤ 0 0 1 0 0 −1 0 0 ⎦. Then AB = ⎣ 1 0 0 ⎦, but BA = ⎣ −1 0 0 ⎦, and so A and B 1 0 0 1 0 0 1

do not commute. In particular, starting from the point (1 0), performing the rotation and then the reflection yields (0 1). However, performing the reflection followed by the rotation produces (0 −1).

(13) (a) False. As explained in Section 8.7 of the textbook, the third coordinate of a vector in homogeneous coordinates must be nonzero. (b) False. Vectors in homogeneous coordinates that are nonzero scalar multiples of each other represent the same pixel. For example, [1 1 1] and [2 2 2], as homogeneous coordinates, both represent the pixel at the point (1 1). (c) False. If A is a matrix for a rotation in homogeneous coordinates, then any nonzero scalar multiple of A also represents this same rotation. (d) True. As explained in Section 8.7 of the textbook, isometries are the composition of translations, rotations and reflections only. However, any similarity can be expressed as a composition of translations, rotations, reflections, and scalings. But, scalings that are not the identity or a reflection are not isometries. (e) True. This is because they do not send the origin to itself, violating part (1) of Theorem 5.1. (f) False. Rotations about the origin and reflections about lines through the origin are linear transformations. However, no other rotations or reflections are linear transformations since they do not send the origin to itself, and therefore violate part (1) of Theorem 5.1.

Section 8.8 (1) (a) Now A =

∙

13 −28 6 −13

¯ ¯ ¸ ¯  − 13 28 ¯¯ , and so A () = ¯¯ = 2 − 1 = ( − 1)( + 1). Hence, −6  + 13 ¯

A has 2 eigenvalues, 1 = 1 and 2 = −1. Next we solve for the corresponding fundamental eigenvectors.

For 1 = 1, we row reduce " # ¯ ¯ ∙ ¸ 1 − 73 ¯ 0 −12 28 ¯¯ 0 ¯ [(1I2 − A) | 0 ] = to obtain  −6 14 ¯ 0 0 0 ¯ 0 producing the fundamental eigenvector [7 3].

For 2 = −1, we row reduce ¯ ¯ ∙ ¸ ∙ ¸ −14 28 ¯¯ 0 1 −2 ¯¯ 0 [(−1I2 − A) | 0 ] = to obtain  −6 12 ¯ 0 0 0 ¯ 0

producing the fundamental eigenvector [2 1].

Therefore, by Theorem 8.9, the general form for the solution to F0 () = AF() is ∙ ¸ ∙ ¸ 7 2  − + 2   F() = 1  3 1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

276

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.8

¯ ¯ ⎤ ¯ −1 1 4 4 −4 −4 ¯¯ ¯  − 2 −2 ¯¯ = 3 − 42 + 3 (c) Now A = ⎣ −1 2 2 ⎦, and so A () = ¯¯ 1 ¯ 1 1 1 −1 −1  − 1 ¯ ⎡

= ( − 1)( − 3). Hence, A has 3 eigenvalues, 1 = 0, 2 = 1 and 3 = 3. Next we solve for the corresponding fundamental eigenvectors.

For 1 = 0, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ −1 −4 −4 ¯¯ 0 1 0 0 ¯¯ 0 [(0I3 − A) | 0 ] = [−A | 0 ] = ⎣ 1 −2 −2 ¯¯ 0 ⎦ to obtain ⎣ 0 1 1 ¯¯ 0 ⎦  −1 −1 −1 ¯ 0 0 0 0 ¯ 0

producing the fundamental eigenvector [0 −1 1].

For 2 = 1, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 0 −4 −4 ¯¯ 0 1 0 −1 ¯¯ 0 1 ¯¯ 0 ⎦  [(1I3 − A) | 0 ] = ⎣ 1 −1 −2 ¯¯ 0 ⎦ to obtain ⎣ 0 1 −1 −1 0 ¯ 0 0 0 0 ¯ 0

producing the fundamental eigenvector [1 −1 1].

For 3 = 3, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 2 −4 −4 ¯¯ 0 1 0 −2 ¯¯ 0 1 −2 ¯¯ 0 ⎦ to obtain ⎣ 0 1 0 ¯¯ 0 ⎦  [(3I3 − A) | 0 ] = ⎣ 1 ¯ −1 −1 2 0 0 0 0 ¯ 0 producing the fundamental eigenvector [2 0 1].

Therefore, by Theorem 8.9, the general form for the solution to F0 () = AF() is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 1 2 F() = 1 ⎣ −1 ⎦ + 2  ⎣ −1 ⎦ + 3 3 ⎣ 0 ⎦  1 1 1

¯ ¯ ⎤ ¯ +5 −5 −6 15 6 −15 ¯¯ ¯  + 5 −15 ¯¯ = 3 − 62 + 9 − 4 = (d) Now A = ⎣ −6 −5 15 ⎦, and so A () = ¯¯ 6 ¯ −6 −6 16 6 6  − 16 ¯ ⎡

(−1)2 (−4). Hence, A has 2 eigenvalues, 1 = 1 and 2 = 4. Next we solve for the corresponding fundamental eigenvectors.

For 1 = 1, we row reduce ¯ ⎡ ⎤ ¯ ⎡ ⎤ 1 1 − 52 ¯¯ 0 6 6 −15 ¯¯ 0 ⎢ ¯ ⎥ 0 ¯ 0 ⎦ [(1I3 − A) | 0 ] = ⎣ 6 6 −15 ¯¯ 0 ⎦ to obtain ⎣ 0 0 ¯ 6 6 −15 ¯ 0 0 0 0 ¯ 0

producing the fundamental eigenvectors [−1 1 0] and [5 0 2].

For 2 = 4, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 9 6 −15 ¯¯ 0 1 0 −1 ¯¯ 0 [(4I3 − A) | 0 ] = ⎣ 6 9 −15 ¯¯ 0 ⎦ to obtain ⎣ 0 1 −1 ¯¯ 0 ⎦  6 6 −12 ¯ 0 0 0 0 ¯ 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

277

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.8

producing the fundamental eigenvector [1 1 1].

Therefore, by Theorem 8.9, the general form for the solution to F0 () = AF() is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −1 5 1 F() = 1  ⎣ 1 ⎦ + 2  ⎣ 0 ⎦ + 3 4 ⎣ 1 ⎦  0 2 1

(There are other possible answers. For example, the first two vectors in the sum could be any basis for the two-dimensional eigenspace corresponding to the eigenvalue 1.)

(2) In each part, we use the technique described in Section 8.8 of the textbook in Theorem 8.12. (a) The characteristic equation is 2 +  − 6 = 0, found by using the coefficients of  00 ,  0 , and , in the given differential equation. Since 2 +  − 6 = ( − 2)( + 3), the characteristic values (the roots of the characteristic equation) are 1 = 2 and 2 = −3. Hence, the general solution for the differential equation is  = 1 2 + 2 −3 . (c) The characteristic equation is 4 − 62 + 8 = 0, found by using the coefficients of  0000 ,  000  00 ,  0 4 2 2 2 and , in the¡given√differential √ ¢equation. Since  − 6 + 8 = ( − 4)( − 2) = ¢¡ (−2)(+2)  − 2  + 2 , the characteristic values (the roots of the characteristic equation) √ √ are 1 = 2, 2 = −2, 3 = 2 and√4 = − √ 2. Hence, the general solution for the differential equation is  = 1 2 + 2 −2 + 3  2 + 4 − 2 . ¯ ¯ ¯  + 11 6 −16 ¯¯ ¯ 4 +1 −4 ¯¯ = (4) (b) First we find the general solution to F0 () = AF(). Now, A () = ¯¯ ¯ 12 6  − 17 ¯ 3 − 52 −  + 5 = ( − 5)(2 − 1) = ( − 5)( − 1)( + 1). Hence, A has 3 eigenvalues, 1 = 5, 2 = 1 and 3 = −1. Next we solve for the corresponding fundamental eigenvectors.

For 1 = 5, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 16 6 −16 ¯¯ 0 1 0 −1 ¯¯ 0 0 ¯¯ 0 ⎦  [(5I3 − A) | 0 ] = ⎣ 4 6 −4 ¯¯ 0 ⎦ to obtain ⎣ 0 1 12 6 −12 ¯ 0 0 0 0 ¯ 0 producing the fundamental eigenvector [1 0 1].

For 2 = 1, we row reduce ¯ ⎡ ⎤ ⎡ 1 12 6 −16 ¯¯ 0 ¯ ⎣ ⎦ ⎣ 4 2 −4 ¯ 0 [(1I3 − A) | 0 ] = to obtain 0 12 6 −16 ¯ 0 0

¯ ⎤ 0 ¯ 0 ¯ 0 1 ¯¯ 0 ⎦  0 0 ¯ 0

1 2

producing the fundamental eigenvector [−1 2 0].

For 3 = −1, we row reduce ¯ ¯ ⎡ ⎤ ⎡ ⎤ 10 6 −16 ¯¯ 0 1 0 −1 ¯¯ 0 [((−1)I3 − A) | 0 ] = ⎣ 4 0 −4 ¯¯ 0 ⎦ to obtain ⎣ 0 1 −1 ¯¯ 0 ⎦  12 6 −18 ¯ 0 0 0 0 ¯ 0 producing the fundamental eigenvector [1 1 1].

Therefore, by Theorem 8.9, the general form for the solution to F0 () = AF() is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 1 F() = 1 5 ⎣ 0 ⎦ + 2  ⎣ 2 ⎦ + 3 − ⎣ 1 ⎦  1 0 1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

278

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.8

Now we want to find the particular solution such that F(0) = [1 −4 0]. Plugging 0 into the general solution yields F(0) = [1  0 1 ] + [−2  22  0] + [3  3  3 ] = [1 − 2 + 3  22 + 3  1 + 3 ].

Setting this equal to [1 −4 0] produces the system ⎧ 1 ⎨ 1 − 2 + 3 = 22 + 3 = −4 ⎩ + 3 = 0 1 ¯ ¯ ⎡ ⎤ ⎡ ⎤ 1 −1 1 ¯¯ 1 1 0 0 ¯¯ 2 2 1 ¯¯ −4 ⎦ to obtain ⎣ 0 1 0 ¯¯ −1 ⎦ shows that the unique solution Row reducing ⎣ 0 1 0 1 ¯ 0 0 0 1 ¯ −2 to the differential equation with the given initial condition is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 1 F() = 25 ⎣ 0 ⎦ −  ⎣ 2 ⎦ − 2− ⎣ 1 ⎦  1 0 1

(5) (b) Base Step ( = 1): A = [−0 ], I1 − A = [ + 0 ], and so A () =  + 0 .

Inductive Step: Assume true for . Prove true for  + 1. For the case  + 1, ⎤ ⎡ 0 1 0 0 ··· 0 ⎢ 0 0 1 0 ··· 0 ⎥ ⎥ ⎢ ⎢ .. .. ⎥  . . . . .. .. .. ..

A=⎢ . . ⎥ ⎥ ⎢ ⎣ 0 0 0 0 ··· 1 ⎦ −0 −1 −2 −3 · · · − Then

⎡

 ⎢ 0 ⎢ ⎢ (I+1 − A) = ⎢ ... ⎢ ⎣ 0 0

−1 0 0  −1 0 .. .. .. . . .

0 0 0 1 2 3

Using a cofactor expansion on the first column yields ¯ ¯  −1 0 · · · ¯ ¯ .. .. .. . . ¯ . . . |I+1 − A| =  ¯ . ¯ 0 0 0 ··· ¯ ¯ 1 2 3 · · ·

0 0 .. .

0 0 .. .

··· ···

 −1

−1  + 

0 .. .  −1

¯ ¯ ¯ ¯ ¯ ¯ −1 ¯¯  +  ¯

279

⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0 .. .

¯ ¯ −1 0 · · · ¯ ¯  −1 · · · ¯ (+1)+1 0 ¯ . + (−1) .. .. ¯ .. . . ¯ ¯ 0 0 ···

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

⎤

··· ··· .. .

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯  −1 ¯ 0 0 .. .

0 0 .. .

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

The first determinant equals |I − B|, where ⎡ 0 1 0 ⎢ 0 0 1 ⎢ ⎢ .. ..

B = ⎢ ... . . ⎢ ⎣ 0 0 0 −1 −2 −3

··· ··· .. .

0 0 .. .

0 0

··· ···

0 −−1

1 −

Section 8.9

⎤

⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Now, using the inductive hypothesis on the first determinant, and Theorems 3.2 and 3.10 on the second determinant (since its corresponding matrix is lower triangular) yields: A () = ( +  −1 + · · · + 2  + 1 ) + (−1)+2 0 (−1) = +1 +   + · · · + 1  + 0 

(7) (a) True. If F() = 0, then F0 () also equals 0. The equation 0 = A0 is clearly satisfied for every matrix A. (b) True. Example 6 in Section 4.1 of the textbook shows that the set of all real-valued functions on R forms a vector space. An analogous proof shows that the set V of all R -valued functions on R is also a vector space. It is easy to show that the subset of continuously differentiable functions that are solutions to F0 () = AF() is closed under addition and scalar multiplication, and so Theorem 4.2 shows that this set is a subspace of V, and is thus a vector space. ∙ ¸ 1 2 (c) True. Since A = is upper triangular, it is easy to see that the eigenvalues for A 0 3 are 1 = 1 and 2 = 3. It is also easy to verify (by direct multiplication) that [1 0] is an for 2 . Hence, by Theorem 8.9, the solution eigenvector for ½ ∙ ¸1 and that ∙ [1¸1]¯ is an eigenvector ¾ ¯ 1 1 ¯    ∈ R , which agrees with the given solution. set is 1  + 2 3 0 1 ¯ 1 2 (d) False. The fact that the given matrix is not diagonalizable only means that the methods of Section 8.8 can not be used to solve the problem. For this particular matrix, complex eigenvalues need to be used, and will produce the general solution F() = [ sin  +  cos  − sin  +  cos ], which you can easily verify satisfies F0 () = AF(). (Note for those who have some experience with complex functions: using the method of Section 8.8 and complex eigenvalues to solve this problem ∙ ¸ ∙ ¸ ∙ ¸ −   sin  +  cos  + 2 − , which equals , will produce the solution F() = 1  1 1 − sin  +  cos  where  = 1 + 2 and  = (−1 + 2 ).)

Section 8.9

¸ ∙ ¸ 21 9 26 and A b = . Thus, to solve 9 11 19 ¯ # " ¯ ∙ ¸ ¯ 23 ¯ 1 0 21 9 26 ¯ 30 ¯ A Av = A b, we row reduce . obtaining ¯ 9 11 ¯ 19 0 1 ¯ 11 10 ¤ £ 11 Therefore, the unique least-squares solution is v = 23 30  10 . Now, ||Av − b|| = ⎡ 29 ⎤ ⎡ ⎤° °⎡ ⎡ ⎤° ° ⎤ ° ° ° 2 ¸° √ ∙ 6 5 ° ° 5 ° 3 " 23 # ° ° ° ° 1 1 1 ° ⎥ ° ⎢ 30 1 °⎣ 1 −1 ⎦ ° −  −  ° = 6 ≈ 0408 ⎣ 0 ⎦° = °⎣ − ⎦ − ⎣ 0 ⎦° − = ° 3 ° ° ° 11 ° ° 6 3 6 ° 6 ° 4 10 25 4 ° 4 ° ° 1

(1) (a) Using the given matrix A, A A =

∙

6

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

280

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.9

Note that °⎡ ⎡ ⎤° °⎡ ⎤ ⎡ ⎤° ⎤ ° 2 ° 5 ° 3 ∙ ¸ 5 ° ° ° ° 5 ° ⎣ 1 −1 ⎦ 1 − ⎣ 0 ⎦° = °⎣ 0 ⎦ − ⎣ 0 ⎦° = k[0 0 1]k = 1 ||Az − b|| = ° ° ° ° ° 1 ° 4 4 ° ° 5 1 4 °

which is larger than ||Av − b||.

⎡

⎤ ⎡ ⎤ 14 8 6 18 7 ⎦. Thus, to solve (c) Using the given matrix A, A A = ⎣ 8 5 9 ⎦ and A b = ⎣ 6 9 75 −35 A Av = A b, we row reduce ¯ ⎤ ⎡ ¯ ⎡ ⎤ 1 0 −7 ¯¯ 17 3 14 8 6 ¯¯ 18 ⎥ ⎢ ¯ ⎣ 8 5 9 ¯ 7 ⎦ obtaining ⎣ 0 1 13 ¯ − 23 3 ⎦ ¯ ¯ 6 9 75 ¯ −35 0 0 0 ¯ 0

Therefore, there is an infinite number of least-squares solutions, all of the form ¤ £ 23 1 v = 7 + 17 3  −13 − 3   . Using  = 0 and  = 3 produces the two particular least-squares ¤ ¤ £ £ 23 1 solutions 17 3  − 3  0 and 8 −12 3 . With v as the first of these vectors, ||Av − b|| = °⎡ °⎡ ⎤ ⎡ ⎤° ⎤ ⎡ 17 ⎤ ⎡ ⎤° ° ° ° 11 ° ¸° √ ∙ 3 3 ° ° 3 ° ° 3 ° 2 1 −1 ° 2 1 1 ° 6 ⎥ ° ⎥ ° °⎣ ⎢ ° ⎢ 23 5 5 ⎦ ⎣ − 3 ⎦ − ⎣ 2 ⎦° = °⎣ 3 ⎦ − ⎣ 2 ⎦° = ° − − ° = ≈ 0816 ° 3 2 ° ° 3 3 3 ° ° ° ° 3 17 6 6 ° ° ° 1 0 −7 ° 0 3

Using the other particular solution we computed for v, or any of the infinite number of least √

squares solutions instead, produces the same value of 36 for ||Av − b||. Note that ||Az − b|| = °⎡ ⎤⎡ ⎤ ⎡ ⎤° °⎡ ⎤ ⎡ ⎤° ° 2 1 −1 ° 1 3 ° 3 ° ° ° ° 4 ° °⎣ 3 2 ⎦ ⎣ 1 ⎦ − ⎣ 2 ⎦° = °⎣ 0 ⎦ − ⎣ 2 ⎦° = k[1 −2 2]k = 3 5 ° ° ° ° ° 1 0 −7 −1 6 ° ° 8 6 ° which is larger than ||Av − b||.

⎡

⎤ ⎡ ⎤ 62 −4 40 29 5 −8 ⎦ and A b = ⎣ −3 ⎦. Thus, to solve (2) (a) Using the given matrix A, A A = ⎣ −4 40 −8 32 20 ¯ 19 ⎤ ⎡ ¯ ⎡ ⎤ 4 ¯ 1 0 7 ¯ 42 62 −4 40 ¯¯ 29 ¯ ⎢ 5 ⎥ 5 −8 ¯¯ −3 ⎦ obtaining ⎣ 0 1 − 87 ¯ − 21 A Av = A b, we row reduce ⎣ −4 ⎦. ¯ 40 −8 32 ¯ 20 ¯ 0 0 0 0

Therefore, there is an infinite number of least-squares solutions, all of the form ¤ £ 8 5 4 19 19 8 5 v = − 47  + 19 42  7  − 21   . Now, setting − 7  + 42 ≥ 0 yields  ≤ 24 . And, setting 7  − 21 ≥ 0

5 ≤ , which also implies  ≥ 0. Hence, the desired solution set is produces 24 ©£ 4 ª ¤ ¯ 5 8 5 19 ¯ − 7  + 19 42  7  − 21   24 ≤  ≤ 24

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

281

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

⎢ (3) (b) First, (0 I3 − C) = ⎣

− 32

3

−11

12

5 − 72

2

Section 8.9

⎤

⎥ −4 ⎦. We start with the homogeneous system 21 2

(0 I3 − C)x = 0 and add the equation 1 + 2 + 3 = 1. This produces the system Ax = b, with ⎤ ⎡ ⎡ ⎤ 3 2 − 32 0 ⎥ ⎢ ⎢ 0 ⎥ 5 − 72 −4 ⎥ ⎢ ⎥ A=⎢ ⎥ and b = ⎢ ⎣ 0 ⎦ ⎣ −11 12 21 ⎦ 2 1 1 1 1 We find a least-squares solution for this system. Now, ⎤ ⎡ 597 ⎡ ⎤ −153 − 275 4 2 1 ⎥ ⎢  665 ⎣ 1 ⎦ 147 and A b = A A = ⎣ −153 ⎦ 4 275 525 1 147 − 2 4

Thus, to solve A Av = A b, we row reduce ¯ ⎡ 597 ⎤ ¯ 1 −153 − 275 4 2 ¯ ⎢ ¯ ⎥ 665 147 ¯ 1 ⎦ obtaining ⎣ −153 4 ¯ 525 ¯ 147 − 275 1 2 4

¯ ⎤ 1 0 0 ¯¯ 046 ⎣ 0 1 0 ¯ −036 ⎦  ¯ 0 0 1 ¯ 090 ⎡

where we have rounded to two places after the decimal point. Hence, the approximate eigenvector 0 is v ≈ [046 −036 090]. Direct computation ° ( I3 − C)v ≈ [003 −004 007], which ° 0 produces ° ° is close to the zero vector. (Note that ( I3 − C)v ≈ 0086, which is of the same order of magnitude as the difference between 0 = 32 and the actual eigenvalue for C nearest to 0 , which √ is 2 ≈ 1414.)

(6) Let b = [1    ], let A be as in Theorem 8.4, and let W = {Ax | x ∈ R } Since projW b ∈ W exists, there must be some v ∈ R such that Av = projW b. Hence, v satisfies part (1) of Theorem 8.13, so (A A)v = A b by part (3) of Theorem 8.13. This shows that the system (A A)X = A b is consistent, which proves part (2) of Theorem 8.4.

Next, let  () = 0 + 1  + · · · +   and z = [0  1       ] A short computation shows that ||Az − b||2 =   the sum of the squares of the vertical distances illustrated in Figure 8.10, just before the definition of a least-squares polynomial in Section 8.3 of the textbook. Hence, minimizing ||Az−b|| over all possible ( + 1)-vectors z gives the coefficients of a degree  least-squares polynomial for the given points (1  1 )     (   ) However, parts (2) and (3) of Theorem 8.13 show that such a minimal solution is found by solving (A A)v = A b thus proving part (1) of Theorem 8.4.

Finally, when A A is row equivalent to I+1 , the uniqueness condition holds by Theorems 2.15 and 2.16, which proves part(3) of Theorem 8.4.

(7) (a) False. The least-squares solution of an inconsistent system Ax = b is, instead, any vector v such that the distance between Av and b is minimized. ¡ ¢ ¡ ¢ = (b) True. If A is an × matrix, then A A is ×, which is square. Also, A A = A A A A, and so A A is symmetric. (c) True. If W is the subspace {Ax | x ∈ R }, then projW b is a well-defined vector in W. Hence, by the definition of W, there must exist a vector v in R with Av = projW b. Theorem 8.13 asserts that such a vector v is a least-squares solution to Ax = b. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

282

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.10

(d) True. By part (1) of Theorem 8.13, both Av1 and Av2 must equal projW b, and so they must equal each other. (e) True. This is illustrated in Example 3 in Section 8.9 of the textbook.

Section 8.10 (1) In each part, we use the fact that the entries for the upper triangular matrix C are given, for  ≤ , by  = (coefficient of   ) in the formula for . Then, the entries of the symmetric matrix A are given, for   , by  =  = 12  , and, on the diagonal,  =  . ∙ ¸ ∙ ¸ 8 12 8 6 (a) Using the formulas given above produces C = and A = .

0 −9 6 −9 ⎤ ⎡ ⎡ ⎤ 5 2 − 32 5 4 −3 ⎢ 5 ⎥ 5 ⎦ and A = ⎣ 2 −2 (c) Using the formulas given above yields C = ⎣ 0 −2 2 ⎦.

5 3 0 0 0 − 0 2

2

(2) (a) The entries for the symmetric matrix A representing  are given, for  6= , by  = 12 (coefficient of   ) in the expression for , and, on the diagonal,  = (the coefficient of 2 ). Hence, ∙ ¸ 43 −24 A= . We need to orthogonally diagonalize A. −24 57

Now A () = ( − 43)( − 57) − 242 = 2 − 100 + 1875 = ( − 75)( − 25). Thus, A has eigenvalues 1 = 75 and 2 = 25. Using row reduction to find particular solutions to the systems [75I2 − A | 0] and [25I2 − A | 0] as in the Diagonalization Method of Section 3.4 produces the basis {[−3 4]} for 75 and {[4 3]} for 25 . Since 75 and 25 are 1-dimensional, the given bases are already orthogonal. To create orthonormal bases, we need to normalize the vectors, yielding { 15 [−3 4]} for 75 and { 15 [4 3]} for 25 . Thus, an orthonormal basis  for R2 is the set ¢ ¡  = 15 [−3 4] 15 [4 3] . Now P is the matrix whose columns are the vectors in , and D is the

diagonal matrix whose main diagonal entries are the eigenvalues 75 and 25. Hence, ∙ ¸ ∙ ¸ 1 −3 4 75 0 and D =  P= 4 3 0 25 5

Next, direct computation using the given formula for  yields ([1 −8]) = 4075. However, we can also calculate ([1 −8]) as [1 −8] D[1 −8] , where ∙ ¸∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ ∙ ¸ 1 −35 1 −3 4 1 −7 1 1 −1  = =  =P = [1 −8] = P 4 3 −8 −4 −8 −8 5 5 −20 Then, ([1 −8]) =

£

−7 −4

¤

∙

75 0 0 25

¸∙

−7 −4

¸

=

£

−7 −4

¤

∙

−525 −100

¸

= 4075

(c) The entries for the symmetric matrix A representing  are given, for  6= , by  = 12 (coefficient of   ) in the expression for , and, on the diagonal,  = (the coefficient of 2 ). Hence, ⎡ ⎤ 18 48 −30 18 ⎦  A = ⎣ 48 −68 −30 18 1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

283

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 8.10

We need to orthogonally diagonalize A. Now A () = 3 + 492 − 4802 = (2 + 49 − 4802). Thus, 1 = 0 is an eigenvalue for A. Using the quadratic formula on the second factor of A () produces the eigenvalues 2 = 49 and 3 = −98. Then, using row reduction to find particular solutions to the systems [0I3 − A | 0], [49I3 − A | 0] and [−98I3 − A | 0] as in the Diagonalization Method of Section 3.4 produces the basis {[2 3 6]} for 0 , {[−6 −2 3]} for 49 and {[3 −6 2]} for −98 . Since 0 , 49 and −98 are 1-dimensional, the given bases are already orthogonal. To create orthonormal bases, we need ª © ª © ª © to normalize the vectors, yielding 17 [2 3 6] for 0 , 17 [−6 −2 3] for 49 and 17 [3 −6 2] for ¡ ¢ −98 . Thus, the orthonormal basis  for R3 is the set  = 17 [2 3 6] 17 [−6 −2 3] 17 [3 −6 2] .

Now P is the matrix whose columns are the vectors in , and D is the diagonal matrix whose main diagonal entries are the eigenvalues 0, 49 and −98. Hence, ⎡ ⎤ ⎡ ⎤ 2 −6 3 0 0 0 1⎣ 3 −2 −6 ⎦  and D = ⎣ 0 49 0 ⎦ P= 7 6 3 2 0 0 −98

Next, direct computation using the given formula for  yields ([4 −3 6]) = −3528. However, we can also calculate ([4 −3 6]) as [4 −3 6] D[4 −3 6] , where ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 3 6 4 35 5 4 4 1 1 [4 −3 6] = P−1 ⎣ −3 ⎦ = P ⎣ −3 ⎦ = ⎣ −6 −2 3 ⎦ ⎣ −3 ⎦ = ⎣ 0 ⎦ = ⎣ 0 ⎦  7 7 42 3 −6 2 6 6 6 6 Then,

([4 −3 6]) =

£

5 0 6

¤

⎡

⎤⎡ ⎤ ⎡ ⎤ 0 0 0 5 0 £ ¤ ⎣ 0 49 0 ⎦⎣ 0 ⎦ = 5 0 6 ⎣ 0 ⎦ = −3528 0 0 −98 6 −588

(4) Yes. If (x) = Σ   , 1 ≤  ≤  ≤ , then x C1 x and C1 upper triangular imply that the ( ) entry for C1 is zero if    and  if  ≤ . A similar argument describes C2 . Thus, C1 = C2 .

(6) (a) True. This is discussed in the proof of Theorem 8.14. (b) False. For this quadratic form, the coefficients of 2 and  2 are zero, which are allowed in the definition of a quadratic form. (c) False. A quadratic form can be represented by many matrices. In particular, every quadratic form is represented by both an upper triangular matrix and a symmetric matrix (which are usually distinct). For example, the quadratic form ([ ]) = 2 can be expressed both as £

 

¤

∙

0 2 0 0

¸∙

 

¸

and as

£

 

¤

∙

0 1 1 0

¸∙

 

¸



(d) True. This is the conclusion of the Principal Axes Theorem (Theorem 8.16). (e) True. This can be inferred from the proof of the Principal Axes Theorem (Theorem 8.16) and the Orthogonal Diagonalization Method from Section 6.3 of the textbook. This fact is utilized in the Quadratic Form Method in Section 8.10 of the textbook.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

284

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.1

Chapter 9 Section 9.1 (1) (a) For the first system, row reduce ¯ ¯ ∙ ¸ ∙ ¸ ¯ 10 5 −2 1 0 ¯¯ 602 ¯ to obtain  5 −1995 ¯ 175 0 1 ¯ 1500

Hence, the solution to the first system is (602 1500). For the second system, row reduce ¯ ¯ ∙ ¸ ∙ ¸ ¯ 10 5 −2 1 0 ¯¯ 302 ¯ to obtain  5 −199 ¯ 175 0 1 ¯ 750

Hence, the solution to the second system is (302 750). These systems are ill-conditioned, because a very small change in the coefficient of  leads to a very large change in the solution.

(2) Answers to this problem may differ significantly from the following if you round at different places in the algorithm. We rounded the original numbers in the system, and directly after performing each row operation. (a) Without partial pivoting, we perform the following sequence of row operations: 1 h1i (new row h1i is [1 −5990 −1370]) h1i ← 000072

h2i ← −231 h1i + h2i (new row h2i is [0 3960 3030]) 1 h2i (new row h2i is [0 1 0765]) h2i ← 3960

h1i ← 5990 h2i + h1i (new row h1i is [1 0 3210])

giving the solution (3210 0765).

With partial pivoting, we perform the following sequence of row operations: h1i ↔ h2i

1 h1i (new row h1i is [1 4280 −567]) h1i ← 231

h2i ← −000072 h1i + h2i (new row h2i is [0 −123 −0944])

1 h2i (new row h2i is [0 1 0767]) h2i ← − 123

h1i ← 4280 h2i + h1i (new row h1i is [1 0 3230])

giving the solution (3230 0767). The actual solution is (3214 0765). (c) Without partial pivoting, we perform the following sequence of row operations: 1 h1i (new row h1i is [1 722 397 −108]) h1i ← 000032

h2i ← 241 h1i + h2i (new row h2i is [0 174000 95700 −26600])

h3i ← −49 h1i + h3i (new row h3i is [0 −35300 −19500 5580]) 1 h2i (new row h2i is [0 1 055 −0153]) h2i ← 174000

h1i ← −722 h2i + h1i (new row h1i is [1 0 −01 247])

h3i ← 35300 h2i + h3i (new row h3i is [0 0 −85 179])

1 h3i (new row h3i is [0 0 1 −211]) h3i ← − 85

h1i ← 01 h3i + h1i (new row h1i is [1 0 0 226])

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

285

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.1

h2i ← −055 h3i + h2i (new row h2i is [0 1 0 101])

giving the solution (226 101 −211).

With partial pivoting, we perform the following sequence of row operations: h1i ↔ h2i

1 h1i (new row h1i is [1 09 00332 239]) h1i ← − 241

h2i ← −000032 h1i + h2i (new row h2i is [0 0231 0127 −00354])

h3i ← −49 h1i + h3i (new row h3i is [0 09 0773 166]) h2i ↔ h3i

1 h2i (new row h2i is [0 1 0859 184]) h2i ← 09

h1i ← −09 h2i + h1i (new row h1i is [1 0 −0740 −163])

h3i ← −0231 h2i + h3i (new row h3i is [0 0 −00714 −425])

1 h3i (new row h3i is [0 0 1 595]) h3i ← − 00714

h1i ← 074 h3i + h1i (new row h1i is [1 0 0 277])

h2i ← −0859 h3i + h2i (new row h2i is [0 1 0 −327])

giving the solution (277 −327 595). The actual solution is (267 −315 573).

(3) Answers to this problem may differ significantly from the following if you round at different places in the algorithm. We rounded directly after performing each row operation. (a) Without partial pivoting, we perform the following sequence of row operations: 1 h1i (new row h1i is [1 −5989 −1368]) h1i ← 000072

h2i ← −231 h1i + h2i (new row h2i is [0 3959 3029]) 1 h2i (new row h2i is [0 1 07651]) h2i ← 3959

h1i ← 5989 h2i + h1i (new row h1i is [1 0 3214])

giving the solution (3214 07651).

With partial pivoting, we perform the following sequence of row operations: h1i ↔ h2i 1 h1i (new row h1i is [1 −4275 −5662]) h1i ← 231

h2i ← −000072 h1i + h2i (new row h2i is [0 −1234 −09438])

1 h2i (new row h2i is [0 1 07648]) h2i ← − 1234

h1i ← 4275 h2i + h1i (new row h1i is [1 0 3213])

giving the solution (3213 07648). The actual solution is (3214 0765). (c) Without partial pivoting, we perform the following sequence of row operations: 1 h1i (new row h1i is [1 7231 3969 −108]) h1i ← 000032

h2i ← 241 h1i + h2i (new row h2i is [0 174100 95640 −26600])

h3i ← −49 h1i + h3i (new row h3i is [0 −35390 −19450 5575]) 1 h2i (new row h2i is [0 1 05493 −01528]) h2i ← 174100

h1i ← −7231 h2i + h1i (new row h1i is [1 0 −02988 2490])

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

286

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.1

h3i ← 35390 h2i + h3i (new row h3i is [0 0 −1027 1674])

1 h3i (new row h3i is [0 0 1 −1630]) h3i ← − 1027

h1i ← 02988 h3i + h1i (new row h1i is [1 0 0 −2380])

h2i ← −05493 h3i + h2i (new row h2i is [0 1 0 8801])

giving the solution (−238 8801 −1630).

With partial pivoting, we perform the following sequence of row operations: h1i ↔ h2i 1 h1i (new row h1i is [1 09004 00332 2390]) h1i ← − 241

h2i ← −000032 h1i + h2i (new row h2i is [0 02311 01270 −003532])

h3i ← −49 h1i + h3i (new row h3i is [0 08804 07732 1661]) h2i ↔ h3i

1 h2i (new row h2i is [0 1 08782 1887]) h2i ← 08804

h1i ← −09004 h2i + h1i (new row h1i is [1 0 −07575 −1675])

h3i ← −02311 h2i + h3i (new row h3i is [0 0 −007595 −4364])

1 h3i (new row h3i is [0 0 1 5746]) h3i ← − 007595

h1i ← 07575 h3i + h1i (new row h1i is [1 0 0 2678])

h2i ← −08782 h3i + h2i (new row h2i is [0 1 0 −3159])

giving the solution (2678 −3159 5746). The actual solution is (267 −315 573).

(4) (a) First we express the given system as (

1 2

= − 15 2 = − 37 1

+

26 5

−

6



Starting with the initial values 1 = 0 and 2 = 0, we plug these in the right side of these equations to obtain the new values 1 = 26 5 and 2 = −6. These are the values listed in the table, below, in the row labeled “After 1 Step.” These new values are again put in the right side to solve for 1 and 2 . After rounding to 3 decimal places, this yields 1 = 6400 and 2 = −8229, the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps After 5 Steps After 6 Steps After 7 Steps After 8 Steps After 9 Steps

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

287

1 0000 5200 6400 6846 6949 6987 6996 6999 7000 7000

2 0000 −6000 −8229 −8743 −8934 −8978 −8994 −8998 −9000 −9000

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.1

After 9 steps we get the same result as we had after 8 steps, and so we are done. The solution obtained for the system is (7 −9).

(c) First we express the given system as ⎧ 1 ⎪ ⎨ 1 = − 7 2 1 2 = 6 1 ⎪ ⎩ 1 3 = 3 1

+ − −

2 7 3 1 6 3 1 6 2

62 7 9 2 13 3

− + −



Starting with the initial values 1 = 0, 2 = 0 and 3 = 0, we plug these in the right side of these 9 13 equations to obtain the new values 1 = − 62 7 , 2 = 2 and 3 = − 3 . These values, rounded to 3 places after the decimal point, are listed in the table, below, in the row labeled “After 1 Step.”

These new values are again put in the right side to solve for 1 , 2 and 3 . After rounding to 3 decimal places, this yields 1 = −10738, 2 = 3746 and 3 = −8036, the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table: 1 2 3 Initial Values 0000 0000 0000 After 1 Step −8857 4500 −4333 After 2 Steps −10738 3746 −8036 After 3 Steps −11688 4050 −8537 After 4 Steps −11875 3975 −8904 After 5 Steps −11969 4005 −8954 After 6 Steps −11988 3998 −8991 After 7 Steps −11997 4001 −8996 After 8 Steps −11999 4000 −8999 After 9 Steps −12000 4000 −9000 After 10 Steps −12000 4000 −9000

After 10 steps we get the same result as we had after 9 steps, and so we are done. The solution obtained for the system is (−12 4 −9).

(5) (a) First we express the given system as (

1 2

= − 15 2 = − 37 1

+

26 5

−

6



We start with the initial values 1 = 0 and 2 = 0. We plug 2 = 0 into the right side of the first 26 equation to obtain the new value 1 = 26 5 . Then, we substitute 1 = 5 into the right side of the 288 second equation yielding 2 = − 35 . After rounding to 3 places after the decimal point, we listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug 2 = −8229 into the right side of the first equation to solve for 1 , which, after rounding to 3 decimal places, yields 1 = 6846. Substituting 1 = 6846 into the right side of the second equation produces 2 = −8934. The two new results are the numbers in the row labeled “After 2 Steps.” We

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

288

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.1

continue in this manner, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps After 5 Steps After 6 Steps

1 0000 5200 6846 6987 6999 7000 7000

2 0000 −8229 −8934 −8994 −9000 −9000 −9000

After 6 steps we get the same result as we had after 5 steps, and so we are done. The solution obtained for the system is (7 −9).

(c) First we express the given system as ⎧ 1 ⎪ ⎨ 1 = − 7 2 1 2 = 6 1 ⎪ ⎩ 1 3 = 3 1

+ − −

2 7 3 1 6 3 1 6 2

− + −

62 7 9 2 13 3



We start with the initial values 1 = 0, 2 = 0 and 3 = 0. We plug 2 = 3 = 0 into the right side of the first equation to obtain the new value 1 = − 62 7 ≈ −8857. Then, we substitute 1 = −8857 and 3 = 0 into the right side of the second equation, yielding 2 ≈ 3024. Finally, we plug 1 = −8857 and 2 = 3024 into the right side of the third equation, yielding 3 ≈ −7790.

We have listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug 2 = 3024 and 3 = −7790 into the right side of the first equation to solve for 1 , which yields 1 ≈ −11515. Substituting 1 = 11515 and 3 = −7790 into the right side of the second equation produces 2 ≈ 3879. Finally, we plug 1 = −11515 and 2 = 3879 into the right side of the third equation, yielding 3 ≈ −8818. The three new results are the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table: 1 2 3 Initial Values 0000 0000 0000 After 1 Step −8857 3024 −7790 After 2 Steps −11515 3879 −8818 After 3 Steps −11931 3981 −8974 After 4 Steps −11990 3997 −8996 After 5 Steps −11998 4000 −8999 After 6 Steps −12000 4000 −9000 After 7 Steps −12000 4000 −9000 After 7 steps we get the same result as we had after 6 steps, and so we are done. The solution obtained for the system is (−12 4 −9).

(6) (a) A satisfies all of the criteria, so it is strictly diagonally dominant. (In row 1, | − 3|  |1|; that is, |11 |  |12 |. In row 2, |4|  | − 2|; that is, |22 |  |21 |.)

(b) A is not strictly diagonally dominant because |22 |  |21 |. (Note that |11 | = |12 |, which also prevents A from being strictly diagonally dominant.) (c) A satisfies all of the criteria, so it is strictly diagonally dominant. (In row 1, |11 |  |12 | + |13 |; in row 2, |22 |  |21 | + |23 |; in row 3, |33 |  |31 | + |32 |.)

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

289

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.1

(d) A is not strictly diagonally dominant because |22  |21 | + |23 |. (e) A is not strictly diagonally dominant because |22 | = |21 + |23 |.

(7) (a) Put the third equation first, and move the other two down. Then express the equations as follows: ⎧ 1 3 25 ⎪ 8 2 − 8 3 + 8 ⎨ 1 = 2 1 2 = − 13 1 − 13 3  ⎪ ⎩ 1 2 26 3 = − 15 1 + 15 2 + 15

We start with the initial values 1 = 0, 2 = 0 and 3 = 0. We plug 2 = 3 = 0 into the right side of the first equation to obtain the new value 1 = 25 8 = 3125. Then, we substitute 1 = 3125 and 3 = 0 into the right side of the second equation, yielding 2 ≈ −0481. Finally, we plug 1 = 3125 and 2 = −0481 into the right side of the third equation, yielding 3 ≈ 1461.

We have listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug 2 = −0481 and 3 = 1461 into the right side of the first equation to solve for 1 , which yields 1 ≈ 2517. Substituting 1 = 2517 and 3 = 1461 into the right side of the second equation produces 2 ≈ −0500. Finally, we plug 1 = 2517 and 2 = −0500 into the right side of the third equation yielding 3 ≈ 1499. The three new results are the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps

1 0000 3125 2517 2500 2500

2 0000 −0481 −0500 −0500 −0500

3 0000 1461 1499 1500 1500

After 4 steps we get the same result as we had after 3 steps, and so we are done. The solution obtained for the system is (25 −05 15).

(c) Put the second equation first, the fourth equation second, the first equation third, and the third equation fourth. Then express the equations as follows: ⎧ 1 1 49  = − 29 2 + ⎪ ⎪ 9 3 − 9 4 + 9 ⎪ 1 ⎪ 86 ⎨ 2 = − 1 1 − 3 3 + 2 4 − 17 17 17 17  1 1 2 120 ⎪ 3 = − 13  −  −  + 1 2 4 ⎪ 13 13 13 ⎪ ⎪ ⎩ 3 1 55 4 = − 17 1 + 14 2 − 14 3 − 7

We start with the initial values 1 = 0, 2 = 0, 3 = 0 and 4 = 0. We plug 2 = 3 = 4 = 0 into the right side of the first equation to obtain the new value 1 = 49 9 ≈ 5444. Then, we substitute 1 = 5444, 3 = 0, and 4 = 0 into the right side of the second equation, yielding 2 ≈ −5379. Next, we substitute 1 = 5444, 2 = −5379, and 4 = 0 into the right side of the third equation, yielding 3 ≈ 9226. Finally, we plug 1 = 5444, 2 = −5379, and 3 = 9226 into the right side of the fourth equation, yielding 4 ≈ −10447. We have listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug 2 = −5379, 3 = 9226, and 4 = −10447 into the right side of the first equation to solve for 1 , which yields 1 ≈ 8826.

Substituting 1 = 8826, 3 = 9226 and 4 = −10447 into the right side of the second equation produces 2 ≈ −8435. Next, substituting 1 = 8826, 2 = −8435 and 4 = −10447 into the right side of the third equation produces 3 ≈ 10808. Finally, we plug 1 = 8826, 2 = −8435,

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

290

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.1

and 3 = 10808 into the right side of the fourth equation, yielding 4 ≈ −11698. The four new results are the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps After 5 Steps After 6 Steps After 7 Steps After 8 Steps

1 0000 5444 8826 9820 9973 9995 9999 10000 10000

2 0000 −5379 −8435 −8920 −8986 −8998 −9000 −9000 −9000

3 0000 9226 10808 10961 10994 10999 11000 11000 11000

4 0000 −10447 −11698 −11954 −11993 −11999 −12000 −12000 −12000

After 8 steps we get the same result as we had after 7 steps, and so we are done. The solution obtained for the system is (10 −9 11 −12).

(8) First we express the given system as ⎧ ⎨ 1 2 ⎩ 3

= 52 = 61 = −71

+ 3 − 23 − 2

+ 16 − 13  + 12

Using the Jacobi Method, we start with the initial values 1 = 0, 2 = 0 and 3 = 0, and plug these in the right side of these equations to obtain the new values 1 = 16, 2 = −13 and 3 = 12. These values are listed in the table, below, in the row labeled “After 1 Step.” These new values are put in the right side to again solve for 1 , 2 and 3 yielding 1 = −37, 2 = 59 and 3 = −87, the numbers in the row labeled “After 2 Steps.” We continue in this manner for 6 steps, producing the results given in the following table: 1 2 3 Initial Values 00 00 00 After 1 Step 160 −130 120 After 2 Steps −370 590 −870 After 3 Steps 2240 −610 2120 After 4 Steps −770 9070 −14950 After 5 Steps 30560 25150 −3560 After 6 Steps 122350 190350 −238950 Note that the numbers are diverging rather than converging to a single solution.

Next, we use the Gauss-Seidel Method. We start with the initial values 1 = 0, 2 = 0 and 3 = 0.

We plug 2 = 3 = 0 into the right side of the first equation to obtain the new value 1 = 16. Then, we substitute 1 = 16 and 3 = 0 into the right side of the second equation, yielding 2 = 83. Finally, we plug 1 = 16 and 2 = 83 into the right side of the third equation, yielding 3 = −183. We have listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug 2 = 830 and 3 = −1830 into the right side of the first equation to solve for 1 , which yields 1 = 248. Substituting 1 = 248 and 3 = −183 into the right side of the second equation produces 2 = 1841. Finally, we plug 1 = 248 and 2 = 1841 into the right side of the third equation, yielding 3 = −3565. The three

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

291

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.2

new results are the numbers in the row labeled “After 2 Steps.” We continue in this manner for four steps, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps

1

2

3

00 160 2480 56560 1246480

00 830 18410 410530 9091410

00 −1830 −35650 −806330 −17816650

Again, the values are diverging rather than converging.

To find the actual solution, we use Gaussian Elimination. We reduce ¯ ⎤ ⎡ ¯ ⎡ ⎤ 1 −5 −1 ¯¯ 16 1 −5 −1 ¯¯ 16 ⎥ 4 ¯ ⎣ 6 −1 −2 ¯ 13 ⎦ to ⎢ 1 29 ¯ − 83 ⎣ 0 29 ⎦  ¯ ¯ 7 1 1 ¯ 12 0 0 1 ¯ 1

4 The last row yields 3 = 1. The second row gives 2 = − 29 (1) − 83 29 = −3, and the first row produces 1 = 5(−3) + 1(1) + 16 = 2. Hence, the actual solution for the system is (2 −3 1).

(10) (a) True. This is the definition of roundoff error given in Section 9.1 of the textbook. (b) False. An ill-conditioned system is a system for which a very small change in the coefficients in the system leads to a very large change in the solution set. (c) False. The method of partial pivoting specifies that one should use row swaps to ensure that each pivot element is as large as possible in absolute value. (d) True. Roundoff errors do not accumulate in iterative methods since each iteration is using the previous iteration as a new starting point. Thus, in some sense, each iteration starts the solution process over again. (e) False. This is done in the Gauss-Seidel Method, not the Jacobi Method. (f) False. The values obtained should be  = 6 and  = 1, because the new value  = 6 is used in the computation of the new value of .

Section 9.2 (1) (a) First, use the following row operations (in the given order) h1i ← 12 h1i 

h2i ← 6 h1i + h2i 

and

h2i ← 15 h2i

to obtain the upper triangular matrix U =

∙

1 −2 0 1

¸ . Using the formulas for the variables 

given in the LDU Decomposition Method, we find 11 = 2, 21 = −6, and 22 = 5. Then, using the formulas for L and D given in that same Method, we obtain " # ∙ ¸ ∙ ¸ 1 0 1 0 2 0 L = −6 =  and D =  −3 1 0 5 1 2 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

292

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.2

(c) First, use the following row operations (in the given order) h1i ← − h1i,

h2i ← −2 h1i + h2i, h3i ← −2 h1i + h3i, h2i ← 12 h2i,

h3i ← −8 h2i + h3i  and h3i ← 13 h3i

⎡

⎤ 1 −4 2 1 −4 ⎦. Using the formulas for the variables to obtain the upper triangular matrix U = ⎣ 0 0 0 1  given in the LDU Decomposition Method, we find 11 = −1, 21 = 2, 31 = 2, 22 = 2, 32 = 8, and 33 = 3. Then, using the formulas for L and D given in that same Method, we obtain ⎤ ⎡ ⎡ ⎤ ⎡ ⎤ 1 0 0 1 0 0 −1 0 0 ⎢ 2 1 0 ⎥ L = ⎣ −1 ⎦ = ⎣ −2 1 0 ⎦  and D = ⎣ 0 2 0 ⎦  8 2 −2 4 1 0 0 3 1 −1

2

(e) First, use the following row operations (in the given order) h1i ← − 13 h1i,

h2i ← −4 h1i + h2i, h3i ← −6 h1i + h3i, h4i ← 2 h1i + h4i, h2i ← − 32 h2i,

h3i ← − h2i + h3i,

h4i ← − 43 h2i + h4i,

and

h3i ← 2 h3i

⎡

1 − 13

− 13

0

1

0

0

⎢ ⎢ 0 to obtain the upper triangular matrix U = ⎢ ⎢ 0 ⎣ 0

1

5 2

1 3 − 11 2

⎤

⎥ ⎥ ⎥. Using the formulas for the 3 ⎥ ⎦ 1

variables  given in the LDU Decomposition Method, we find 11 = −3, 21 = 4, 31 = 6,

41 = −2, 22 = − 23 , 32 = 1, 42 = 43 , 33 = 12 , 43 = 0 and 44 = 1. Then, using the formulas for L and D given in that same Method, we obtain ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 0 1 0 0 0 −3 0 0 0 ⎢ 4 ⎢ ⎥ ⎢ ⎥ 1 0 0 ⎥ ⎢ −3 ⎥ ⎢ − 43 ⎢ 0 − 23 0 0 ⎥ 1 0 0 ⎥ ⎥=⎢ ⎥  and D = ⎢ ⎥ L=⎢ 1 ⎢ 6 ⎢ ⎥ ⎢ 0 3 1 0 ⎥ 0 12 0 ⎥ ⎣ −3 (−23) ⎦ ⎣ −2 − 2 1 0 ⎦ ⎣ ⎦ (43) 2 −2 0 −2 0 1 0 0 0 1 1 3 −3 (−23) (12) (3) (a) The coefficient matrix for the system AX = B is A = c 2016 Elsevier Ltd. All rights reserved.

Copyright °

293

∙

−1 5 2 −13

¸ ∙ ¸ −9 , with B = . We 21

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.2

need to find the KU decomposition for A. To do this, we use the following row operations (in the given order) h1i ← − h1i,

h2i ← −2 h1i + h2i  and h2i ← − 13 h2i

to obtain the upper triangular matrix U =

∙

1 −5 0 1

¸ . Using the formulas for the variables 

given in the LDU Decomposition Method, we find 11 = −1, 21 = 2, and 22 = −3, yielding the ∙ ¸ −1 0 lower triangular matrix K = . Now, AX = B is equivalent to K(UX) = B. Letting 2 −3

Y = UX, we must first solve KY = B. This is equivalent to the system ½ = −9 −1  21 − 32 = 21

whose solution is easily seen to be 1 = 9, 2 = −1; that is, Y = [9 −1]. Finally, we find X by solving the system UX = Y, which is ½ 1 − 52 = 9  2 = −1 Back substitution gives the solution 2 = −1, 1 = 4. Hence, the solution set for the original system is {(4 −1)}. ⎡ ⎤ ⎡ ⎤ −1 3 −2 −13 (c) The coefficient matrix for the system AX = B is A = ⎣ 4 −9 −7 ⎦, with B = ⎣ 28 ⎦. −2 11 −31 −68 We need to find the KU decomposition for A. To do this, we use the following row operations (in the given order) h1i ← − h1i,

h2i ← −4 h1i + h2i,

h3i ← 2 h1i + h3i,

h2i ← 13 h2i,

h3i ← −5 h2i + h3i 

and

h3i ← − 12 h3i

⎡

⎤ 1 −3 2 1 −5 ⎦. Using the formulas for the variables to obtain the upper triangular matrix U = ⎣ 0 0 0 1

 given in the LDU Decomposition Method, we find 11 = −1, 21 = 4, 31 = −2, 22 = 3, ⎡ ⎤ −1 0 0 0 ⎦. Now, AX = B 32 = 5, and 33 = −2, yielding the lower triangular matrix K = ⎣ 4 3 −2 5 −2 is equivalent to K(UX) = B. Letting Y = UX, we must first solve KY = B. This is equivalent

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

294

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

to the system

⎧ ⎨ −1 41 ⎩ −21

+ 32 + 52

− 23

Section 9.3

= −13 = 28  = −68

whose solution is easily seen to be 1 = 13, 2 = 13 (−4(13) + 28) = −8, 3 = − 12 (2(13) − 5(−8) − 68) = 1; that is, Y = [13 −8 1]. Finally, we find X by solving the system UX = Y, which is ⎧ ⎨ 1 − 32 + 23 = 13 2 − 53 = −8  ⎩ 1 3 =

Back substitution gives the solution 3 = 1, 2 = 5(1) − 8 = −3, 1 = 3(−3) − 2(1) + 13 = 2.

Hence, the solution set for the original system is {(2 −3 1)}.

(4) (a) False. Nonsingular matrices that require row swaps to reduce to I may not have an LDU ∙ ¸ 0 1 Decomposition. For example, has no LDU Decomposition, as shown in Exercise 2 of 1 0 Section 9.2. (b) True. The method for doing this is given in the proof of Theorem 9.1. (c) False. The given row operation adds a multiple of row 3 to row 2, which is above row 3. This is not allowed for lower Type (II) operations. (d) False. First solve for Y in KY = B, then solve for X in UX = Y.

Section 9.3 (1) In each part, we only print  digits after the decimal point for all computations, even though more digits are actually used and computed. We write all vectors as row vectors to conserve space, even when correct notation calls for a column vector. Also, let A represent the given matrix. √ (a) We begin with u0 = [1 0]. Then, w1 = Au0 = [2 36], and so kw1 k = 22 + 362 ≈ 3606. Using w1 1 ≈ 3606 [2 36] ≈ [006 100]. These results are displayed in the first line of this gives u1 = kw 1k the table, below. Continuing in this manner for a total of 9 iterations produces the following:

 1 2 3 4 5 6 7 8 9

w = Au−1 [2 36] [3606 2496] [2214 4269] [3288 3699] [2824 4111] [3081 3934] [2958 4030] [3021 3984] [2990 4008]

kw k 3606 4385 4809 4949 4987 4997 4999 5000 5000

w u = kw k [006 100] [082 057] [046 089] [066 075] [057 082] [062 079] [059 081] [060 080] [060 080]

Since u8 = u9 , we stop here. Hence, the Power Method produces the eigenvector [060 080].

Since the vectors u8 and u9 are equal (rather than being the negations of each other), the sign of the corresponding eigenvalue is positive, and so the eigenvalue is kw9 k = 50. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

295

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

(c) We begin with u0 = [0 1 0]. Then, w1 = Au0 = [3 0 1], and so kw1 k =

Section 9.3

√ 32 + 02 + 12 ≈ 316.

w1 1 ≈ 316 [3 0 1] ≈ [095 000 032]. These results are displayed in the Using this gives u1 = kw 1k first line of the table, below. Continuing in this manner for a total of 7 iterations produces the following: w  w = Au−1 kw k u = kw k 1 [3 0 1] 316 [095 000 032] 2 [158 126 316] 375 [042 034 084] 3 [101 126 244] 293 [034 043 083] 4 [115 118 230] 283 [041 042 081] 5 [125 122 245] 301 [042 041 081] 6 [123 123 247] 302 [041 041 082] 7 [122 123 245] 300 [041 041 082]

Since u6 = u7 , we stop here. Hence, the Power Method produces the eigenvector [041 041 082].

Since the vectors u6 and u7 are equal (rather than being negations of each other), the sign of the corresponding eigenvalue is positive, and so the eigenvalue is kw7 k = 30.

(e) We begin with u0 = [3 8 2 3]. Then, w1 = Au0 = [17 40 14 19], and so √ w1 1 ≈ 49457 [17 40 14 19] ≈ kw1 k = 172 + 402 + 142 + 192 ≈ 49457. Using this gives u1 = kw 1k [0344 0809 0283 0384]. These results are displayed in the first line of the table, below. Continuing in this manner for a total of 15 iterations produces the following:  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

w = Au−1 [17 40 14 19] [2123 4448 2143 2285] [1825 4224 1392 1928] [1935 4468 1440 2010] [1878 4471 1232 1929] [1893 4537 1190 1929] [1878 4553 1116 1903] [1879 4575 1084 1896] [1874 4585 1053 1886] [1873 4593 1036 1881] [1871 4598 1022 1876] [1870 4602 1013 1874] [1869 4605 1007 1872] [1869 4606 1003 1871] [1868 4607 1000 1870]

kw k 49457 5840 5179 5461 5363 5413 5397 5406 5403 5405 5405 5405 5405 5405 5405

w u = kw k [0344 0809 0283 0384] [0364 0762 0367 0391] [0352 0816 0269 0372] [0354 0818 0264 0368] [0350 0834 0230 0360] [0350 0838 0220 0356] [0348 0844 0207 0353] [0347 0846 0201 0351] [0347 0848 0195 0349] [0346 0850 0192 0348] [0346 0851 0189 0347] [0346 0851 0187 0347] [0346 0852 0186 0346] [0346 0852 0185 0346] [0346 0852 0185 0346]

Since u14 = u15 , we stop here. Hence, the Power Method produces the eigenvector [0346 0852 0185 0346]. Since the vectors u14 and u15 are equal (rather than being negations of each other), the sign of the corresponding eigenvalue is positive, and so the eigenvalue is kw15 k = 5405.

(3) (b) Let 1      be the eigenvalues of A with |1 |  | |, for 2 ≤  ≤ . Let {v1      v } be a basis of fundamental eigenvectors for R corresponding to 1       , respectively. Suppose the initial vector in the Power Method is u0 = 01 v1 + · · · + 0 v and the th iteration yields

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

296

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.4

u = 1 v1 + · · · +  v . Thus, for 2 ≤  ≤ ,  6= 0, and 0 6= 0, we want to prove that ¯ ¯ |1 | ¯¯ 1 ¯¯ |01 | =  | | ¯  ¯ |0 |

Proof: The Power Method states that each u is derived from u−1 as u =  Au−1 , where  is a normalizing constant. A proof by induction shows that u =  A u0 for some constant  . (Base Step ( = 0): u0 = I u0 = 1A0 u0 , with 0 = 1. Inductive Step: u =  Au−1 =  A(−1 A−1 u0 ) =  −1 A u0 , with  =  −1 .) Now, if u0 = 01 v1 + · · · + 0 v , then u =  A u0 = 01  A v1 +· · ·+0  A v . But because v is an eigenvector for A corresponding to the eigenvalue  , A v =  v . Therefore u = 01  1 v1 + · · · + 0   v . Since we also have u = 1 v1 + · · · +  v , the uniqueness of expression of a vector with respect to a basis shows that  = 0   . Hence, ¯ ¯ ¯01  1 ¯ ¯¯ 1 ¯¯ |01 | |1 | ¯ =¯ ¯ = ¯ . ¯0   ¯ ¯  ¯ |0 | | |

(4) (a) False. The eigenvalue could be − kAu−1 k. This occurs when u = −u−1 . For an example, ∙ ¸ −2 1 consider applying the Power Method to the matrix , starting with u0 = [1 0]. Then 0 1

w1 = [−2 0], and kw1 k = 2. Hence, u1 = [−1 0] = −u0 , and the Power Method yields the eigenvector [−1 0]. The eigenvalue corresponding to this eigenvector is −2. However, kAu0 k = kw1 k = 2.

(b) True. Each iteration of the Power Method will just repeatedly produce the initial vector.

(c) True. Starting with u0 = [1 0 0 0], the Power Method will produce u = 12 [1 1 1 1] for all  ≥ 1 and w = [2 2 2 2] for all  ≥ 2. Note that k[2 2 2 2]k = 4.

(d) False. The dominant eigenvalue in this case is −3, not 2, because | − 3|  |2|. Hence, the Power Method will produce an eigenvector corresponding to the eigenvalue −3. (Note: The Power Method converges here since A is diagonalizable and there is a dominant eigenvalue.)

Section 9.4 (1) (a) Following the QR Factorization Method, we label the columns of A as w1 = [2 −2 1] w2 = [6 0 6] and w3 = [−3 −9 −3].

Step 1: Beginning the Gram-Schmidt Process, we obtain v1 v2

= w1 = [2 −2 1] and ³ ´ 2 ·v1 = w2 − w v1 ·v1 v1

[606]·[2−21] = [6 0 6] − [2−21]·[2−21] [2 −2 1]

= [6 0 6] − 18 9 [2 −2 1] = [2 4 4] 

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

297

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.4

Multiplying this vector by a factor of 2 = 12 to simplify, we let v2 = [1 2 2] Finally, ³ ³ ´ ´ w3 ·v2 3 ·v1 v3 = w3 − w v1 ·v1 v1 − v2 ·v2 v2

[−3−9−3]·[122] = [−3 −9 −3] − [−3−9−3]·[2−21] [2−21]·[2−21] [2 −2 1] − [122]·[122] [1 2 2]

= [−3 −9 −3] − 99 [2 −2 1] − −27 9 [1 2 2] = [−2 −1 2]  Thus we obtain v3 = [−2 −1 2]

Step 2: Normalizing v1  v2  v3 , we get ¤ ¤ ¤ £ £ £ u1 = 23  − 23  13  u2 = 13  23  23  u3 = − 23  − 13  23 

Step 3: From the QR Factorization Method, we know that these vectors are the columns of Q. Hence, ⎡ ⎤ 2 1 −2 1⎣ −2 2 −1 ⎦  Q= 3 1 2 2

Step 4:

R = Q A

⎡

⎤⎡ ⎤ ⎡ ⎤ 2 −2 1 2 6 −3 3 6 3 1 ⎣ 1 2 2 ⎦ ⎣ −2 0 −9 ⎦ = ⎣ 0 6 −9 ⎦  3 −2 −1 2 1 6 −3 0 0 3

=

(c) Following the QR Factorization Method, we label the columns of A as w1 = [1 −2 1] w2 = [5 −4 5] and w3 = [−3 −2 −5].

Step 1: Beginning the Gram-Schmidt Process, we obtain v1 v2

= w1 = [1 −2 1] and ³ ´ 2 ·v1 = w2 − w v1 ·v1 v1

= [5 −4 5] − [5−45]·[1−21] [1−21]·[1−21] [1 −2 1] = [5 −4 5] − 18 6 [1 −2 1] = [2 2 2]  Multiplying this vector by a factor of 2 = 12 to simplify, we let v2 = [1 1 1] Finally, ³ ³ ´ ´ w3 ·v2 3 ·v1 − v v3 = w3 − w 1 v1 ·v1 v2 ·v2 v2

[−3−2−5]·[111] = [−3 −2 −5] − [−3−2−5]·[1−21] [1−21]·[1−21] [1 −2 1] − [111]·[111] [1 1 1]

−10 = [−3 −2 −5] − −4 6 [1 −2 1] − 3 [1 1 1]

= [1 0 −1]  Thus we obtain v3 = [1 0 −1]

Step 2: Normalizing v1  v2  v3 , we get ¸ ¸ ¸ ∙ ∙ ∙ 1 1 1 2 1 1 1 1 u1 = √  − √  √  u2 = √  √  √  u3 = √  0 − √  6 6 6 3 3 3 2 2

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

298

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.4

Rationalizing the denominators produces "√ "√ √ √ # "√ √ √ # √ # 6 2 6 6 3 3 3 2 2 −     0 − u1 =  u2 =  u3 =  6 6 6 3 3 3 2 2 Step 3: From the QR Factorization Method, we know that these vectors are the columns of Q. Hence, √ √ √ ⎤ ⎡ 6 2 3 3 2 √ √ 1⎢ ⎥ Q = ⎣ −2 6 2 3 0 ⎦ 6 √ √ √ 6 2 3 −3 2 Step 4:

R = Q A

=

√ √ ⎤⎡ ⎡ √ ⎤ ⎡ √6 3√6 − 2 √6 ⎤ 6 −2 6 6 3 1 5 −3 √ ⎥ √ √ ⎥ √ √ 1 ⎢ 10 ⎣ −2 −4 −2 ⎦ = ⎢ 3 2 3 2 3 3 − 3 ⎦  2 0 2 ⎣ ⎦ ⎣ 6 3 √ √ √ 1 5 −5 0 −3 2 2 3 2 0 0

(2) (a) In this problem, we want to find a least-squares solution for the linear system AX = B, where ⎡ ⎤ ⎡ ⎤ 3 10 −8 A = ⎣ 4 −4 ⎦ and B = ⎣ 30 ⎦ 12 27 10

using a QR Factorization for A, as shown in Example 2 in Section 9.4.

Following the QR Factorization Method, we label the columns of A as w1 = [3 4 12] and w2 = [10 −4 27].

Step 1: Beginning the Gram-Schmidt Process, we obtain = w1 = [3 4 12] and ³ ´ 2 ·v1 = w2 − w v1 ·v1 v1

v1 v2

= [10 −4 27] − [10−427]·[3412] [3412]·[3412] [3 4 12] = [10 −4 27] − 338 169 [3 4 12] = [4 −12 3]  Hence, we let v2 = [4 −12 3]

Step 2: Normalizing v1  v2 , we get ¤ £ 3 4 12 ¤ £4 3 u1 = 13  13  13  u2 = 13  − 12 13  13 

Step 3: From the QR Factorization Method, we know that these vectors are the columns of Q. Hence, ⎡ ⎤ 3 4 1 ⎣ 4 −12 ⎦  Q= 13 12 3

Step 4:

R = Q A

=

1 13

∙

3 4 12 4 −12 3

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

299

¸

⎡

⎤ ∙ ¸ 3 10 ⎣ 4 −4 ⎦ = 13 26  0 13 12 27

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.4

Now,

∙ ¸ 1 216 Q B=  and hence, 13 −362 ∙ ¸∙ ¸ ∙ ¸ 1 13 26  216 RX = =  0 13  13 −362 Since R is upper triangular, we can quickly find the solution by using back substitution. The last 362 equation asserts 13 = − 362 13 , which leads to  = − 169 ≈ −2142. Finally, the first equation gives 216 362 13 + 26 = 13 . Substituting − 169 for  and solving for  leads to  = 940 169 ≈ 5562. (c) In this problem, we want to find a least-squares solution for the linear system AX = B, where ⎡ ⎤ ⎡ ⎤ 1 15 1 7 ⎢ 4 −4 ⎢ ⎥ 18 ⎥ ⎥ and B = ⎢ 11 ⎥ A=⎢ ⎣ 0 ⎦ ⎣ 8 −22 −5 ⎦ −8 10 −1 12 

using a QR Factorization of A, as shown in Example 2 in Section 9.4.

Following the QR Factorization Method, we label the columns of A as w1 = [1 4 0 −8] w2 = [15 −4 8 10] and w3 = [1 18 −22 −1].

Step 1: Beginning the Gram-Schmidt Process, we obtain v1 v2

= w1 = [1 4 0 −8] and ³ ´ 2 ·v1 = w2 − w v1 ·v1 v1

= [15 −4 8 10] − [15−4810]·[140−8] [140−8]·[140−8] [1 4 0 −8] = [15 −4 8 10] − −81 81 [1 4 0 −8] = [16 0 8 2]  Multiplying this vector by a factor of 2 = 12 to simplify, we let v2 = [8 0 4 1] Finally, ³ ³ ´ ´ w3 ·v2 3 ·v1 v3 = w3 − w v1 ·v1 v1 − v2 ·v2 v2

[118−22−1]·[8041] = [1 18 −22 −1] − [118−22−1]·[140−8] [8 0 4 1] [140−8]·[140−8] [1 4 0 −8] − [8041]·[8041]

−81 = [1 18 −22 −1] − 81 81 [1 4 0 −8] − 81 [8 0 4 1]

= [8 14 −18 8] Multiplying this vector by a factor of 2 = 12 to simplify, we obtain v3 = [4 7 −9 4]

Step 2: Normalizing v1  v2  v3 , we get " √ √ √ √ # ¸ ¸ ∙ ∙ 8 4 1 1 4 8 2 2 7 2 2 2 2   0 −  u2 =  0   u3 =  −  u1 =  9 9 9 9 9 9 9 18 2 9

Step 3: From the QR Factorization Method, we know that these vectors are the columns of Q. Hence, ⎡ √ ⎤ 2 16 4 2 √ ⎥ ⎢ 8 0 7 2 ⎥ 1 ⎢ ⎢ √ ⎥ Q= 18 ⎢ 0 8 −9 2 ⎥ ⎦ ⎣ √ −16 2 4 2 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

300

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.4

Step 4:

R = Q A

Now,

=

⎡ ⎤ ⎤ ⎤ ⎡ 1 15 1 2 8 0 −16 9 −9 9 ⎢ ⎥ 4 −4 18 1 ⎣ 16 ⎥ = ⎣ 0 18 −9 ⎦  0 8 2 ⎦⎢ 18 ⎣ 0 √ √ √ √ √ 8 −22 ⎦ 0 0 18 2 4 2 7 2 −9 2 4 2 −8 10 −1 ⎡

⎤ ⎡ −15 1 ⎦  and hence, Q B = ⎣ 16 √ 3 33 2 ⎤⎡ ⎤ ⎤ ⎡ ⎡ −15 9 −9 9  ⎦ ⎣  ⎦ = 1 ⎣ 16 ⎦  RX = ⎣ 0 18 −9 √ √ 3  33 2 0 0 18 2

Since R is upper triangular, we √ √ can quickly find the solution by using back substitution. The last equation asserts 18 2 = 11 2, which leads to  = 11 18 ≈ 0611. The middle equation states that 11 65 .

Substituting  = into this equation yields  = 108 ≈ 0602. Finally, the first 18 − 9 = 16 3 18 65 11 equation gives 9 − 9 + 9 = −5. Substituting 108 for  and 18 for  and solving for  leads to 61 ≈ −0565.  = − 108 (5) (a) True. Suppose A is a nonsingular matrix having size  × . Then the columns of A form a basis for R .by part (c) of Exercise 15 in Section 4.5. Then by Theorem 9.2, Q is also an  ×  matrix, and the columns of Q span R as well. Since the columns of Q are also an orthonormal set of vectors by Theorem 9.2, they form an orthonormal basis for R . Therefore Q is an orthogonal matrix by Theorem 6.7. (b) True. Suppose A is a singular matrix having size  × . Since A is singular, not all of the  columns of A are linearly independent, and so the more general QR Factorization Method (described at the end of Section 9.4) must be used here. Note that the matrix Q created by this more general method is also  ×  and its columns form a linearly independent set. Thus Q is a nonsingular matrix. Now suppose R has all of its main diagonal entries nonzero. Then |R| 6= 0 (by Theorem 3.2), and R is nonsingular. Then A = QR is nonsingular, a contradiction. (c) True. In the QR Factorization Method the columns of Q are found by performing the GramSchmidt Process on the columns of A and then normalizing. Since A is an upper triangular matrix, the Gram-Schmidt Process produces vectors that are multiples of the standard basis vectors for R , and once these vectors are normalized, each resulting vector is either a standard basis vector in R or its opposite. Thus, the matrix Q that is created by the QR Factorization Method is a diagonal matrix.

Note that the QR Factorization Method produces a matrix R with all main diagonal entries positive. Therefore, if A has all positive entries on the main diagonal, then Q equals I . However, if A has some negative entries on the diagonal, then the corresponding diagonal entries in Q would equal −1 instead of 1. (d) False. By Theorem 9.3, X = R−1 Q B, not R Q−1 B. For a counterexample, consider A = ∙ ¸ ∙ ¸ 1 −1 3 and B = . Then the QR Factorization Method gives A = QR with Q = I2 and 0 1 2 R = A. The solution for the system is X = [5 2], but R Q−1 B = [3 −1].

(e) True. If QR is the QR Factorization of A, then Q is a matrix having an orthonormal set of columns. Hence, Q Q = I . Therefore, R = I R = Q QR = Q A. c 2016 Elsevier Ltd. All rights reserved.

Copyright °

301

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.5

Section 9.5 (1) For each part, one possibility is given. ∙ ¸ 34 12 (a) First, A A = , and A A () = ( − 34)( − 16) − (−12)2 = 2 − 50 + 400 = 12 16 √ √ √ ( − 40)( − 10). Hence, 1 = 40 and 2 = 10, implying  1 = 40 = 2 10 and  2 = 10. ∙ ¸ 1 −2 , giving the fundamental eigenvector [2 1].

Row reducing (40I2 − A A) produces 0 0 " # 1 12 Similarly, row reducing (10I2 − A A) produces , giving the fundamental eigenvector 0 0 o n [−1 2]. Normalizing these vectors yields the set √15 [2 1] √15 [−1 2] of right singular vectors for A. Next, we find the left singular vectors for A by computing u1 = 11 Av1 and u2 = 12 Av2 , where v1 and v2 are the right singular vectors. ∙ ¸ µ ∙ ¸¶ ∙ ¸ ∙ ¸ 3 4 2 10 1 1 1 1 1 1 √ = 10√2 = √2 .

Now, u1 = 1 Av1 = 2√10 5 5 0 1 10 1 ∙ ¸µ ∙ ¸¶ ∙ ¸ ∙ ¸ 3 4 −1 5 1 1 1 1 1 1 √ √ √ √ = 5 2 = 2 .

Similarly, u2 = 2 Av2 = 10 5 5 0 2 −5 −1

Since we have two left singular vectors, and that is all we need for a basis for R2 , the set {u1  u2 } of left singular vectors is complete.

Therefore, using the right singular vectors as columns for V, the left singular vectors as columns for U, and the singular values as diagonal entries for Σ, we can express A as UΣV with U = √12

∙

1 1 1 −1

¸

 Σ=

¸ ∙ √ ∙ ¸ 2 −1 2 10 √0  and V = √15  1 2 10 0

A short computation verifies that µ ∙ ¸µ ∙ ¸¶ ∙ √ ¸¶ ∙ ¸ 1 1 2 10 √0 2 1 3 4 √1 √1 = = A 2 5 1 −1 −1 2 5 0 10 0 ⎡

⎤ 130 140 −200 400 −340 ⎦, and A A () = 3 −9002 +72900 = (−810)(−90). (c) First, A A = ⎣ 140 −200 −340 370 √ √ √ √ Hence, 1 = 810, 2 = 90, and 3 = 0, implying  1 = 810 = 9 10,  2 = 90 = 3 10, ⎡ ⎤ 1 0 12 ⎢ ⎥ and 3 = 0. Row reducing (810I3 − A A) produces ⎣ 0 1 1 ⎦, giving the fundamen0 0 0 ⎡ ⎤ 1 0 2 tal eigenvector [−1 −2 2]. Similarly, row reducing (90I3 − A A) produces ⎣ 0 1 −2 ⎦, 0 0 0 giving the fundamental eigenvector [−2 2 1].

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

302

Finally, row reducing (0I3 − A A) produces

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

1 0

−1

Section 9.5

⎤

⎢ ⎥ ⎢ 0 1 − 1 ⎥, giving the fundamental eigenvector [2 1 2]. Normalizing these vectors yields 2 ⎣ ⎦ 0 0 0 ª © the set 13 [−1 −2 2] 13 [−2 2 1] 13 [2 1 2] of right singular vectors for A. Next, we find the left singular vectors for A by computing u1 = 11 Av1 , and u2 = 12 Av2 , where v1 , and v2 are the first two right singular vectors. (Since Av3 = 0, we do not use v3 to compute a left singular vector.) ⎛ ⎡ ⎤⎞ ∙ ¸ ∙ ¸ ∙ ¸ −1 7 20 −17 −81 −3 1 ⎝ 1 ⎣ −2 ⎦⎠ = √ √1 = .

Now, u1 = 11 Av1 = 9√110 3 27 10 10 −9 0 9 27 1 2 ⎛ ⎡ ⎤⎞ ∙ ¸ ∙ ¸ ∙ ¸ −2 7 20 −17 9 1 1 1 1 ⎣ 1 1 ⎝ ⎦ ⎠ √ √ √ 2 = 10 . = 9 10 Similarly, u2 = 2 Av2 = 3 10 3 −9 0 9 27 3 1

Since we have two left singular vectors, and that is all we need for a basis for R2 , the set {u1  u2 } of left singular vectors is complete.

Therefore, using the right singular vectors as columns for V, the left singular vectors as columns for U, and the singular values as diagonal entries for Σ, we can express A as UΣV with ⎡ ⎤ ¸ ∙ ¸ ∙ √ −1 −2 2 −3 1 9 10 0 0 √ 2 1 ⎦  and V = 13 ⎣ −2  Σ= U = √110 1 3 0 3 10 0 2 1 2

A short computation verifies that

⎛ ⎡ ⎤⎞ ¸ ¸¶ ∙ √ ∙ ¸ −1 −2 2 −3 1 9 10 √0 7 20 −17 0 ⎝1 ⎣ ⎦⎠ = √1 −2 2 1 = A 3 10 1 3 −9 0 9 0 3 10 0 2 1 2 ∙ ¸ 5 3  , and A A () = 2 −10+16 = (−8)(−2). Hence, 1 = 8, and 2 = 2, (f) First, A A = 3 5 ∙ ¸ √ √ √ 1 −1  implying  1 = 8 = 2 2, and  2 = 2. Row reducing (8I2 − A A) produces , giving 0 ¸ ∙0 1 1 the fundamental eigenvector [1 1]. Similarly, row reducing (2I2 −A A) produces , giving 0 0 o n the fundamental eigenvector [−1 1]. Normalizing these vectors yields the set √12 [1 1] √12 [−1 1] of right singular vectors for A. Next, we find the left singular vectors for A by computing u1 = 11 Av1 , and u2 = 12 Av2 , where v1 , and v2 are the two right singular vectors. ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ µ ∙ ¸¶ 10 14 24 6 1 1 ⎝1 ⎣ 1 ⎣ ⎦⎠ √1 ⎦ = 1 ⎣ 3 ⎦.

12 0 12 = Now, u1 = 11 Av1 = 2√ 7 28 7 2 2 1 1 7 8 2 ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ µ ∙ ¸¶ 10 14 4 2 −1 1 ⎣ −12 ⎦ = 17 ⎣ −6 ⎦.

Similarly, u2 = 12 Av2 = √12 ⎝ 17 ⎣ 12 0 ⎦⎠ √12 = 14 1 1 7 6 3 µ

∙

Since we only have two left singular vectors, we need one more for a basis for R3 . Thus, we need

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

303

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.5

to expand the set {u1  u2 } to an orthonormal basis.

Row reducing the matrix whose columns are u1  u2  e1  e2  e3 shows, by the Independence Test Method, that the set {u1  u2  e1 } is linearly independent. Performing the Gram-Schmidt Process 1 [9 −6 −18], which on this set has no effect on u1 and u2 , but does produce a new third vector 49 1 normalizes to u3 = 7 [3 −2 −6]. This completes the set {u1  u2  u3 } of left singular vectors.

Therefore, using the right singular vectors as columns for V, the left singular vectors as columns for U, and the singular values as diagonal entries for Σ, we can express A as UΣV with ⎡

⎤ ⎡ √ ⎤ ∙ ¸ 6 2 3 2 2 √0 1 −1 1⎣ 1 ⎦ ⎣ ⎦ U = 7 3 −6 −2  Σ =  2  and V = √2 0 1 1 2 3 −6 0 0

A short computation verifies that ⎛ ⎡ ⎤⎞ ⎡ √ ⎡ ⎤ ⎤ ¸¶ µ ∙ 6 2 3 2 2 √0 10 14 1 1 ⎝ 1 ⎣ 3 −6 −2 ⎦⎠ ⎣ 0 = 17 ⎣ 12 0 ⎦ = A 2 ⎦ √12 7 −1 1 2 3 −6 1 7 0 0

¸ (2) (a) First, A A = , and A A () = 2 −305+22500 = (−180)(−125). Hence, 1 = √ √ √ √ 180, and 2 = 125, implying  1 = 180 = 6 5, and  2 = 125 = 5 5. Row reducing (180I2 − ∙ ¸ 1 −2  , giving the fundamental eigenvector [2 1]. Similarly, row reducing A A) produces 0 0 " # 1 12  , giving the fundamental eigenvector [−1 2]. Normalizing (125I2 − A A) produces 0 0 o n these vectors yields the set √15 [2 1] √15 [−1 2] of right singular vectors for A. 

∙

169 22 22 136

Next, we find the left singular vectors for A by computing u1 = 11 Av1 , and u2 = 12 Av2 , where v1 , and v2 are the two right singular vectors. ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ µ ∙ ¸¶ 94 −128 60 2 2 1 ⎝ 1 ⎣ 1 ⎣ 1 ⎣ 95 110 ⎦⎠ √15 300 ⎦ = 15 10 ⎦. = 450 Now, u1 = 11 Av1 = 6√ 15 5 1 142 46 330 11 ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ µ ∙ ¸¶ 94 −128 −350 −14 −1 1 ⎝ 1 ⎣ 1 ⎣ 1 ⎣ 95 110 ⎦⎠ √15 125 ⎦ = 15 5 ⎦.

Also, u2 = 12 Av2 = 5√ = 375 15 5 2 142 46 −50 −2

Since we only have two left singular vectors, we need one more for a basis for R3 . Thus, we need to expand the set {u1  u2 } to an orthonormal basis.

Row reducing the matrix whose columns are u1  u2  e1  e2  e3 shows, by the Independence Test Method, that the set {u1  u2  e1 } is linearly independent. Performing the Gram-Schmidt Process on this set has no effect on u1 and u2 , but does produce a new third vector 19 [1 2 −2], which 1 [5 10 −10]. This completes the set {u1  u2  u3 } of left singular normalizes to u3 = 13 [1 2 −2] = 15 vectors.

Therefore, using the right singular vectors as columns for V, the left singular vectors as columns

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

304

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.5

for U, and the singular values as diagonal entries for Σ, we can express A as UΣV with ⎡ ⎤ ⎡ √ ⎤ ∙ ¸ 2 −14 5 6 5 √ 0 2 −1 1 ⎣ 1 ⎦ ⎣ ⎦ 5 10  Σ = U = 15 10  0 5 5  and V = √5 1 2 11 −2 −10 0 0 ⎡ ⎤ 1 √ 0 0 6 5 ⎦, and so The matrix Σ+ = ⎣ 1 √ 0 0 5 5 A+

∙

¸¶

2 −1 1 2 ∙ ¸ 104 70 122 1  2250 −158 110 31

= VΣ+ U = =

µ

√1 5

⎡ ⎣

1 √ 6 5

0

0

1 √ 5 5

⎤⎞ 2 10 11 ⎦ ⎝ 1 ⎣ −14 5 −2 ⎦⎠ 15 0 5 10 −10

0

⎡

⎤⎛

⎤ 3 1 ⎣ 27 ⎦ A least-squares solution to Ax = b is given by v = A+ b = 2250 28 ∙ ¸ ∙ ∙ ¸ ∙ ¸ ¸ 5618 6823 24968 45486 1 1   = 2250 = = . Note that both A Av and A b equal 15 .

3364 3874 14951 25826 ⎡ ⎤ 229 −108 6 1 ⎣ −108 349 114 ⎦, and A A () = 3 − 142 + 49 − 36 (c) First, A A = 49 6 114 108 √ = ( − 9)( − 4)( − 1). Hence, 1 = 9, 2 = 4, and 3 = 1, implying  1 = 9 = 3, ⎡ ⎤ 3 1 0 2 √ √ ⎢ ⎥  2 = 4 = 2, and  3 = 1 = 1. Row reducing (9I3 − A A) produces ⎣ 0 1 −3 ⎦, 0 0 0 ∙

104 70 122 −158 110 31

¸

⎡

giving the fundamental eigenvector [−3 6 2]. Similarly, row reducing (4I3 − A A) produces ⎤ ⎡ 1 0 −2 ⎥ ⎢ ⎢ 0 1 − 2 ⎥, giving the fundamental eigenvector [6 2 3]. Finally, row reducing (1I3 − A A) 3 ⎦ ⎣ 0 0 0 ⎡ ⎤ 1 0 13 ⎢ ⎥ 1 ⎥ produces ⎢ ⎣ 0 1 2 ⎦ , giving the fundamental eigenvector [−2 −3 6]. Normalizing these vec0 0 0 ª © tors yields the set 17 [−3 6 2] 17 [6 2 3] 17 [−2 −3 6] of right singular vectors for A. Next, we find the left singular vectors for A by computing u1 = 11 Av1 , u2 = 12 Av2 , and u3 = 13 Av3 , where v1 , v2 , and v3 are the three right singular vectors. ⎛ ⎡ ⎡ ⎡ ⎤⎞ ⎤ ⎤ ⎤⎞ ⎛ ⎡ 23 −11 −6 −147 −1 −3 ⎜1 ⎢ 5 ⎢ ⎢ ⎟ ⎥ ⎥ 25 6 ⎥ ⎢ ⎥⎟ ⎝ 1 ⎣ 6 ⎦⎠ = 1 ⎢ 147 ⎥ = 1 ⎢ 1 ⎥.

Now, u1 = 11 Av1 = 13 ⎜ 294 ⎣ −147 ⎦ 2 ⎣ −1 ⎦ ⎝ 14 ⎣ 19 −17 6 ⎦⎠ 7 2 1 19 18 147 1

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

305

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

⎛

Section 9.5

⎡

⎡ ⎡ ⎤ ⎤⎞ ⎤ ⎛ ⎡ ⎤⎞ 23 −11 −6 98 1 6 ⎜ ⎢ ⎢ ⎢ ⎥ ⎟ ⎥ ⎥ 5 25 6 98 1 ⎢ 1 ⎢ 1 ⎥ ⎥⎟ ⎝ 1 ⎣ 2 ⎦⎠ = 1 ⎢ ⎥ Similarly, u2 = 12 Av2 = 12 ⎜ 196 ⎣ 98 ⎦ = 2 ⎣ 1 ⎦. ⎝ 14 ⎣ 19 −17 6 ⎦⎠ 7 3 1 19 18 98 1 ⎛ ⎡ ⎡ ⎡ ⎤⎞ ⎤ ⎤ ⎤⎞ ⎛ ⎡ 23 −11 −6 −49 −1 −2 ⎜1 ⎢ 5 ⎢ ⎢ ⎟ ⎥ ⎥ 25 6 ⎥ ⎢ ⎥⎟ ⎝ 1 ⎣ −3 ⎦⎠ = 1 ⎢ −49 ⎥ = 1 ⎢ −1 ⎥.

Also, u3 = 13 Av3 = 11 ⎜ 98 ⎣ 2 ⎣ ⎝ 14 ⎣ 19 −17 6 ⎦⎠ 7 49 ⎦ 1 ⎦ 6 1 19 18 49 1

Since we only have three left singular vectors, we need one more for a basis for R4 . Thus, we need to expand the set {u1  u2  u3 } to an orthonormal basis.

Row reducing the matrix whose columns are u1  u2  u3  e1  e2  e3  e4 shows, by the Independence Test Method, that the set {u1  u2  u3  e1 } is linearly independent. Performing the GramSchmidt Process on this set has no effect on u1 , u2 and u3 , but does produce a new fourth vector 1 1 4 [1 −1 −1 1], which normalizes to u4 = 2 [1 −1 −1 1]. This completes the set {u1  u2  u3  u4 } of left singular vectors.

Therefore, using the right singular vectors as columns for V, the left singular vectors as columns for U, and the singular values as diagonal entries for Σ, we can express A as UΣV with ⎡

⎤ ⎡ −1 1 −1 1 3 ⎢ ⎥ ⎢ 1 1 −1 −1 ⎥ Σ = ⎢ 0 U = 12 ⎢ ⎣ −1 1 ⎣ 0 1 −1 ⎦ 1 1 1 1 0 ⎡ 1 ⎤ 0 0 0 3 ⎢ ⎥ 1 ⎥ The matrix Σ+ = ⎢ ⎣ 0 2 0 0 ⎦, and so 0 0 1 0 A+

0 2 0 0

⎡ ⎤⎞ 1 3 −3 6 −2 ⎢ = VΣ+ U = ⎝ 17 ⎣ 6 2 −3 ⎦⎠ ⎢ ⎣ 0 2 3 6 0 ⎡ ⎤ 36 24 12 0 1 ⎣ 12 36 −24 0 ⎦  = 84 −31 −23 41 49 ⎛ ⎡

⎤ ⎡ ⎤ 0 −3 6 −2 0 ⎥ ⎥  and V = 1 ⎣ 6 2 −3 ⎦  7 1 ⎦ 2 3 6 0

⎤⎛ ⎡

−1 1 −1 ⎥ ⎜1 ⎢ 1 1 1 1 ⎜ ⎢ 0 0 ⎥ 2 ⎦ ⎝ 2 ⎣ −1 −1 1 1 −1 −1 0 1 0 0 0 0

⎤⎞ 1 ⎟ 1 ⎥ ⎥⎟ ⎦ 1 ⎠ 1

⎡ ⎤ ⎤ 3 36 24 12 0 ⎢ ⎥ 2 ⎥ 1 ⎣ 12 36 −24 0 ⎦ ⎢ A least-squares solution to Ax = b is given by v = A+ b = 84 ⎣ 9 ⎦ −31 −23 41 49 4 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 264 44 127 1 ⎣ 1 ⎣ −108 ⎦ = 14 −18 ⎦. Note that both A Av and A b equal 17 ⎣ −30 ⎦. = 84 426 71 60 ∙ ¸ ∙ ¸ 1 1 1 1 √ √ (3) (a) Using the results from the answers for part (a) of Exercise 1, u1 = 2 , u2 = 2 , 1 −1 c 2016 Elsevier Ltd. All rights reserved.

Copyright °

306

⎡

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

¸ ∙ ¸ √ √ −1 1 √ , v2 = 5 ,  1 = 2 10 and  2 = 10. Hence, A =  1 u1 v1 +  2 u2 v2 v1 = 2 ¸¶ ³ µ ∙ ¸¶ ³ µ ∙ √ £ ¤´ √ £ ¤´ 1 1 √1 √1 √1 2 1 −1 2 = 2 10 √12 10 + .

5 2 5 1 −1 ⎡ ⎤ ⎡ ⎤ 6 2 (c) Using the results from the answers for part (f) of Exercise 1, u1 = 17 ⎣ 3 ⎦, u2 = 17 ⎣ −6 ⎦, 2 3 ∙ ¸ ∙ ¸ √ √ 1 −1 , v2 = √12 , 1 = 2 2 and  2 = 2. Hence, A =  1 u1 v1 +  2 u2 v2 v1 = √12 1 1 ⎤⎞ ⎛ ⎡ ⎤⎞ ⎛ ⎡ 6 2 ³ £ ³ £ ´ √ √ ¤ ¤´ = 2 2 ⎝ 17 ⎣ 3 ⎦⎠ √12 1 1 + 2 ⎝ 17 ⎣ −6 ⎦⎠ √12 −1 1 .

2 3 √1 5

∙

Section 9.5

2 1

(8) (a) The th column of V is the right singular vector v , which is a unit eigenvector corresponding to the eigenvalue  of A A. But −v is also a unit eigenvector corresponding to the eigenvalue  of A A. Thus, if any vector v is replaced with its opposite vector, the set {v1      v } is still an orthonormal basis for R consisting of eigenvectors for A A. Since the vectors are kept in the same order, the  do not increase, and thus {v1      v } fulfills all the necessary conditions to be a set of right singular vectors for A. For  ≤ , the left singular vector u = 1 Av , so when we change the sign of v , we must adjust U by changing the sign of u as well. For   , changing the sign of v has no effect on U, but still produces a valid Singular Value Decomposition. (b) If the eigenspace  for A A has dimension higher than 1, then the corresponding right singular vectors can be replaced by any orthonormal basis for  , for which there is an infinite number of choices. Then the associated left singular vectors u1      u must be adjusted accordingly.

(9) (a) Each right singular vector v , for 1 ≤  ≤ , must be an eigenvector for A A. Performing the Gram-Schmidt Process on the rows of A, eliminating zero vectors, and normalizing will produce an orthonormal basis for the row space of A, but there is no guarantee that it will consist of ∙ ¸ 1 1  , performing the Gram-Schmidt Process on eigenvectors for A A. For example, if A = 0 1 ¤ £ the rows of A produces the two vectors [1 1] and − 12  12 , neither of which is an eigenvector for ∙ ¸ 1 1 A A = .

1 2

(b) The right singular vectors v+1      v are an orthonormal basis for the eigenspace 0 of A A. Any orthonormal basis for 0 will do. By part (5) of Theorem 9.5, 0 equals the kernel of the linear transformation  whose matrix with respect to the standard bases is A. A basis for ker() can be found by using the Kernel Method. That basis can be turned into an orthonormal basis for ker() by applying the Gram-Schmidt Process and normalizing.

(10) If A = UΣV , as given in the exercise, then A+ = VΣ+ U , and so A+ A = VΣ+ U UΣV = VΣ+ ΣV . Note that Σ+ Σ is an  ×  diagonal matrix whose first  diagonal entries equal 1, with the remaining diagonal entries equal to 0. Note also that since the columns v1      v of V are orthonormal, V v = e , for 1 ≤  ≤ . (a) If 1 ≤  ≤ , then A+ Av = VΣ+ ΣV v = VΣ+ Σe = Ve = v . c 2016 Elsevier Ltd. All rights reserved.

Copyright °

307

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.5

(b) If   , then A+ Av = VΣ+ ΣV v = VΣ+ Σe = V (0) = 0.

(11) If A = UΣV , as given in the exercise, then A+ = VΣ+ U . Note that since the columns u1      u of U are orthonormal, U u = e , for 1 ≤  ≤ . ³ ´ (a) If 1 ≤  ≤ , then A+ u = VΣ+ U u = VΣ+ e = V 1 e = 1 v . Thus, AA+ u = A (A+ u ) ³ ´ = A 1 v = 1 Av = u . (b) If   , then A+ u = VΣ+ U u = VΣ+ e = V (0) = 0. Thus, AA+ u = A (A+ u ) = A (0) = 0.

(14) (a) A =  1 u1 v1 +  2 u2 v2 +  3 u3 v3 +  4 u4 v4 + 5 u5 v5 ⎛ ⎡ ⎤⎞ ⎛ ⎡ ⎤⎞ 1 2 ⎜ ⎢ 0 ⎥⎟ ⎜ ⎢ 0 ⎥⎟ ⎜ 1 ⎢ ⎥⎟ ¡ 1 ⎜1 ⎢ ⎥⎟ ¡ 1 ¢ ¢ ⎜ ⎢ ⎢ ⎥ ⎟ ⎥⎟ = 150 ⎜ 3 ⎢ 2 ⎥⎟ 2 [1 0 1 −1 0 −1] + 30 ⎜ ⎜ 3 ⎢ 1 ⎥⎟ 2 [1 0 −1 1 0 −1] ⎝ ⎣ 0 ⎦⎠ ⎝ ⎣ 0 ⎦⎠ 2 −2 ⎛ ⎡ ⎛ ⎡ ⎤⎞ ⎤⎞ 2 0 ⎜ ⎢ 3 ⎥⎟ ⎜ ⎢ 0 ⎥⎟ ⎜ ⎢ ⎜1 ⎢ ⎥⎟ ¡ ⎥⎟ ¡ ¢ ¢ ⎢ −2 ⎥⎟ 1 [1 −1 0 0 1 1] + 6 ⎜ 1 ⎢ 0 ⎥⎟ 1 [1 1 0 0 −1 1] + 15 ⎜ ⎜5 ⎢ ⎜3 ⎢ ⎥⎟ 2 ⎥⎟ 2 ⎝ ⎣ −4 ⎦⎠ ⎝ ⎣ 0 ⎦⎠ 1 0 ⎛ ⎡ ⎤⎞ 0 ⎜ ⎢ 4 ⎥⎟ ⎜ 1 ⎢ ⎥⎟ ¡ 1 ¢ ⎢ ⎥⎟ + 3⎜ ⎜ 5 ⎢ 0 ⎥⎟ 2 [0 1 1 1 1 0] ⎝ ⎣ 3 ⎦⎠ 0 ⎡ ⎤ ⎡ ⎤ 25 0 25 −25 0 −25 10 0 −10 10 0 −10 ⎢ 0 0 0 ⎢ 0 0 0 ⎥ 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥+⎢ ⎥ 50 0 50 −50 0 −50 5 0 −5 5 0 −5 =⎢ ⎢ ⎥ ⎢ ⎥ ⎣ 0 0 0 0 0 0 ⎦ ⎣ 0 0 0 0 0 0 ⎦ 50 0 50 −50 0 −50 −10 0 10 −10 0 10 ⎡ ⎡ ⎤ ⎤ 10 −10 0 0 10 10 0 0 0 0 0 0 ⎢ ⎢ 0 0 0 0 0 0 ⎥ 9 9 0 0 −9 9 ⎥ ⎢ ⎢ ⎥ ⎥ 1 ⎢ 1 ⎢ ⎥ 10 0 0 −10 −10 ⎥ + 5 ⎢ 0 0 0 0 0 0 ⎥ + 2 ⎢ −10 ⎥ ⎣ ⎣ −12 −12 0 0 12 −12 ⎦ 0 0 0 0 0 0 ⎦ 5 −5 0 0 5 5 0 0 0 0 0 0 ⎡ ⎤ 0 0 0 0 0 0 ⎢ 0 12 12 12 12 0 ⎥ ⎢ ⎥ 1 ⎢ ⎥ + 10 ⎢ 0 0 0 0 0 0 ⎥ ⎣ 0 9 9 9 9 0 ⎦ 0 0 0 0 0 0

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

308

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition ⎡

Section 9.5

⎤ 40 −5 15 −15 5 −30 ⎢ 18 3 12 12 −06 18 ⎥ ⎢ ⎥ ⎢ 50 5 45 −45 −5 −60 ⎥ =⎢ ⎥ ⎣ −24 −15 09 09 33 −24 ⎦ 425 −25 60 −60 25 −375 ⎛ ⎡ ⎤⎞ ⎡ ⎤ 25 0 25 −25 0 −25 1 ⎜ ⎢ 0 ⎥⎟ ⎢ 0 0 0 ⎥ ⎜ 1 ⎢ ⎥⎟ ¡ 1 ⎥ ¢ ⎢ 0 0 0 ⎢ 2 ⎥⎟ [1 0 1 −1 0 −1] = ⎢ 50 0 50 −50 0 −50 ⎥ ; (b) A1 = 1 u1 v1 = 150 ⎜ ⎜ 3 ⎢ ⎥⎟ 2 ⎢ ⎥ ⎝ ⎣ 0 ⎦⎠ ⎣ 0 0 0 0 0 0 ⎦ 50 0 50 −50 0 −50 2 ⎛ ⎡ ⎤⎞ 2 ⎜ ⎢ 0 ⎥⎟ ⎜1 ⎢ ⎥⎟ ¡ 1 ¢    ⎢ ⎥⎟ A2 = 1 u1 v1 + 2 u2 v2 = A1 +  2 u2 v2 = A1 + 30 ⎜ ⎜ 3 ⎢ 1 ⎥⎟ 2 [1 0 −1 1 0 −1] ⎝ ⎣ 0 ⎦⎠ −2 ⎡ ⎤ ⎡ ⎤ 25 0 25 −25 0 −25 10 0 −10 10 0 −10 ⎢ 0 0 0 ⎢ 0 0 0 ⎥ 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 5 0 −5 5 0 −5 ⎥ = ⎢ 50 0 50 −50 0 −50 ⎥ + ⎢ ⎥ ⎣ 0 0 0 0 0 0 ⎦ ⎣ 0 0 0 0 0 0 ⎦ 50 0 50 −50 0 −50 −10 0 10 −10 0 10 ⎡ ⎤ 35 0 15 −15 0 −35 ⎢ 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ = ⎢ 55 0 45 −45 0 −55 ⎥ ⎥; ⎣ 0 0 0 0 0 0 ⎦ 40 0 60 −60 0 −40 ⎛ ⎡ ⎤⎞ 2 ⎜ ⎢ 0 ⎥⎟ ⎜1 ⎢ ⎥⎟ ¡ 1 ¢     ⎢ ⎥⎟ A3 =  1 u1 v1 +  2 u2 v2 + 3 u3 v3 = A2 +  3 u3 v3 = A2 + 15 ⎜ ⎜ 3 ⎢ −2 ⎥⎟ 2 [1 −1 0 0 1 1] ⎝ ⎣ 0 ⎦⎠ 1 ⎤ ⎡ ⎡ ⎤ 35 0 15 −15 0 −35 10 −10 0 0 10 10 ⎢ 0 0 0 ⎢ ⎥ 0 0 0 0 0 0 0 0 0 ⎥ ⎢ ⎢ ⎥ ⎥ 1 ⎢ ⎢ ⎥ 10 0 0 −10 −10 ⎥ = ⎢ 55 0 45 −45 0 −55 ⎥ + 2 ⎢ −10 ⎥ ⎣ 0 0 0 ⎣ 0 0 0 ⎦ 0 0 0 0 0 0 ⎦ 40 0 60 −60 0 −40 5 −5 0 0 5 5 ⎡ ⎤ 40 −5 15 −15 5 −30 ⎢ 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ 5 45 −45 −5 −60 ⎥ = ⎢ 50 ⎥; ⎣ 0 0 0 0 0 0 ⎦ 425 −25 60 −60 25 −375 A4 = 1 u1 v1 + 2 u2 v2 +  3 u3 v3 +  4 u4 v4 = A3 + 4 u4 v4

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

309

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.5

⎛ ⎡

⎤⎞ 0 ⎜ ⎢ 3 ⎥⎟ ⎜1 ⎢ ⎥⎟ ¡ 1 ¢ ⎢ ⎥⎟ = A3 + 6 ⎜ ⎜ 5 ⎢ 0 ⎥⎟ 2 [1 1 0 0 −1 1] ⎝ ⎣ −4 ⎦⎠ 0 ⎡ ⎡ ⎤ 40 −5 15 −15 5 −30 0 0 0 ⎢ ⎢ ⎥ 0 0 0 0 0 0 9 9 0 ⎢ ⎢ ⎥ ⎥+ 1⎢ 50 5 45 −45 −5 −60 0 0 0 =⎢ 5⎢ ⎢ ⎥ ⎣ ⎣ −12 −12 0 0 0 0 0 0 0 ⎦ 425 −25 60 −60 25 −375 0 0 0 ⎡ ⎤ 40 −5 15 −15 5 −30 ⎢ 18 18 0 0 −18 18 ⎥ ⎢ ⎥ ⎢ 50 5 45 −45 −5 −60 ⎥ =⎢ ⎥ ⎣ −24 −24 0 0 24 −24 ⎦ 425 −25 60 −60 25 −375

⎤ 0 0 0 0 −9 9 ⎥ ⎥ 0 0 0 ⎥ ⎥ 0 12 −12 ⎦ 0 0 0

(c) By Exercise 13, the singular values of A are  1        , and the singular values of A − A are  +1      5 . p √ √ 21 + · · · + 25 = 1502 + 302 + 152 + 62 + 32 = 23670 Now, by Exercise 12,  (A) = p √ √ ≈ 15385. Similarly,  (A − A1 ) =  22 + · · · + 25 = 302 + 152 + 62 + 32 = 1170 ≈ 3421, p p √ √  (A − A2 ) =  23 + · · · + 25 = 152 + 62 + 32 = 270 ≈ 1643,  (A − A3 ) =  24 +  25 p √ √ √ √ = 62 + 32 = 45 ≈ 671, and  (A − A4 ) =  25 = 32 = 9 = 3. Hence,  (A − A1 ) (A) ≈ 342115385 ≈ 02223,  (A − A2 ) (A) ≈ 164315385 ≈ 01068,  (A − A3 ) (A) ≈ 67115385 ≈ 00436, and  (A − A4 ) (A) ≈ 315385 ≈ 00195.

Of course, instead of using Exercises 12 and 13 to help compute these values, you could have computed each  (A − A ) directly by finding each matrix A − A (using your answers from part (b)), summing the squares of its entries, and then taking the square root. You would, of course, have obtained the same results. (d) The method described in the text for the compression of digital images takes the matrix describing the image and alters it by zeroing out some of the lower singular values. This exercise illustrates how the matrices A that use only the first  singular values for a matrix A get closer to approximating A as  increases. The matrix for a digital image is, of course, much larger than the 5 × 6 matrix considered in this exercise. Also, you can frequently get a very good approximation of the image using a small enough number of singular values so that less data needs to be saved. Using the outer product form of the Singular Value Decomposition, only the singular values and the relevant singular vectors are needed to construct each A , so not all  entries of A need to be kept in storage.

(15) These are the steps to process a digital image in MATLAB, as described in the textbook: — Enter the command: edit — Use the text editor to enter the following MATLAB program: function totalmat = RevisedPic(U, S, V, k) T = V’ totalmat = 0; for i = 1:k

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

310

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.5

totalmat = totalmat + U(:,i)*T(i,:)*S(i,i); end — Save this program under the name RevisedPic.m — Enter the command: A = imread(’picturefilename’) where “picturefilename” is the name of the file containing the picture, including its file extension (preferably .tif or .jpg).

— Enter the command: ndims(A) — If the response is “3”, then MATLAB is treating your picture as if it is in color, even if it appears to be black-and-white.

If this is the case, enter the command: B = A(:,:,1); (to use only the first color of the three used for color pictures) followed by the command: C = double(B); (to convert the integer format to decimal) Otherwise, just enter: C = double(A); — Enter the command: [U,S,V] = svd(C); (This computes the Singular Value Decomposition of C.) — Enter the command: W = RevisedPic(U,S,V,100); (Here, the number “100” represents the number of singular values you are using.

You may change this to any value you like.)

— Enter the command: R = uint8(round(W)); (This converts the decimal output to the correct integer format.)

— Enter the command: imwrite(R,’Revised100.tif’,’tif’) (This will write the revised picture out to a file named “Revised100.tif”.

Of course, you can use any file name you would like, or use the .jpg extension instead of .tif.)

— You may repeat the steps: W=RevisedPic(U,S,V,k); R = uint8(round(W)); imwrite(R,’Revisedk.tif’,’tif’) with different values for k to see how the picture gets more refined as more singular values are used.

— Output can be viewed using any appropriate software you may have on your computer for viewing such files.

(16) (a) False. First, if A is an  ×  matrix, AA is  ×  and A A is  × , so they can not be equal if  6= . Second, AA∙ might ¸not equal A A even for square matrices. You can check that they 1 2 are not equal for A = .

3 4

(b) True. The singular values of A are the nonnegative square roots of the eigenvalues of A A. These eigenvalues were proven to be nonnegative in the text at the beginning of this section. ∙ ¸ 1 2 (c) False. For a counterexample, consider A = , v = i, and w = j. Clearly, v · w = 0, but 3 4 (Av) · (Aw) = [1 3] · [2 4] = 14 6= 0. This does not contradict part (2) of Lemma 9.4 which states that (Av ) ⊥ (Av ), since that only applies to right singular vectors v and v (when  6= ).

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

311

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Section 9.5

(d) False. Only the first  left singular vectors are determined by the right singular vectors, where  = rank(A). The remaining left singular vectors can be chosen in any way so that {u1      u } is an orthonormal basis for R . There could be an infinite number of ways to make the choices for u+1      u , depending on the values of  and . (See Exercise 8.) (e) False. By part (5) of Theorem 9.5, {v+1      v } forms an orthonormal basis for ker(), not for (ker())⊥ . (f) True. We established this method for proving that matrices are equal at the beginning of the proof of Theorem 9.6. (g) False. See Exercises 4 and 8. However, the matrix Σ in the Singular Value Decomposition of a matrix A is uniquely determined by A. (h) True. The product VΣ U certainly equals A and has the correct form for a Singular Value Decomposition. Exercise 5 can then be used to show that this implies that it is a Singular Value Decomposition for A . (i) False. Because every matrix has a Singular Value Decomposition, every matrix also has a pseudoinverse, which is computed using the Singular Value Decomposition. (j) True. In part (c) of Exercise 10, we prove that A+ A is the matrix for the orthogonal projection onto (ker())⊥ with respect to the standard basis for R . But, for a nonsingular  ×  matrix A, (ker())⊥ = R . Thus, A+ A = I , and so A+ = A−1 .

A different approach to proving this can be made using Theorem 9.8 and the fact that for a nonsingular matrix A, Ax = b has the unique solution A−1 b (part (1) of Theorem 2.16), which must equal the least-squares solution. Hence A+ b can be shown to equal A−1 b for every vector b, proving that A+ = A−1 . (k) True. Only the first  right singular vectors are used, corresponding to the  nonzero singular values. The remaining  −  right singular vectors are irrelevant, since they all correspond to  +1 = · · · =  = 0.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

312

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Appendix B

Appendices Appendix B (1) (a)  is not a function because it is not defined for every element of its stated domain. In particular,  is not defined for   1. (c)  is not a function because two values are assigned to each  6= 1.

(e)  is not a function because it is not defined for every element of its stated domain. In particular,  is not defined at  = 2 , where  is an odd integer. (f)  is a function. Its range is the set of all prime numbers. The range is contained in the set of prime numbers because the definition of  defines an image under  to be a prime number. The range contains every prime number, because, if  is prime, () = , and so  is in the range. The image of 2 = (2) = 2. The pre-image of 2 = {0 1 2}, the set of all natural numbers less than or equal to 2.

(2) (a) The pre-image of {10 20 30} is the set of all integers whose absolute value is 5, 10, or 15, since 2 times the absolute value must be 10, 20, or 30. This is the set {−15 −10 −5 5 10 15}.

(c) The pre-image of the set of multiples of 4 is the set of all integers which, when doubled, have absolute values that are multiples of 4. This is the set of integers whose absolute value is a multiple of 2; that is, the set {    −8 −6 −4 −2 0 2 4 6 8   }. q √ ¢ q ¡ 5−1 ¢2 ¡ 3 1 2 2 = (3) ( ◦  )() = ( ()) =  5−1 3 + 2 = 4 4 16 (25 − 10 + 1) + 2 = 4 75 − 30 + 35. √ ¡√ ¢ ( ◦ )() =  (()) =  32 + 2 = 14 (5 32 + 2 − 1). µ∙ ¸¶ µ µ∙ ¸¶¶ µ∙ ¸∙ ¸¶ ∙ ¸ µ∙ ¸∙ ¸¶   3 −2  −4 4 3 −2  (4) ( ◦  ) =  = =   1 4  0 2 1 4  µ∙ ¸∙ ¸¶ ∙ ¸ ∙ ¸∙ ¸ −4 4 3 −2  −8 24  = = .

0 2 1 4  2 8  µ∙ ¸¶ µ µ∙ ¸¶¶ µ∙ ¸∙ ¸¶ ∙ ¸ µ∙ ¸∙ ¸¶   −4 4  3 −2 −4 4  ( ◦ ) =  = =   0 2  1 4 0 2  ¸∙ ¸ µ∙ ¸∙ ¸¶ ∙ ¸ ∙  3 −2 −4 4  −12 8 . = = −4 12  1 4 0 2  (8) The function  is not one-to-one because  ( + 1) =  ( + 3) = 1, which shows that two different domain elements have the same image. The function  is not onto because there is no pre-image for  . For  ≥ 3 the pre-image of P2 is P3 .

(10) The function  is one-to-one, because  (A1 ) =  (A2 ) =⇒ B−1 A1 B = B−1 A2 B −1 =⇒ B(B A1 B)B−1 = B(B−1 A2 B)B−1 =⇒ (BB−1 )A1 (BB−1 ) = (BB−1 )A2 (BB−1 ) =⇒ I A1 I = I A2 I =⇒ A1 = A2 .

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

313

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Appendix B

The function  is onto, because for any C ∈ M ,  (BCB−1 ) = B−1 (BCB−1 )B = C.

Also,  −1 (A) = BAB−1 .

(12) (a) False. Functions may have two different elements of the domain mapped to the same element of the codomain. For example,  : R → R given by  () = 2 is such a function, since  maps both 2 and −2 to 4. (b) True. Since each element of  is assigned to exactly one element of  ,  is a function by definition. However, a function  having this property is not onto, since not every element of  has a corresponding element of .

(c) False. Since  (5) =  (6),  is not one-to-one. Hence,  −1 does not exist. For example, the constant function  : R → R given by  () = 7 satisfies  (5) =  (6), but has no inverse, and so  −1 (5) is not defined. However, for any function  ,  −1 ({5}) is defined and represents the set of all pre-images of 5. If  is the constant function  : R → R given by  () = 7, then  −1 ({5}) equals the empty set, while if  : R → R is given by  () = 5, then  −1 ({5}) = R.

(d) False. For  to be onto, the codomain of  must equal the range of  , not the domain of  . For a specific counterexample, consider  : R →R given by  () = 0. Then the domain and codomain of  are both R; however,  is not onto because no nonzero element of the codomain R has a pre-image. (e) False. Instead,  is one-to-one if  (1 ) =  (2 ) implies 1 = 2 . Notice, by the way, that every function, by definition, has the property that 1 = 2 implies  (1 ) =  (2 ). Hence, any function that is not one-to-one (such as  : R → R given by  () = 0) provides a specific counterexample to the given statement. (f) False. Consider  : {0 1} → {0 1} given by  () = , and : {0 1 2} → {0 1} given by (0) = 0, and (1) = (2) = 1. Then simple computation shows that ( ◦  )() actually equals  (). Now, both  and  ◦  are easily seen to be one-to-one, but  is not one-to-one because (1) = (2).

(g) False. By Theorem B.2, in order for a function  to have an inverse, if must be both one-toone and onto. Hence, any function that is one-to-one but not onto (for example,  : R → R given by  () = arctan()) or that is onto but not one-to-one (for example, : R → R given by () = 3 − ) provides a counterexample.

(h) False. Instead, by Theorem B.4, ( ◦  )−1 =  −1 ◦  −1 . For a specific counterexample to the statement in the problem, consider  : {0 1 2} → {0 1 2} given by  (0) = 1,  (1) = 2 and  (2) = 0, and

: {0 1 2} → {0 1 2} given by (0) = 1, (1) = 0 and (2) = 2.

Then, it is easily seen that

 −1 : {0 1 2} → {0 1 2} is given by  −1 (0) = 2,  −1 (1) = 0 and  −1 (2) = 1, and  −1 : {0 1 2} → {0 1 2} is given by  −1 (0) = 1,  −1 (1) = 0 and  −1 (2) = 2.

Also,

 ◦  : {0 1 2} → {0 1 2} is given by ( ◦  )(0) = 0, ( ◦  )(1) = 2 and ( ◦  )(2) = 1.

Hence,

(◦ )−1 : {0 1 2} → {0 1 2} is given by (◦ )−1 (0) = 0, (◦ )−1 (1) = 2 and (◦ )−1 (2) = 1.

However,

 −1 ◦  −1 : {0 1 2} → {0 1 2} is given by ( −1 ◦  −1 )(0) = 2, ( −1 ◦  −1 )(1) = 1 and c 2016 Elsevier Ltd. All rights reserved.

Copyright °

314

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Appendix C

( −1 ◦  −1 )(2) = 0.

Therefore, ( ◦  )−1 6=  −1 ◦  −1 .

Appendix C (1) (a) (6 − 3) + (5 + 2) = (6 + 5) + (−3 + 2) = 11 −  ³ ´ ³ ´ (c) 4 (8 − 2) − (3 + ) = 4 (8 − 3) + (−2 − 1) = 4(5 − 3) = 20 − 12 ³ ´ ³ ´ (e) (5 + 3)(3 + 2) = (5)(3) − (3)(2) + (5)(2) + (3)(3)  = 9 + 19 ³ ´ ³ ´ (g) (7 − )(−2 − 3) = (7)(−2) − (−1)(−3) + (7)(−3) + (−1)(−2)  = −17 − 19 (i) 9 − 2 = 9 + 2

³ ´ ³ ´ (k) (6 + )(2 − 4) = (6)(2) − (1)(−4) + (6)(−4) + (1)(2)  = 16 − 22 = 16 + 22 p √ (m) |−2 + 7| = (−2)2 + (7)2 = 53 ¢2 ¡√ 2 2 + 2 = 2 + 2 .

(2) In each part we use the formula 1 = || 2 and the fact that, for  =  + , || = 2

3 1 (a) If  = 6 − 2, then  = 6 + 2. Also, || = 62 + (−2)2 = 40. Hence, 1 = 6+2 40 = 20 + 20 .

4 1 (c) If  = −4 + , then  = −4 − . Also, ||2 = (−4)2 + (−1)2 = 17. Hence, 1 = −4− 17 = − 17 − 17 .

(3) In all parts, let 1 = 1 + 1  and 2 = 2 + 2 . (a) Part (1): 1 + 2

= = = = =

(1 + 1 ) + (2 + 2 ) (1 + 2 ) + (1 + 2 ) (1 + 2 ) − (1 + 2 ) (1 − 1 ) + (2 − 2 ) 1 + 2 

Part (2): (1 2 ) = = = = = =

(1 + 1 )(2 + 2 ) (1 2 − 1 2 ) + (1 2 + 2 1 ) (1 2 − 1 2 ) − (1 2 + 2 1 ) (1 2 − (−1 )(−2 )) + (1 (−2 ) + 2 (−1 )) (1 − 1 )(2 − 2 ) 1 2 

(b) If 1 6= 0, then 11 exists. Hence, 11 (1 2 ) = 11 (0), implying 2 = 0.

(c) Part (4): 1 = 1 ⇐⇒ 1 + 1  = 1 − 1  ⇐⇒ 1  = −1  ⇐⇒ 21  = 0 ⇐⇒ 1 = 0 ⇐⇒ 1 is real.

Part (5): 1 = −1 ⇐⇒ 1 + 1  = −(1 − 1 ) ⇐⇒ 1 + 1  = −1 + 1  ⇐⇒ 1 = −1 ⇐⇒ 21 = 0 ⇐⇒ 1 = 0 ⇐⇒ 1 is pure imaginary.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

315

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Appendix D

2

(5) (a) False. Instead,  = || . For example, if  = 1 + , then  = 1 −  and  = (1 + )(1 − ) = 2, √ √ but || = 12 + 12 = 2. (b) False. Every real number  (such as  = 3) has the property that  = . (c) True. A pure imaginary number is of the form  = 0+ = , whose conjugate is  = 0− = −. (d) True. The additive inverse of  =  +  is − = − − . (e) False. The complex number 0 has no multiplicative ³ ´ inverse. However, every nonzero complex number  has the multiplicative inverse 1 =

1 ||2

.

Appendix D (1) In each part, let E represent the given matrix. (a) The operation (III): h2i ↔ h3i was performed on I3 to obtain E. The inverse operation is (III): h2i ↔ h3i. Since this row operation is its own inverse, the matrix E is its own inverse. (b) The operation (I): h2i ← −2 h2i was performed on I3 to obtain E. The inverse operation is ⎡ ⎤ 1 0 0 (I): h2i ← − 12 h2i. Hence, E−1 = ⎣ 0 − 12 0 ⎦, the matrix obtained by performing the inverse 0 0 1 row operation on I3 . (e) The operation (II): h3i ← −2 h4i + h3i was performed on I4 to obtain E. The inverse operation ⎡ ⎤ 1 0 0 0 ⎢ 0 1 0 0 ⎥ ⎥ is (II): h3i ← 2 h4i + h3i. Hence, E−1 = ⎢ ⎣ 0 0 1 2 ⎦, the matrix obtained by performing the 0 0 0 1 inverse row operation on I4 .

(2) In each part, let A represent the given matrix. We list the row operations needed to convert A to reduced row echelon form, the resulting matrix after each step, the inverse of the row operation used, and the elementary matrix corresponding to each inverse operation: (a) Row Operation h1i ← 14 h1i h2i ← −3 h1i + h2i h2i ← 4 h2i h1i ← − 94 h2i + h1i

Resultant Matrix " # 1 94 " " ∙

3

7

1 0

9 4 1 4

1

9 4

0

1

1 0 0 1

#

#

¸

Inverse Row Operation h1i ← 4 h1i h2i ← 3 h1i + h2i h2i ← 14 h2i h1i ← 94 h2i + h1i

Then A = F4 F3 F2 F1 I2 . c 2016 Elsevier Ltd. All rights reserved.

Copyright °

316

Elementary Matrix for Inverse Operation ∙ ¸ 4 0 F4 = 0 1 ∙ ¸ 1 0 F3 = 3 1 " # 1 0 F2 = 0 14 " # 1 94 F1 = 0 1

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Appendix D

(c) Row Operation

h1i ↔ h2i

h1i ← − 13 h1i

h4i ← −3 h1i + h4i

h2i ↔ h3i

h2i ← 16 h2i

h3i ← 15 h3i

h2i ← 53 h3i + h2i

h1i ← − 23 h4i + h1i

h2i ← 16 h4i + h2i

Resultant Matrix ⎡ ⎤ −3 0 0 −2 ⎢ 0 0 5 0 ⎥ ⎢ ⎥ ⎣ 0 6 −10 −1 ⎦ 3 0 0 3 ⎡ 2 ⎤ 1 0 0 3 ⎢ 0 0 5 0 ⎥ ⎢ ⎥ ⎣ 0 6 −10 −1 ⎦ ⎡

3 0 1 0

0 0

3

2 3

⎤

⎢ 0 0 5 0 ⎥ ⎢ ⎥ ⎣ 0 6 −10 −1 ⎦ 0 0 0 1 ⎡ 2 ⎤ 1 0 0 3 ⎢ 0 6 −10 −1 ⎥ ⎢ ⎥ ⎣ 0 0 5 0 ⎦ 0 0 0 1 ⎡ 2 ⎤ 1 0 0 3 ⎢ ⎥ ⎢ 0 1 − 53 − 16 ⎥ ⎢ ⎥ ⎣ 0 0 5 0 ⎦ 0 0 0 1 ⎡ 2 ⎤ 1 0 0 3 ⎢ ⎥ 5 ⎢ 0 1 − 3 − 16 ⎥ ⎢ ⎥ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ 2 ⎤ 1 0 0 3 ⎢ ⎥ ⎢ 0 1 0 − 16 ⎥ ⎢ ⎥ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ ⎥ ⎢ 0 1 0 − 16 ⎥ ⎢ ⎥ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ 0 1 0 0 ⎥ ⎢ ⎥ ⎣ 0 0 1 0 ⎦ 0 0 0 1

Then A = F9 F8 F7 F6 F5 F4 F3 F2 F1 I4 . c 2016 Elsevier Ltd. All rights reserved.

Copyright °

317

Inverse Row Operation h1i ↔ h2i

h1i ← −3 h1i

h4i ← 3 h1i + h4i

h2i ↔ h3i

h2i ← 6 h2i

h3i ← 5 h3i

h2i ← − 53 h3i + h2i

h1i ← 23 h4i + h1i

h2i ← − 16 h4i + h2i

Elementary Matrix for Inverse Operation ⎡ ⎤ 0 1 0 0 ⎢ 1 0 0 0 ⎥ ⎥ F9 = ⎢ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ −3 0 0 0 ⎢ 0 1 0 0 ⎥ ⎥ F8 = ⎢ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ 0 1 0 0 ⎥ ⎥ F7 = ⎢ ⎣ 0 0 1 0 ⎦ 3 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ 0 0 1 0 ⎥ ⎥ F6 = ⎢ ⎣ 0 1 0 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ 0 6 0 0 ⎥ ⎥ F5 = ⎢ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ 0 1 0 0 ⎥ ⎥ F4 = ⎢ ⎣ 0 0 5 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ ⎥ ⎢ 0 1 − 53 0 ⎥ F3 = ⎢ ⎥ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 23 ⎢ 0 1 0 0 ⎥ ⎥ F2 = ⎢ ⎣ 0 0 1 0 ⎦ ⎡

0 0 0 1 0 0

1

0

⎤

⎢ ⎥ ⎢ 0 1 0 − 16 ⎥ F1 = ⎢ ⎥ ⎣ 0 0 1 0 ⎦ 0 0 0 1

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 5th edition

Appendix D

(7) A is nonsingular if and only if rank(A) =  (by Theorem 2.15) if and only if A is row equivalent to I (by the definition of rank) if and only if A = E · · · E1 I = E · · · E1 for some elementary matrices E1      E (by Theorem D.3).

(10) (a) True. An elementary matrix is formed by performing a single row operation on I . Since I is square, the result of any row operation performed on it will also be square. (b) False. The equation B = EA is true only if B can be obtained from A using a single row operation — namely the row operation corresponding to the elementary matrix E. However, it may take several row operations to convert one matrix A to a row equivalent matrix B. For example, I2 and 5I2 are row equivalent, but it takes at least two row operations to convert I2 to 5I2 . −1 · · · E−1 (c) False. The inverse of E1 E2 · · · E is E−1 2 E1 , not E · · · E2 E1 . For example, if ∙ ¸ ∙ ¸ 2 0 1 0 and E2 = , then E1 E2 = 2I2 , and so (E1 E2 )−1 = 12 I2 . However, E1 = 0 1 0 2

E2 E1 = 2I2 .

(d) True. If A is nonsingular, then A−1 is also nonsingular. The given statement then follows from Corollary D.4. (e) True. The proof of Theorem D.2 shows that the elementary matrix corresponding to the reverse of a row operation  is the inverse of the elementary matrix corresponding to . The given statement follows from this fact and Theorem D.1.

c 2016 Elsevier Ltd. All rights reserved.

Copyright °

318