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CompleteSolutions Manual Linear Algebra A Modern Introduction FOURTH EDITION

Poole Trent University

David

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1 Contents 1 Vectors 3 1.1 The Geometry and Algebra of Vectors 3 1.2 Length and Angle: The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Exploration: Vectors and Geometry 25 1.3 Lines and Planes 27 Exploration: The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 1.4 Applications 44 Chapter Review 48 2 Systems of Linear Equations 53 2.1 Introduction to Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 2.2 Direct Methods for Solving Linear Systems 58 Exploration: Lies My Computer Told Me 75 Exploration: Partial Pivoting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Exploration: An Introduction to the Analysis of Algorithms . . . . . . . . . . . . . . . . . . . . . . 77 2.3 Spanning Sets and Linear Independence 79 2.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 2.5 Iterative Methods for Solving Linear Systems 112 Chapter Review 123 3 Matrices 129 3.1 Matrix Operations 129 3.2 Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 3.3 The Inverse of a Matrix 150 3.4 The LU Factorization 164 3.5 Subspaces, Basis, Dimension, and Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 3.6 Introduction to Linear Transformations 192 3.7 Applications 209 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 4 Eigenvalues and Eigenvectors 235 4.1 Introduction to Eigenvalues and Eigenvectors 235 4.2 Determinants 250 Exploration: Geometric Applications of Determinants 263 4.3 Eigenvalues and Eigenvectors of n × n Matrices 270 4.4 Similarity and Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 4.5 Iterative Methods for Computing Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 4.6 Applications and the Perron-Frobenius Theorem 326 Chapter Review 365

2 CONTENTS 5 Orthogonality 371 5.1 Orthogonality in Rn 371 5.2 Orthogonal Complements and Orthogonal Projections 379 5.3 The Gram-Schmidt Process and the QR Factorization 388 Exploration: The Modified QR Process 398 Exploration: Approximating Eigenvalues with the QR Algorithm 402 5.4 Orthogonal Diagonalization of Symmetric Matrices 405 5.5 Applications 417 Chapter Review 442 6 Vector Spaces 451 6.1 Vector Spaces and Subspaces 451 6.2 Linear Independence, Basis, and Dimension 463 Exploration: Magic Squares 477 6.3 Change of Basis 480 6.4 Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491 6.5 The Kernel and Range of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . 498 6.6 The Matrix of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 Exploration: Tiles, Lattices, and the Crystallographic Restriction 525 6.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527 Chapter Review 531 7 Distance and Approximation 537 7.1 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537 Exploration: Vectors and Matrices with Complex Entries 546 Exploration: Geometric Inequalities and Optimization Problems 553 7.2 Norms and Distance Functions 556 7.3 Least Squares Approximation 568 7.4 The Singular Value Decomposition 590 7.5 Applications 614 Chapter Review 625 8 Codes 633 8.1 Code Vectors 633 8.2 Error-Correcting Codes 637 8.3 Dual Codes 641 8.4 Linear Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647 8.5 The Minimum Distance of a Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 650

Chapter 1 Vectors

1.1 The Geometry and Algebra of Vectors

3 H-2, 3L 3 2 1 H2, 3L H3, 0L 1 2 3 H3, -

1. -2 -1 -1 2L -2 2. Since 2 + 3 3 0 5 = , 3 2 + 3 2 3 = 4 0 , 2 3 + 2 3 0 = 0 , 2 + 3 3 = 2 5 , 5 plotting those vectors gives 1 2 3 4 5 -1 c b -2 a -3 d -4 -5

In standard position, the vectors are

# » (a) AB = [4 1,2 ( 1)]= [3,3].

# » (b) AB = [2 0, 1 ( 2)]= [2,1]

4 CHAPTER 1. VECTORS 1.1. THE GEOMETRY AND ALGEBRA OF VECTORS 4 c d a 1 2 3 4 b 3 2 1 c d a b 1 1 y 3. 2 z c 1 b -2 -1 0 02 1 0 a 3 2 x -1 d -2

3 0 3 3 3 0 3 1 2 3 1 4 2 2 = 0 , 2 2 = 0 , 2 2 = 4 , 2 1 = 3 1 0 1 # » 1 1 0 1 1 0 1 2 3

4. Since the heads are all at (3, 2,1), the tails are at

3 2 1 -1 -2

5. The four vectors AB are

3 3 = 3 , 3 2 2 2 2

» (c) AB = 1 2,

1 1 , 1 1 = 1 , 1 6 3 2 3 6 6 -1 1 2 3

# » (d) AB =

6. Recall the notationthat [a, b] denotes a move of a units horizontally and b units vertically Then during the first part of the walk, the hiker walks 4 km north, so a = [0,4]. During the second part of the walk, the hiker walks a distance of 5 km northeast. From the components, we get √ √

Thus the net displacement vector is

GEOMETRY

ALGEBRA

5 5 CHAPTER

VECTORS

1.1. THE

AND

OF VECTORS

b = [5cos45◦ , 5sin 45◦] =

" 5 2 , 2 5 2 # . 2 √ √ 3 2 3 + 2 c = a + b = 5 " 5 2 2 3 , 4 + 5 2 # 2 7. a + b = 0 + 3 = 0 + 3 = 3 . 2 a + b 1 b a 8. b c = 2 3 2 3 = 2 ( 2) = 3 3 4 0 1 2 3 4 5 3 2 b -c 1 b - c 9. d c = 3 2 2 3 5 = . 5 1 2 3 4 1 2 3 4 5 -1 d -2 -3 d - c -c -4 10. a + d = 6 3 0 + 3 = 2 3 + 3 = 0 + ( 2) -5 a 1 2 3 4 5 6 -1 d a + d 2 -2 11. 2a + 3c = 2[0,2,0]+ 3[1, 2,1] = [2 0,2 2,2 0]+ [3 1,3 ( 2),3 1] = [3, 2,3].

3b 2c + d = 3[3,2,1] 2[1, 2,1]+ [ 1, 1, 2] = [3·3,3 ·2,3 ·1]+ [ 2 · 1, 2 · ( 2), 2 · 1]+ [ 1, 1, 2] = [6,9, 1].

12.

13. u =

so that

# »

14. (a) AB = b a # » # » # » # »

(b) Since OC = AB, we have BC = OC b = (b a) b = a

# » (c) AD = 2a

# » # » # » (d) CF = 2OC = 2AB = 2(b a) = 2(a b).

# » # » # »

(e) AC = AB + BC = (b a) + ( a)= b 2a # » # » # » # »

(f) Note that F A and OB are equal, and that DE = AB Then

# » # » # » # » # »

BC + DE + F A = a AB + OB = a (b a) + b = 0.

15. 2(a 3b) + 3(2b+ a) property e. distributivity = (2a 6b)+ (6b+ 3a) property b. associativity = (2a + 3a) + ( 6b + 6b) = 5a

16.

3(a c) + 2(a + 2b)+ 3(c b)

property e. distributivity = ( 3a + 3c) + (2a + 4b)+ (3c 3b) property b. associativity = ( 3a + 2a) + (4b 3b)+ (3c + 3c) = a + b+ 6c.

17. x a = 2(x 2a) = 2x 4a ⇒ x 2x = a 4a ⇒ x = 3a ⇒ x = 3a

18.

x + 2a b = 3(x + a) 2(2a b) = 3x + 3a 4a + 2b ⇒ x 3x = a 2a + 2b + b ⇒ 2x = 3a + 3b ⇒ 3 3 x = 2 a 2 b.

19. We have 2u+ 3v = 2[1, 1]+3[1,1] = [2·1+3·1,2·( 1)+ 3·1] = [5,1]. Plots of all three vectors are 2

VECTORS 1.1. THE GEOMETRY AND ALGEBRA OF VECTORS 6

CHAPTER 1.

h 1 , √ 3 i

210◦

◦

h √ 3 , 1

2 2 2 2 " 1 u + v = 2 √ 3 √ 3 , 2 2 1 # " 1 2 , u v = 2 + √ 3 √ 3 1 # , + 2 2 2

[cos60◦ ,sin 60◦] =

, and v = [cos

,sin 210

] =

1 w v v -1 1 2 3 4 5 6 v u -1 v u -2

20. We have u 2v = [ 2,1] 2[2, 2] = [ ( 2) 2 2, 1 2 ( 2)] = [ 2,3]. Plots of all three vectors are

21. From the diagram, we see that w = 6 2u + 4v.

22. From the diagram, we see that w = 2u+ 3v

23. Property (d) states that u+ ( u) = 0. The first diagram below shows u along with u. Then, as the diagonal of the parallelogram, the resultant vector is 0

Property (e) states that c(u + v) = cu + cv The second figure illustrates this.

ALGEBRA

7 7 CHAPTER

-u -v 3 -u 2 w -v 1 u - 1 2 v 9 8 7 6 5 4 3 2 u 1 u w u v v v - 2 -1 1 2 3 4 5 6 7

1.1. THE GEOMETRY AND

OF VECTORS

1. VECTORS

2 -1 -1 -2

-u 5 -u 4 w v 3 v 2 v 1 v -1 1 2 3 4 5 6 u -1

24. Let u = [u1,u2, ...,un] and v = [v1,v2, ...,vn], and let c and d be scalars in R.

Property (d):

Property (e):

u + ( u) = [u1,u2, ,un] + ( 1[u1,u2, ,un])

= [u1,u2, ,un] + [ u1, u2, , un]

= [u1 u1, u2 u2, ...,un un]

= [0,0,...,0] = 0.

c(u+ v) = c([u1, u2, ...,un] + [v1,v2, ...,vn])

= c([u1 + v1, u2 + v2, ,un + vn])

= [c(u1 + v1), c(u2 + v2), ,c(un + vn)]

= [cu1 + cv1, cu2 + cv2, ,cun + cvn]

= [cu1,cu2, ,cun] + [cv1,cv2, ,cvn]

= c[u1,u2, ,un] + c[v1,v2, ,vn]

= cu+ cv.

Property (f):

Property (g):

(c + d)u = (c + d)[u1, u2, ...,un]

= [(c + d)u1, (c + d)u2, ...,(c + d)un]

= [cu1 + du1, cu2 + du2, ...,cun + dun]

= [cu1,cu2, ,cun] + [du1,du2, ,dun]

= c[u1,u2, ,un] + d[u1,u2, ,un]

= cu + du.

c(du)= c(d[u1,u2, ,un])

= c[du1,du2, ,dun]

= [cdu1,cdu2, ,cdun]

= [(cd)u1, (cd)u2, ...,(cd)un]

= (cd)[u1, u2, ...,un]

= (cd)u

25. u + v = [0,1]+ [1,1] = [1,0].

26. u + v = [1,1,0]+ [1,1,1] = [0,0,1].

8 CHAPTER 1. VECTORS 1.1. THE GEOMETRY AND ALGEBRA OF VECTORS 8 cv u cu v c

u + vL c

u u u + v u cv -u v

27. u + v = [1,0,1,1]+ [1,1,1,1] = [0,1,0,0].

28. u + v = [1,1,0,1,0]+ [0,1,1,1,0] = [1,0,1,0,0].

29.

30. 31. 2 + 2 + 2 = 6 = 0 in Z3.

32. 2 ·2 ·2 = 3 ·2 = 0 in Z3

33. 2(2 + 1 + 2) = 2 2 = 3 1 + 1 = 1 in Z3

34. 3 + 1 + 2 + 3 = 4 ·2 + 1 = 1 in Z4

35. 2 ·3 ·2 = 4 ·3 + 0 = 0 in Z4.

36. 3(3 + 3 + 2) = 4 ·6 + 0 = 0 in Z4

37. 2 + 1 + 2 + 2 + 1 = 2 in Z3, 2 + 1 + 2 + 2 + 1 = 0 in Z4, 2 + 1 + 2 + 2 + 1 = 3 in Z5

38. (3 + 4)(3 + 2 + 4 + 2) = 2 ·1 = 2 in Z5

39. 8(6 + 4 + 3) = 8 ·4 = 5 in Z9

40. 2100 = 210 10 = (1024)10 = 110 = 1 in Z

41. [2,1,2]+ [2,0,1] = [1,1,0] in Z3

42. 2[2, 2, 1] = [2·2, 2 ·2, 2 ·1] = [1, 1, 2] in Z3

43. 2([3,1,1,2]+ [3,3,2,1]) = 2[2,0,3,3] = [2·2, 2 ·0, 2 ·3, 2 ·3] = [0,0,2,2] in Z4 . 3] = [2,3,1,1] in Z4 5

44. x = 2 + ( 3)= 2 + 2 = 4 in Z5.

45. x = 1 + ( 5)= 1 + 1 = 2 in Z6

46. x = 2 1 = 2 in Z3

47. No solution. 2 times anything is always even, so cannot leave a remainder of 1 when divided by 4.

48. x = 2 1 = 3 in Z5

49. x = 3 14 = 2 4 = 3 in Z5

50. No solution. 3 times anything is always a multiple of 3, so it cannot leave a remainder of 4 when divided by 6 (which is also a multiple of 3).

51. No solution. 6 times anything is always even, so it cannot leave an odd number as a remainder when divided by 8.

1.1.

GEOMETRY AND ALGEBRA

9 9 CHAPTER

VECTORS + 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2 · 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 + 0 1 2 3 4 0 0 1 2 3 4 1 1 2 3 4 0 2 2 3 4 0 1 3 3 4 0 1 2 4 4 0 1 2 3 · 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1 3 3 4 3, 2 · 11

THE

OF VECTORS

52. x = 8 19 = 7 9 = 8 in Z11

53. x = 2 1(2 + ( 3))= 3(2 + 2) = 2 in Z5

54. No solution. This equation is the same as 4x = 2 5 = 3 = 3 in Z6 But 4 times anything is even, so it cannot leave a remainder of 3 when divided by 6 (which is also even).

55. Add 5 to both sides to get 6x = 6, so that

= 1 or x = 5 (since 6 ·1 = 6 and 6 ·5 = 30 = 6 in Z8).

56. (a) All values. (b) All values. (c) All values.

57. (a) All a = 0 in Z5 have a solution because 5 is a prime number.

(b) a = 1 and a = 5 because they have no common factors with 6 other than 1.

1.2 Length and Angle: The Dot Product

1. Following Example 1.15, u

2. Following Example 1.15, u

7. Finding a unit vector v in the same direction as a given vector u is called normalizing the vector u Proceed as in Example 1.19:

so a unit vector v in the same direction as

VECTORS 1.2. LENGTH AND ANGLE: THE DOT PRODUCT 10 2

CHAPTER 1.

· v =

· v = 1 2 3 2 · 3 1 4 6 = ( 1)· 3 + 2 ·1 = 3 + 2 = 1. = 3 ·4 + ( 2)· 6 = 12 12 = 0. 1 2 3. u · v = 2 · 3 = 1 ·2 + 2 ·3 + 3 ·1 = 2 + 6 + 3 = 11. 3 1 4. u · v = 3.2 ·1.5 + ( 0.6)·4.1 + ( 1.4)· ( 0.2)= 4.8 2.46 + 0.28 = 2.62. 1 4 √ √ 5. u · v = 2 · 2 = 1 ·4 + √ 2· ( √ 2) + √ 3· 0 + 0 ·( 5)= 4 2 = 2. √ 3 0 0 5 1 12 2.29 6. u · v = 3.25 · 1.72 = 1 12 ·2.29 3.25 ·1.72 + 2.07 ·4.33 1.83 ·( 1 54) = 3 6265. 2.07 1 83 4.33 1 54

kuk= p ( 1)2 + 22 = √ 5,

1 v = 1 " 1 # u = √ " 1 # = √ 5 kuk 5 2 √ 5

8. Proceed as in Example 1.19:

so a unit vector v in the direction of

CHAPTER 1. VECTORS 1.2. LENGTH AND ANGLE: THE DOT PRODUCT 11

kuk= p 32 + ( 2)2 = √ 9+ 4 = √ 13,

1 v = kuk 1 u = √ 13 " 3 # = 2 " 3 # √ 13 . 2√ 13

9. Proceed as in Example

10.

11. Proceed as in Example

12. Proceed as in Example

13. Following Example 1.20, we compute:

1.2. LENGTH AND ANGLE: THE DOT PRODUCT 11 11 CHAPTER 1. VECTORS 6 3 3 1 2

1.19: kuk= p 12 + 22 + 32 = √ 14, so a unit vector v in the direction of u is 1 1 1 1 √ 14 v = u = √ 2 = 2 kuk 14 √ 14 3 3√ 14

1.19: kuk= p 3 22 + ( 0.6)2 + ( 1.4)2 = √ 10.24 + 0.36 + 1.96 = √ 12.56≈ 3 544, so a unit vector v in the direction of u is 1 1 1 5 0.903 v = u = 0.4 ≈ 0 169

Proceed as in Example

1.19: kuk r 3 544 2 2.1 √ 2 0.395 √ kuk = 12 + √ 2 + 3 + 02 = 6, so a unit vector v in the direction of u is 1 1 √ 6 1 √ 6 √ 6 6 1 1 √ 2 √ 2 1 √ 3 √ √ 3 3 v = u = √ √ = √6 = = √ kuk √ √ 6 0 0 2 0 2 0

1.19: kuk= p 1 122 + ( 3 25)2 + 2.072 + ( 1 83)2 = √ 1.2544 + 10.5625 + 4.2849 + 3.3489 = √ 19.4507 ≈ 4 410, so a unit vector v in the direction of u is 1 v u = kuk 1 4 410 1.12 3.25 2.07 1.83 ≈ 0.254 0.737 0.469 0.415 .

u v = 1 2 q 3 1 2 = 4 1 , so d(u,v) = ku vk = 3 ( 4) 4 + 12 = √ 17 1

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