A Concise Introduction to Linear Algebra, First Edition Géza Schay SOLUTION MANUAL

SOLUTION MANUAL

Instructors Solution Manual for Introduction to Linear Algebra

Geza Schay and Dennis Wortman

1.1.2. r = (5, 5), p − q = (−3, 1), q − p = (3, −1).

1.1.4. The midpoint of the edge from P (0, 1, 1) to Q(1, 1, 1) has position vector m = (p + q)/2 = (1/2, 1, 1). The center of the face through (0, 0, 1), (0, 1, 1), (1, 1, 1) and (1, 0, 1) has position vector f = [(0, 0, 1) + (1, 1, 1)]/2 = (1/2, 1/2, 1). Similarly, the center of the whole cube has position vector c = [(0, 0, 0) + (1, 1, 1)]/2 = (1/2, 1/2, 1/2).

1.1.6. Let p = (p1 , p2 ) and q = (q1 , q2 ).

Proof of Property 4: cp = 0 is equivalent to (cp1 , cp2 ) = (0, 0), that is, to cp1 = 0 and cp2 = 0.

Now, these last two relations hold if and only if either c = 0 with arbitrary p1 and p2 or c = 0 but p1 = p2 = 0, that is, p = 0.

Proof of Property 5: (−c)p = (−cp1 , −cp2 ) = c(−p1 , −p2 ) = c(−p) and (−c)p = (−cp1 , −cp2 ) = −(cp1 , cp2 ) = −(cp).

Proof of Property 6: c(p − q) = c(p1 − q1 , p2 − q2 ) = (cp1 − cq1 , cp2 − cq2 ) = (cp1 , cp2 ) − (cq1 , cq2 ) = cp − cq.

Proof of Property 7: (c − d)p = ((c − d)p1 , (c − d)p2 ) = (cp1 , cp2 ) − (dp1 , dp2 ) = c(p1 , p2 ) − d(p1 , p2 ) = cp − dp. −−→ 1.1.8. Let p = OO . Then for each i, r i = ri −p. If r denotes the position vector of the center of mass relative to O , then

r

n n n 1 1 1 = r−p = mi ri −p = mi ri − mi p M i=1 M i=1 M i=1

n n 1 1 = mi (ri −p) = mi r i .

M i=1 M i=1

1.1.10. By the solution of Exercise 1.1.9, the centroid P of the face formed by B, C, and D has position vector p = 13 (b + c + d). The position vector of the point that is one fourth of the way from P to A is 1 3 1 1 1 1 p + (a − p ) = p + a = (b + c + d) + a = (a + b + c + d) = p.

4 4 4 4 4 4 1.2.2. If p = (2, −2, 1) and q = (2, 3, 2), then p + q = (4, 1, 3) and 2

p − q = (0, −5, −1). Thus |p + q|2 = |p − q|2 = 26, |p|2 = 9, |q|2 = 17.

Hence |p + q|2 = |p|2 + |q|2 . The vectors p and q span a parallelogram with diagonals of equal length, that is, they span a rectangle. Also, the equation |p + q|2 = |p|2 + |q|2 expresses the Pythagorean Theorem for the triangle with edges p, q and p + q, which shows again that p and q are perpendicular to each other.

1.2.4. cos θ =

26 p·q √ = −0.997; θ ≈ 3.065 radians. = −√ |p||q| 20 · 34

1.2.6. Between two space diagonals, p = (1, 1, 1) and q = (1, 1, −1): cos θ =

p·q 1+1−1 1 = √ √ = , θ ≈ 70.5 ◦ . |p||q| 3 3· 3

Between two face diagonals, p = (1, 0, −1) and q = (0, 1, −1): cos θ =

p·q 0+0+1 1 = √ √ = , θ = 60 ◦ . |p||q| 2 2· 2

Between a space diagonal and an edge, p = (1, 1, 1) and q = (1, 0, 0): cos θ =

p·q 1+0+0 1 = √ √ = √ , θ ≈ 54.7 ◦ . |p||q| 3· 1 3

Between a space diagonal and a face diagonal, p = (1, 1, 1) and q = (1, 1, 0): cos θ =

p·q 1+1+0 2 = √ √ = √ , θ ≈ 35.3 ◦ . |p||q| 3· 2 6

The other types of angles are obvious: 45 ◦ or 90 ◦ . −→ −−→ 1.2.8. a) Let p = AB = (1, 1) and q = BC = (2, −5). The projection of p onto q is then

p·q 2−5 6 15 p1 = q= (2, −5) = − , . |q|2 4 + 25 29 29 3

b) If D denotes the foot of the projection of p onto q, then

−−→ 6 15 35 14 AD = p − p1 = (1, 1) − − , = , , 29 29 29 29 and

−−→ 1√ 2 7 dist A from BC = |AD| = 35 + 142 = √ .

29 29

c)

1 −−→ −−→ 1√ 2 7 7 Area of ∆ = |BC||AD| = 2 + 52 · √ = .

2 2 2 29 1.2.10. |p + q|2 = (p + q) · (p + q) = p2 + 2p · q + q2 , and |p − q|2 = (p − q) · (p − q) = p2 − 2p · q + q2 .

Thus |p + q|2 + |p − q|2 = 2p2 + 2q2 = 2|p|2 + 2|q|2 .

Geometrically, this result says that the sum of the squares of the lengths of the diagonals of a parallelogram equals the sum of the squares of the lengths of the four sides.

1.2.12.The result is trivially true if q = 0, so we may assume that q = 0.

Following the hint, let f(λ) = |p − λq|2 = |p|2 − (2p · q)λ + λ2 |q|2 for any scalar λ. Now, f is a quadratic function of λ and f (λ) ≥ 0 for all λ. Hence the graph of f is a parabola opening upwards and the minimum value will occur p·q at the vertex, where f (λ) = 2|q|2 λ − 2(p · q) = 0. Thus λ0 = |q| 2 yields p·q 2 the minimum value of f , which is f (λ0 ) = |p| − |q|2 . Cauchy’s inequality follows from the fact that f (λ0 ) ≥ 0. Finally, equality occurs if and only if q = 0 or |p − λq| = 0, that is, if and only if q = 0 or p = λq, for any λ, and thus, if and only if p and q are parallel.

1.2.14. a. Substituting r = p − q into the Triangle Inequality, |r + q| ≤ |r| + |q| yields the inequality |p| − |q| ≤ |p − q|. Similarly, |q| − |p| ≤ |q − p| = |p − q|. Combining these two inequalities yields the desired result.

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b. Equality occurs when the vectors p and q are parallel and point in the same direction; that is, p = λq, with λ ≥ 0, or q = λp, with λ ≥ 0. This follows from the fact that equality occurs in the Triangle Inequality if and only if the vectors are parallel and point in the same direction.

1.2.16. a. From Figure 1.12 we see that p3 cos α3 = . |p| On the other hand, up · k =

p·k (p1 , p2 , p3 ) · (0, 0, 1) p3 = = . |p| |p| |p|

and so uP · k = cos α3 . The other two relations can be obtained similarly. b.. By expressing each of the cosines as in the first equation above, we obtain cos2 α1 + cos2 α2 + cos2 α3 =

p21 p22 p23 p21 + p22 + p23 + + = = 1. |p|2 |p|2 |p|2 |p|2

This is the three-dimensional analog of the formula sin2 α + cos2 α = 1. c. p = (p1 , p2 , p3 ) = (|p| cos α1 , |p| cos α2 , |p| cos α3 ) = |p|(cos α1 , cos α2 , cos α3 ).

1.2.18. The angle between ea = (cos α, sin α ) and eb = (cos β, sin β) is α − β and thus ea · eb = |ea ||eb | cos(α − β) = cos(α − β) Also ea · eb = cos α cos β + sin α sin β.

The result follows from combining the two equations.

1.2.20. a. Let α denote the angle between the axes labelled ξ and η. Then p1 = p1 + p2 cos α and p2 = p2 + p1 cos α, from which it follows that p1 =

p1 − p2 cos α , 1 − cos2 α 5

and p2 − p1 cos α .

1 − cos2 α If uξ and uη denote the unit vectors in the directions of the axes, then p = p1 uξ + p2 uη and q = q 1 uξ + q 2 uη and thus p2 =

p · q = p1 q 1 + p2 q 2 + p1 q 2 cos α + p2 q 1 cos α.

Substituting the above expressions for p1 and p2 into the last equation and simplifying yields p · q = p1 q 1 + p2 q 2 By a symmetric argument, it also follows that p · q = p1 q1 + p2 q2 . b. From the first expression for p · q above we can read off g11 = g22 = 1 and g12 = g21 = cos α.

1.3.2. The scalar parametric form is x = 1, y = −2 + t, z = 4 and the nonparametric equations are: x = 1 and z = 4 (y is arbitrary). −−→ 1.3.4. v = P0 P1 = (0, 8, −7), p = p0 +tv = (1, −2, 4) + t(0, 8, −7). In scalar form: x = 1, y = −2 + 8t, z = 4 − 7t. The nonparametric equations = z−4 . are: x = 1 and y+2 8 −7 1.3.6. v = (3, −4, 3) and the vector parametric form of the equation of the line is p = p0 + tv = (5, 4, −8) + t(3, −4, 3) and thus the scalar parametric form is x = 5 + 3t, y = 4 − 4t, z = −8 + 3t, and the nonparametric form is x−5 = y−4 = z+8 .

3 −4 3 1.3.8. The direction vector n of the desired line is orthogonal to the normal vectors u = (3, −4, 3) and v = (−1, 3, 4) of the two planes. Proceeding as in Example 1.3.7 in the text, we can express the equation of the plane through the origin determined by the vectors u and v in the form p = su+tv, where s and t are parameters, or, in scalar form, as x = 3s − t, y = −4s + 3t, z = 3s + 4t. Eliminating the parameters results in the equation 5x + 3y − z = 0, and thus n = (5, 3, −1). Therefore, the vector parametric form of the equation of the line is p = p0 + tn = (1, −2, 4) + t(5, 3, −1), and thus the scalar parametric form is x = 1 + 5t, y = −2 + 3t, z = 4 − t, and the nonparametric form is x−1 = y+2 = z−4 5 3 −1 1.3.10. The equation of the line can be written as p = b + t(a − b), with p0 = b and v = a − b. This simplifies to p = ta + (1 − t)b.

For any point P between A and B, the parameter t is between 0 and 1 and means the fraction dist(P, B)/dist(A, B), since dist(P, B) = |p − b| = 6

|t(a − b)| = tdist(A, B). Similarly, 1 − t means the fraction dist(P, A)/dist(A, B).

If P is outside the interval [A, B] on the side of A, then t is greater than 1 and is still dist(P, B)/dist(A, B). On the other hand, if P is outside the interval [A, B] on the side of B, then t is less than 0 and equals −dist(P, B)/dist(A, B).

1.3.12. Both points P0 (1, −2, 4) and P1 (3, −2, 1) must lie in the required plane, and so we may choose u = p1 −p0 = (2, 0, −3). For v we may choose the same v that we have for the given line v = (2, 1, −3), and for the fixed point of the plane we may choose either P0 or P1 . Thus a parametric vector equation can be written as p = (1, −2, 4) + s(2, 0, −3) + t(2, 1, −3) or as p = (3, −2, 1) + s(2, 0, −3) + t(2, 1, −3).

The nonparametric equation can be obtained by eliminating s and t from x = 1 + 2s + 2t, y = −2 + t, z = 4 − 3s − 3t. This elimination results in 3x + 2z = 11.

1.3.14. In this case it is easier to start with the nonparametric form.

Since the required plane is to be orthogonal to the line given by p = (3, −2, 1) + t(2, 1, −3), we may take its normal vector to be the direction vector of the line, that is, let n = (2, 1, −3). We may take p0 = 0, and so n · (p − p0 ) = 0 becomes (2, 1, −3) · (x, y, z) = 0, that is, 2x + y − 3z = 0.

To write parametric equations, we may choose any two nonparallel vectors (x, y, z) whose components satisfy the above equation to be u and v.

For instance, we may take u = (1, 1, 1) and v = (0, 3, 1), and obtain p = s(1, 1, 1) + t(0, 3, 1) as a parametric vector equation of the plane.

1.3.16. In this case it is easier to start with the nonparametric form.

Since the normal vectors of parallel planes must be parallel, we may take the n of the two planes to be the same, and read it off the given equation as n = (3, −4, 3). Because P0 (1, −2, 4) is to be a point of the required plane, we may take p0 = (1, −2, 4), and so n · (p − p0 ) = 0 becomes (3, −4, 3)(x − 1, y + 2, z − 4) = 0, that is, 3x − 4y + 3z = 23.

To write parametric equations, choose any two distinct vectors a and b, other than p0 , whose components satisfy the above equation, and let u = a − p0 and v = b − p0 , because such vectors a and b represent points of the plane and so u and v are then nonzero vectors lying in the plane. For instance, take a = (9, 1, 0) and b = (0, 1, 9). Then u = (9, 1, 0) − (1, −2, 4) = 7

(8, 3, −4) and v = (0, 1, 9) − (1, −2, 4) = (−1, 3, 5). Thus we obtain p = (1, −2, 4) + s(8, 3, −4) + t(−1, 3, 5) as a parametric vector equation of the plane. (As a check, notice that s = 1, t = 0 gives p = a, and s = 0, t = 1 gives p = b.)

1.3.18. The vectors u = p1 −0 = (1, 6, −3) and v = p2 −0 = (7, −2, 5) are two nonparallel vectors lying in the required plane. Since a vector parametric equation of the plane is p = s(1, 6, −3) + t(7, −2, 5), the corresponding scalar form of the equation is x = s + 7t, y = 6s − 2t, z = −3s + 5t. Eliminating the parameters results in the nonparametric equation 12x − 13y − 22z = 0.

1.3.20. We can proceed as in Example 1.3.3 in the text to decompose the vector equation (5, 1, 1) + s(−2, 1, 6) = (3, −2, 1) + t(2, 1, −3) into three scalar equations: 5 − 2s = 3 + 2t, 1 + s = −2 + t, 1 + 6s = 1 − 3t Solving this system yields s = −1 and t = 2. Hence the lines intersect at the point (7, 0, −5).

1.3.22. In terms of scalar components, the equation of the given line is x = 5 − 2s, y = 1 + s, z = 1 + 6s.

Substituting these equations into that of the given plane results in 7(5 − 2s) + (1+s)+2(1+6s) = 8; the solution is s = 30. Hence the point of intersection is (−55, 31, 181).

1.3.24. Rewriting the equation of the given line in component form and replacing s by r yield x = 3 − 3r, y = −2 + 5r, z = 6 + 7r and rewriting the equation of the given plane in component form gives x = 4 − 2s + t, y = −2 + s + 3t, z = 1 + 3s + 2t.

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Combining those sets of equations and simplifying result in the system 3r − 2s + t = −1 5r − s − 3t = 0 7r − 3s − 2t = −5 The solution of this system is r = 4 (and s = 59/7 and t = 27/7), so the point of intersection is (−9, 18, 34). −−→ 1.3.26. Pick a point P on the plane, say, P (1, 0, 0). Then P P0 = (0, 2, −4).

From the given equation, n = (3, 2, −2). Thus, as in Example 1.3.6, −−→ |P P0 · n| |0 · 3 + 2 · 2 + (−4) · (−2)| 12 √ = D= =√ . |n| 9+4+4 17 1.3.28. First find the plane through 0 that is parallel to the direction vectors of the given lines, as in Example 1.3.6: p = s(−3, 5, 7) + t(−2, 1, 6), or, in scalar parametric form, x = −3s − 2t, y = 5s + t, z = 7s + 6t.

Eliminating the parameters yields the equation of the plane in scalar form: 23x + 4y + 7z = 0, from which we obtain a normal vector, n = (23, 4, 7), to the plane, and hence also to the two given lines.

The point P (3, −2, 6) lies on the first line and Q(5, 1, 1) lies on the second −→ one. Thus P Q = (2, 3, −5), and as in Example 1.3.6, −→ |P Q · n| |2 · 23 + 3 · 4 + (−5) · 7| 23 √ = =√ .

D= |n| 529 + 16 + 49 594 1.3.30. Following the hint, let Q be the point (3, −2, 6) on the line L , so −−→ 5, 7), and thus the QP0 = (0, 6, −6). The direction of the line L is u = (−3, −−→ −−→ u u component of QP0 parallel to L is given by QP0 · |u| |u| = 12 (−3, 5, 7).

83 −−→ Then the component of QP0 orthogonal to L is (0, −6, 6) −

12 1 (−3, 5, 7) = (36, −558, 414), 83 83 9

and therefore the distance from P0 to P is

1

d = (36, −558, 414) ≈ 8.38.

83

1.3.32. The vector n is orthogonal to the line containing P0 and P , since −−→ P P0 = p0 − p and so Equation 1.83 is an equation of the line through the point P0 that is orthogonal to n. −−→ 1.3.34. Applying the result of Exercise 1.3.33, with q = OP0 = (1, −2, 4), √ n = (3,2,−2) , p = (1, 0, 0) and d = n · p = √317 , we have f (q) = n · q − d = 17 − √1217 . Hence D = √1217 . −→ −→ 1.3.36. Let q = OQ. Then P Q = q − p = q − p0 − tv, and so −→ 2 −→ −→

P Q = P Q · P Q = [(q − p0 ) − tv] · [(q − p0 ) − tv] = (q − p0 )2 − 2t(q − p0 ) · v + t2 v2

2 (q − p0 ) · v ((q − p0 ) · v)2 2 2 = v t− + (q − p0 ) − v2 v2

,

−→ −→ and P Q · v = (q − p0 ) · v − tv2 . Thus 2 P Q is orthogonal −→ to v if and only if − →

(q − p0 ) · v − tv2 = 0, and also P Q (and hence P Q , too) is minimized −→ 0 )·v = 0. Hence P Q is orthogonal to v if and only if the if and only if t − (q−p v2 distance between P and Q is minimized.

2.1.2. The same row operations as in Exercise 2.1.1 give

    2 2 − 3 0 2 2 −3 0  1 5 2 0  →  0 4 7/2 0  .

−4 0 6 0 0 0 − 7/2 0

The last row corresponds to the equation − 72 x3 = 0, so x3 = 0. The second row corresponds to the equation 4x2 + 72 x3 = 0 and thus x2 = − 78 ·0 = 0. Finally, the first row corresponds to 2x1 + 2x2 − 3x3 = 0, from which we obtain x1 = −x2 + 32 x3 = 0. Thus, in vector form the solution is x = 0.

2.1.4. This equation constitutes a system, which is already in echelon form. The variables x2 and x3 are free, and so we set x2 = s and x3 = t.

10

Hence 2x1 + 2s − 3t = 0, and x1 = −s + 32 t. Thus, in vector form the solution is     −1 3/2 x =  1  s +  0  t.

0 1 2.1.6. 

1  −2 −6

  − 1 1 1

  −1 0 0 → 0 −2 0  1 →  0 0

0 3 6

0 3 6 0 3 0

 − 1 1 −3 2  −6 4

 − 1 1 −3 2  .

0 0

Column 3 is free. Thus we set x3 = t. Then the second row of the reduced matrix gives 3x2 − 3t = 2 and so x2 = t + 23 , and from the first row we get x1 = t + 1. Hence     1 1 x = t  1  +  2/3  .

1 0 2.1.8.



3  −1 6 

−1  0 → 0

−6 2 −8

−1 2 −3 2 0 4

2 5 9

  1 12 −1

  3 1 r1 ↔ r2 3 −2 9 6

  3 1 −1

  10 15 r2 ↔ r3 0 16 15 0

2 −6 −8

2 −1 −3

2 4 0

2 9 5

 3 1 1 12  −2 9

 3 1 16 15  .

10 15

The last matrix is in echelon form and the forward elimination is finished.

The fourth column has no pivot and so x4 is free and we set x4 = t. Then the last row corresponds to the equation 5x3 +10t = 15, which gives x3 = 3−2t.

The second row yields 4x2 + 9 (3 − 2t) + 16t = 15, whence x2 = 12 t − 3.

Finally, the first row gives x1 = −1 + 2 12 t − 3 + 2 (3 − 2t) + 3t = −1. In 11

vector form the solution is 

2.1.10.  2 4  0 1   3 3 1 2  1 2  0 1   0 −3 0 0

   −1 0  −3   1/2     x =  3  +  −2  t.

0 1

  1 7 1

  3 7  0 r ↔ r4   3 − 1 9  1 2 3 11

  3 11 1

  3 7  0 →   0 − 10 −24  −5 −15 0  1  0 →   0 0

2 1 3 4 2 1 0 0 2 1 0 0

 3 11 3 7  → − 1 9  1 7

 3 11 3 7   − 1 −3  −5 −15

 3 11 3 7  . − 1 −3  0 0

Thus x3 = 3, x2 = −2, and x1 = 6 and in vector form,   6 x =  −2  .

3 2.1.12.



3  −1 4 

−6 2 −8

−1  0 → 0

−1 2 −3 2 0 0

2 5 5

  1 7 −1

  3 1 r1 ↔ r2 3 −2 6 4

  3 1 −1

  10 10 → 0 10 10 0

2 −6 −8

2 −1 −3

2 0 0

2 5 0

 3 1 1 7  −2 6

 3 1 10 10  .

0 0

The last matrix is in echelon form and the forward elimination is finished.

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The second and fourth columns have no pivot and so we set x2 = s and x4 = t. Then the second row yields 5x3 + 10t = 10, whence x3 = 2 − 2t.

Finally, the first row gives x1 = −1 + 2s + 2 (2 − 2t) + 3t = 2s − t + 3. Thus, in vector form the solution is       3 2 −1  0   1   0       x =  2  +  0  s +  −2  t.

0 0 1

2.1.14. In the second step of the row-reduction, two elementary row operations are performed simultaneously so that r2 and r3 are both replaced by essentially the same expression, apart from a minus sign. Geometrically this action means that the line of intersection of two planes is replaced by a plane containing that line, rather than by two new planes intersecting in the same line. Indeed, if we substitute s = 14 − 2t in the wrong solution, we 5 obtain the correct one. Algebraically the mistake is that the second step is not reversible, it enlarges the solution set, and so the new augmented matrix is not row-equivalent to the original one.

2.1.16. The three planes have equations x + 2y = 2 3x + 6y − z = 8 x + 2y + z = 4

(0.1) (0.2) (0.3)

The line of intersection of planes (0.1) and (0.2) is p = (2, 0, −2) + t(−2, 1, 0), that of planes (0.1) and (0.3) is p = (2, 0, 2) + t(−2, 1, 0), and that of planes (0.2) and (0.3) is p = (3, 0, 1) + t(−2, 1, 0).

Hence all three lines of intersection of pairs of the planes have the same direction vector, (−2, 1, 0), and thus are parallel.

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2.2.2. 

∗ p2 0

  ∗ p1 ∗ ,  0 p3 0

∗ p2 0

  ∗ p1 ∗ ,  0 0 0

∗ 0 0

  ∗ p1 p2  ,  0 0 0



p1 0 0

  0 ∗ p2  ,  0 0 0

p1 0 0

  ∗ 0 0 ,  0 0 0

0 0 0

  p1 0 0 ,  0 0 0

p1  0 0 0  0 0

 ∗ 0 , 0

∗ 0 0

 0 0 .

0

0 0 0

2.2.4. Reduce [A|b] until A is reduced to echelon form:

    1 −2 b1 1 − 2 b1  2 − 4 b2  →  0 0 b2 − 2b1  .

−6 12 b3 0 0 b3 + 6b1

Hence the conditions for consistency are b2 − 2b1 = 0 and b3 + 6b1 = 0.

2.2.6. Reduce [A|b] until A is reduced to echelon form:

    1 0 −1 b1 1 0 −1 b1  −2   3 − 1 b2  3 − 3 b2 + 2b1  → 0 .

 3 −3   0 b3 0 0 0 b3 + b2 − b1  2 0 −2 b4 0 0 0 b4 − 2b1

Hence the conditions for consistency are b3 + b2 − b1 = 0 and b4 − 2b1 = 0.

2.3.2.         1 0 0 1 0 ∗ 1 ∗ 0 1 ∗ ∗  0 1 0 ,  0 1 ∗ ,  0 0 1 ,  0 0 0 , 0 0 1 0 0 0 0 0 0 0 0 0 

0  0 0

1 0 0

  0 0   1 , 0 0 0

1 0 0

  ∗ 0   0 , 0 0 0

0 0 0

  1 0   0 , 0 0 0

0 0 0

 0 0 .

0

2.3.4. To apply the method of Gauss-Jordan elimination to Exercise 2.1.8, 14

we continue its reduction:

   −1 2 2 3 1 −1

 0   4 9 16 15 0 → 0 0 5 10 15 0

   −1 2 0 −1 −5 −1  0 4 0 − 2 −12  →  0 0 0 1 2 3 0  1 →  0 0

2 4 0

2 9 1

2 1 0

0 0 1

0 1 0

0 0 1

 3 1 16 15  → 2 3

 −1 −5 − 1/2 −3  2 3

 0 −1 − 1/2 −3  .

2 3

Next, apply back substitution to the last matrix, which is in reduced echelon form. Again, we set x4 = t. Then the third row corresponds to the equation x3 + 2t = 3, which gives x3 = 3 − 2t. The second row yields x2 − 12 t = −3, whence x2 = 12 t − 3. Finally, the first row gives x1 = −1.

2.3.6. To apply the method of Gauss-Jordan elimination to Exercise 2.1.12, we continue its reduction:

    −1 2 2 3 1 −1 2 2 3 1  0 0 5 10 10  →  0 0 1 2 2  → 0 0 0 0 0 0 0 0 0 0

  1 −2 0 1 3  0 0 1 2 2  .

0 0 0 0 0 Next, apply back substitution to the last matrix, which is in reduced echelon form. The second and fourth columns have no pivot and so we set x2 = s and x4 = t. Then the second row gives x3 = 2 − 2t and the first row gives x1 = 2s − t + 3.

2.3.8. First represent the system as an augmented matrix and row-reduce it to echelon form.

2 2 − 3 − 2 4 2 2 − 3 − 2 4 → 6 6 3 6 0 0 0 12 12 −12 The second row corresponds to the equation x3 + x4 = −1, and the first row to the equation 2x1 + 2x2 − 3x3 − 2x4 = 4. We can get a particular solution 15

by setting the free variables to x2 = 0 and x4 = 0. Then we obtain x3 = −1, and x1 = 1/2. Hence   1/2  0   xb =   −1  0 is a particular solution of Ax = b. Similarly, setting x2 = 1 and x4 = 0 gives x3 = −1, and x1 = −1/2. Thus   −1/2  1   x b =   −1  0

is another particular solution of Ax = b.

For the homogeneous equation Ax = 0 we can use the same echelon matrix as above, except that the entries of its last column must be replaced by zeros. Then we get the equations x3 + x4 = 0 and 2x1 + 2x2 − 3x3 − 2x4 = 0.

The general solution is obtained by setting x4 = t and x2 = s, and solving for the other variables, obtaining x3 = −t and x1 = −s − 12 t. Thus the general solution of Ax = 0 is 

   −1 −1/2  1   0     v = s  0  + t  −1  .

0 1

Hence the general solution of the inhomogeneous equation Ax = b can be written either as 

     1/2 −1 −1/2  0       + s 1  + t 0  x=  −1   0   −1  0 0 1 16

or as 

     −1/2 −1 −1/2      1   + s 1  + t 0 . x=  −1   0   −1  0 0 1

The two equation represent the same plane, since replacing s by s + 1 in the first of these results in the second one.

2.4.2. By Equation (2.69) (cA)x = c(Ax) for all in Definition 2.4.2, n x ∈ R . Now, (Ax)i = j aij xj and [(cA)x]i = j (cA)ij xj . Thus, from Equation (2.69) we get

(cA)ij xj = c aij xj = (caij )xj . j

j

j

Since this equation must hold for all xj , the coefficients of the xj must be the same on both sides (just choose one xj = 1 and the rest 0, for j = 1, 2, . . . , n), that is, we must have (cA)ij = caij for all i, j.

2.4.4.

AB =



3  BA = 2 1

2.4.6.

2 1

3 −2

5 3

 −4 2 2  1 −3 

3  BA = 2 1

3  2 1



−4 17 2 = 2 −3

3 −2



5 3

−6 −4 −2

and AB does not exist.

17

9 6 3



2  6 = −1  − 12 −8  . −4

− 17 −17 17 2 9

,

 3 16  . −4

2.4.8. AB is undefined and   3 −4  2 2  3  2 BA =   1 −3  1 − 2 −2 5

5 3

2.4.10.

Rα Rβ = = =

cos α − sin α sin α cos α

cos (α + β) sin (α + β)



2  6 =  −1 1

cos β − sin β sin β cos β

cos α cos β − sin α sin β cos α sin β + cos β sin α

17 2 9 − 16

 3 16  . −4  5

− cos α sin β − cos β sin α cos α cos β − sin α sin β

− sin (α + β) = Rα+β . cos (α + β)

2.4.12. If AB = AC, then A(B − C) = O. Using the given solution of Exercise 2.4.11, we may choose

0 0 0 1 A= and B − C = .

0 0 0 1 This choice still leaves a lot of room for choosing B and C. For instance

1 2 1 1 B= and C = 3 4 3 4 will do. Then AB = AC =

0 0

0 1

0 0

0 1

1 3

2 4

1 3

1 4

= =

0 3

0 4

0 3

0 4

Thus, indeed, AB = AC but B = C. 2.4.14. From the definition, An = (A · · ·n times A) with the understanding that (A · · ·0 times A) means I. Then 18

and .

Ak Al = (A · · ·k times A)(A · · ·l times A) = (A · · ·(k+l) times A) = Ak+l .

Similarly, (Ak )l = (A · · ·k times A)(A · · ·k times A) · · ·l times (A · · ·k times A) = (A · · ·kl times A) = Akl .

2.4.16. A possible choice is 

0 A= 0 0

1 0 0

 0 1 .

0

Then 

0 A2 =  0 0

0 0 0

  1 0 0  and A3 =  0 0 0

0 0 0

 0 0 .

0

2.4.18.

A(B + C) = = = = =

A [(b1 + c1 ) (b2 + c2 ) · · · (bn + cn )] [A(b1 + c1 ) A(b2 + c2 ) · · · A(bn + cn )] [(Ab1 + Ac1 ) (Ab2 + Ac2 ) · · · (Abn + Acn )] [Ab1 Ab2 · · · Abn ] + [Ac1 Ac2 · · · Acn ] AB + AC.

2.4.20.The matrix product AC makes sense only if the number of columns in A is equal to the number of rows in C, say k, and similarly the product BD makes sense only if the number of columns in B is equal to the number of rows in D, say l. Then applying the formula developed in Exercise (2.4.13a) 19

in each of the last two steps below, we obtain 

 c1  c2   .   .   . 

 k  C  c  [A B] = [a1 a2 · · · ak b1 b2 · · · bl ]  1  D  d   d2     .   ..  dl

= (a1 c1 + a2 c2 + · · · + ak ck ) + (b1 d1 + b2 d2 + · · · + bl dl ) = AC + BD.

2.4.22. Applying the result of Exercise 2.4.21, we obtain:

   1 − 2 1 0 1 − 2 1 0

  3 1  1  4 0 0 −3   2  = X11  −1 X21 0 0 0  0 0 2 3 

0 −1 0 0 0 0 7 4

X12 X22

,

,

where

X11 =

1 3

−2 4

1 2

−2 1 + 0 0

0 1

0 0

0 0

=

−3 11

−2 −6

X12 =

1 3

−2 4

1 −3

0 1 + 1 0

0 1

2 7

3 4

=

9 −2

1 8

X21 =

−1 0

0 −1

1 2

−2 0 + 0 0

0 0

0 0

0 0

=

−1 0

0 −1

1 −3

0 0 + 1 0

0 0

2 7

3 4

=

,

−1 −2

2 0

,

−1 3

0 −1

and X22 =

20

.

Thus 

1  3   −1 0  −3  11 =   −1 −2

−2 4 0 −1 −2 −6 2 0

1 0 0 0 9 −2 −1 3

2.5.2.

−1

A

1 = 14

2.5.4.

2.5.6.

 0  1   0  0  1 8  .

0  −1

4 −3



1  −2 −6 A−1 = 10 −13 

2  1 0 A−1 =   0 2 −2

1 2 0 0

−2 0 0 0

−2 5

 0 1   3  4

 −2 −6  . − 18

6 8 24

1 1 −1 −2

1

−3

2

7

1 1 1 0

 0 0  .

0  2

2.5.8. a) Augment the given matrix by the 2 × 2 unit matrix, and reduce it as follows:

2 0 4 1 0 2 0 4 1 0 → .

4 −1 1 0 1 0 − 1 − 7 −2 1 We may write the equations corresponding to the reduced matrix as we did in Example 2.5.1:  

x11 x12

2 0 4  1 0 x21 x22  = .

4 −1 1 −2 1 x31 x32 21

This matrix equation corresponds to a system of two scalar equations for the unknown x s in the first column and another system of two scalar equations for those in the second column. The unknowns x31 and x32 are free. Choosing x31 = s and x32 = t, we get the systems of equations 2x11 + 4s = 1 −2x21 − 7s = −2

2x12 + 4t = 0 , − 2x22 − 7t = 1

from which x11 = 12 − 2s, x21 = 2 − 7s, x12 = −2t, and x22 = −1 − 7t.

Thus  1  − 2s −2t 2 X =  2 − 7s −1 − 7t  s t is a right inverse of A for any s, t, and every right inverse is of this form. b) From    

y11 y12 1 0 0 0 4  y21 y22  2 1 0 . = 0 4 −1 1 y31 y32 0 0 1 we obtain the equations

2y11 + 4y12 = 1 −y12 = 0 4y11 + y12 = 0 The augmented matrix of this system can be reduced as 

    2 4 1 2 4 1 2  0 −1 0 → 0 −1 0 → 0

4 1 0 0 −7 −2 0

 4 1 − 1 0  .

0 −2

The last row corresponds to the contradiction 0 = −2, and so there is no solution matrix Y , that is, no left inverse of A. 2.5.10.∗ .a) If A is 3 × 2, then it has no right-inverse: To find one, we would have to solve AX = I, where I must be 3 × 3, and X 2 × 3. Thus we would have nine equations for six unknowns, and should expect no solutions.

Indeed, if AX = I could be solved, then, as multiplying both sides by an arbitrary vector b ∈ R3 shows, AXb = b would have a solution x = Xb of 22

Ax = b for every such b. But the vectors y = Ax = x1 a1 + x2 a2 constitute a plane (or a line or a point), which cannot contain all b ∈ R3 . Thus Ax = b cannot be solved for every b ∈ R3 , and so AX = I cannot have a solution.

If A is 2 × 3, then it has a right-inverse if and only if its rank is 2: Now in AX = I the I must be 2 × 2, and X 3 × 2. If the rank of A is 2, then Ax = b can be solved for any b ∈ R2 , since one of the variables is free and can be selected at will, and in any echelon form the other two variables are multiplied by nonzero pivots and so can be solved for. Choosing in turn b = e1 and b = e2 gives the solution of AX = I. If A is 2 × 3, but its rank is less than 2, then AX = I has no solution.

This can be seen by reducing A to echelon form, say to U . Then the last row of U, and so of U X as well, must be an all-zero row. But no elementary row operations can change I to a matrix with a zero row. Hence at least one of the systems Axi = ei must be inconsistent.

If A is 2 × 3, then it has no left-inverse, and if A is 3 × 2, then it has a left-inverse if and only if its rank is 2. These statements follow from the previous two by transposition. b) The following statements are straightforward generalizations of the ones above, with analogous proofs: For any m × n matrix A we have: 1.

2.

3.

4.

If m > n, then A has no right-inverse.

If m ≤ n, then A has a right-inverse if and only if its rank is m.

If m < n, then A has no left-inverse.

If m ≥ n, then A has a left-inverse if and only if its rank is n.

c) From Statement 2 above, we see that an m × n matrix A has a rightinverse if and only if m ≤ n and the rank of A is m. But if m is less than n, then the equations Axi = ei have free variables and infinitely many solutions.

If, however, m = n and A has full rank, then the right inverse of A becomes the unique, two-sided inverse of A. Similarly, the case of A with m = n and full rank is the only case with a unique left-inverse.

Thus for any matrix A the following statements are equivalent: 1. A has a unique right-inverse.

2. A has a unique left-inverse.

3. A is square and has full rank.

23

4. A has a (unique) two-sided inverse.

2.5.12.. Writing ai for the ith row of any 3 × n matrix A, we can evaluate the product EA as 

1  EA = c 0

 1    0 a a1 0   a2  =  ca1 + a2  .

1 a3 a3

0 1 0

The inverse matrix E −1 can be obtained by row reduction:

  1 0 0 1 0 0  c 1 0 0 1 0 

0 0 1 0 0 1

 1 0 0 1 0  1 0 −c 1 r2 ← r2 − cr1 0 0 0 1 0 0

 0 0 .

1

Thus



1 −1  E = −c 0

and 

1 −1  E A = −c 0

0 1 0

0 1 0

 0 0 , 1

 1    0 a1 a 0   a2  =  −ca1 + a2  , 1 a3 a3

that is, the effect of multiplying A by E −1 is the elementary row operation of changing the second row into itself minus c times the first row.

2.5.14. Writing ei for the ith row of I, we must have  1   3    e e 0 0 1 1 0 .

P = P I = P  e2  =  e2  =  0 3 1 e e 1 0 0 24

But then P also produces the desired row interchange for any 3 × n matrix A:   1   3  0 0 1 a a 2      1 0 a PA = 0 = a2  .

3 1 0 0 a a1 The inverse matrix P −1 can be obtained by row reduction:

  0 0 1 1 0 0  0 1 0 0 1 0 

1 0 0 0 0 1

  1 0 0 0 0 1 1 0 0 1 0 . r3 ↔ r1  0

0 0 1 1 0 0

Thus P −1 = P . This result could also have been obtained without any computation, by observing that an application of P to itself would interchange the first and third rows of P , and so restore I. 2.5.16. Applying the definition A−n = (A−1 )n , and using the properties of positive integer powers of a matrix from Exercise 2.4.14, we obtain An (A−1 )n = (A(n−1) A)(A−1 (A−1 )n−1 ) = A(n−1) (AA−1 )(A−1 )n−1 = A(n−1) I(A−1 )n−1 = ∂A(n−1) (A−1 )n−1 = · · · = AA−1 = I.

Thus A−n = (A−1 )n is the inverse of An , that is, A−n = (An )−1 . (Remark: this derivation can be made more rigorous through the use of mathematical induction.)

Again applying the definition A−n = (A−1 )n , together with the properties of positive integer powers of a matrix from Exercise 2.4.14, we have A−m A−n = (A−1 )m (A−1 )n = (A−1 )(m+n) = A−(m+n) = A−m−n .

2.5.18. An n × n matrix A is singular if and only if any (and hence each) of the following conditions holds: 1. A is not invertible.

2. The rank of A is less than n.

3. A is not row equivalent to I. 25

4. Ax = b has no solution for some b.

5. For no b does Ax = b have a unique solution .

6. The homogeneous equation Ax = 0 has nontrivial solutions.

2.5.20. If we write D = BC and apply Theorem 2.5.6 to both AD and D, it follows that D and thus also AD are each invertible, and D−1 = C −1 B −1 and (ABC)−1 = (AD)−1 = D−1 A−1 = C −1 B −1 A−1 .

3.1.2. The set of all solutions (x, y) of the equation 2x + 3y = 0 is a vector space, for the following reason: Let u = (x1 , y1 ) and v = (x2 , y2 ) be two arbitrary solutions of 2x + 3y = 0, that is, let 2x1 + 3y1 = 0 and 2x2 + 3y2 = 0 hold and let c be an arbitrary scalar. Then 2(x1 + x2 ) + 3(y1 + y2 ) = (2x1 + 3y1 ) + (2x2 + 3y2 ) = 0 + 0 = 0, and so u + v is also a solution and obviously cu is as well. Thus, U is closed under both operations and is clearly nonempty. The axioms hold automatically, because the operations are the same as in R2 . Hence the set of all solutions is a vector space.

3.1.4. The set S of all twice differentiable functions f for which f (x) + 2f(x) = 0 holds is a vector space. Indeed, letting f and g be two arbitrary functions of S, we obtain (f + g) (x) + 2 (f + g) (x) = [f (x) + 2f (x)] + [g (x) + 2g(x)] = 0 + 0 = 0, and for an arbitrary scalar c, (cf ) (x) + 2 (cf) (x) = c [f (x) + 2f(x)] = 0.

Thus, S is closed under both operations and is clearly nonempty.

3.1.6. The set P of all polynomials in a single variable x is a vector space, since it is clearly closed under both operations and it is nonempty.

3.1.8. This set is not a vector space, because addition is not commutative: For example, let (p1 , p2 ) = (1, 2) and (q1 , q2 ) = (1, 3). Then, by the given addition rule, we have (1, 2)+(1, 3) = (1+3, 0) = (4, 0), but (1, 3)+(1, 2) = (1 + 2, 0) = (3, 0).

3.1.10. This set is not a vector space, because Axiom 7 of Definition 3.1.1 fails to hold in it: Let −a = b = 0, and p = 0. Then (a + b)p = 0(p1 , p2 ) = (0, 0) = 0, but ap+bp = (|a|p1 , |a|p2 )+(|b|p1 , |b|p2 ) = (2|a|p1 , 2|a|p2 ) = 0.

26

3.1.12. This is Part 3 of Theorem 3.1.1 with q = p.

3.1.14. Assume Axiom 3 instead of Axioms 3 and 4 of Definition 3.1.1.

Then we have 0 = 0p. Add p to both sides: p + 0 = p + 0p = 1p + 0p by Axiom 5 of Definition 3.1.1, = (1 + 0)p by Axiom 7, = 1p = p because of 1 + 0 = 1 and Axiom 5. This proves Axiom 3.

To prove Axiom 4, define −p = (−1)p. Then p + (−p) = p + (−1)p by the definition of −p, = 1p + (−1)p by Axiom 5, = [1 + (−1)]p by Axiom 7, = 0p by adding the two numbers, and = 0 by Axiom 3’.

3.2.2. This U is not a subspace of R3 : Take for instance u = [1, 1, 0]T and v = [4, 2, 0]T . Then u, v ∈ U , but u + v = [5, 3, 0]T ∈ / U, because 32 = 5. Thus U is not closed under addition.

3.2.4. This U is a subspace of R4 by Theorem 3.2.2: If we choose n = 4 and A = [1, 1, 1, 0], then the defining equation x1 + x2 + x3 = 0 of U can be written as Ax = 0, and the theorem is directly applicable.

3 3.2.6. This U is a subspace of R by Theorem 3.2.2: If we choose n = 3 1 −1 0 and A = , then the defining equations of U can be written 0 0 1 as Ax = 0, and the theorem is directly applicable.

3.2.8. This U is not a subspace of R3 : Take for instance u = [1, 0, 0]T and v =√[0, 1, 0]T . Then u, v ∈ U , but x = u + v = [1, 1, 0]T ∈ / U , because |x| = 2, but |x1 | + |x2 | = 2. Thus U is not closed under addition.

3.2.10. For instance, let X = R2 , U = {u|u = ue1 } and V = {v|v = ve2 }.

Then U ∪ V contains just the multiples of e1 and of e2 but not their sum if uv = 0. Hence, U ∪ V is not closed under addition and so is not a subspace of R2 .

On the other hand, let U and V be arbitrary subspaces in an arbitrary vector space X. Then U ∪ V is nonempty, since 0 ∈ U ∪ V. It is also closed under multiplication by scalars: If x ∈ U ∪ V, then either x ∈ U or x ∈ V, and, for all c, in the first case, cx ∈ U and so cx ∈ U ∪ V and in the second 27

case, cx ∈ V and so cx ∈ U ∪ V. Furthermore, U ∪ V is closed under addition only if V ⊂ U or U ⊂ V . For, otherwise, assume that neither V ⊂ U nor U ⊂ V hold, that is, that there exist u ∈ U, u ∈ / V and v ∈ V, v ∈ / U. Then u and v both belong to U ∪ V, but u + v ∈ / U ∪ V, because, by the closure of U, u + v ∈ U would imply v ∈ U, and, by the closure of V, u + v ∈ V would imply u ∈ V. Thus, U ∪ V is a subspace of X only if V ⊂ U or U ⊂ V . It is easy to see that the converse is also true. Hence, U ∪ V is a subspace of X if and only if V ⊂ U or U ⊂ V .

3.2.12. U = {x|x ∈ Rn , a · x = 0} is a subspace of Rn : a) 0 ∈ U , and so U is nonempty. b) U is closed under addition: Let u, v ∈ U . Then a · u = 0 and a · v = 0.

Hence a · (u + v) = 0, which shows that u + v ∈ U . c) U is closed under multiplication by scalars: Let u ∈ U and c ∈ R. Then a · u = 0, and therefore c (a · u) = a · (cu) = 0, which shows that cu ∈ U. Alternatively, a · x = aT x, and with A = aT , Theorem 3.2.2 proves the statement.

3.3.2. Suppose c has a decomposition

c=

n

si ai ,

i=1

and b the two decompositions b=

n

ti ai and b =

i=1

n

ui ai .

i=1

Then c = c+b−b =

n

i=1

si ai +

n

i=1

ti ai −

n

i=1

n

ui ai = (si + ti − ui )ai . i=1

Since the two decompositions of b were different, for some i we must have si + ti − ui = si . Thus the last expression provides a new decomposition of c.

28

3.3.4. We have to solve  4 4  2 −3 1 2

    1 s1 7 3   s2  =  16  . −1 −3 s3

Since this system is the same as that of Exercise 3.3.3 without the second row, which did not contribute anything to the problem, it must have the same solution. Thus b = 2a1 − a2 + 3a3 .

3.3.6. We have to solve      4 4 0 s1 4  2 −3     5 s2 = 7  .

1 2 −1 s3 0 We solve this system in augmented matrix form by row reduction as:     4 4 0 4 1 2 − 1 0  2 −3 5 7  →  2 − 3 5 7 

1 2 −1 0 4 4 0 4     1 2 − 1 0 1 2 − 1 0 7 7  →  0 − 7 7 7  . →  0 −7 0 −4 4 4 0 0 0 0

The variable s3 is free. Choosing s3 = 1, we find that the second row of the reduced matrix gives s2 = 0, and the first row leads to s1 = 1. Thus b = a1 + a3 is a possible solution.

3.3.8. The three vectors a1 = (4, 2, 1)T , a2 = (4, −3, 2)T , a3 = (1, 3, −1)T from Exercise 3.3.4 are independent, because s1 a1 + s2 a2 + s3 a3 = 0, that is, 

4  2 1

4 −3 2

    1 s1 0     3 s2 = 0  −1 0 s3

has only the trivial solution, as can be seen by row reduction:     4 4 1 0 1 2 − 1 0  2 −3 3 0  →  2 − 3 3 0 

1 2 −1 0 4 4 1 0 29



1 2  → 0 −7 0 −4

  − 1 0 1

  5 0 → 0 5 0 0

 2 − 1 0 −7 5 0  .

0 15/7 0

Then, by back substitution, we obtain s3 = 0, s2 = 0, and s1 = 0.

This result also follows from the equivalence of Parts 5 and 6 of Theorem 2.5.5, which implies that if Ax = b has a unique solution for some b, as happens for the b of Exercise 3.3.4 above, then Ax = 0 has only the trivial solution.

3.3.10. These vectors are dependent, because in R2 any two noncollinear vectors span the whole plane.

3.3.12. Let a1 and a2 be independent vectors in a vector space X, and v another vector of X, not in Span{a1 , a2 }. Assume that the vectors v, a1 , a2 are dependent. Then s0 v + s1 a1 + s2 a2 = 0 has a nontrivial solution: If s0 = 0, then v = − ss10 a1 − ss20 a2 ∈ Span{a1 , a2 } in contradiction to the definition of v. If, on the other hand, s0 = 0 but s1 and s2 are not both 0, then a1 and a2 are not independent, in contradiction to their definition. Thus v, a1 , a2 must be independent of each other.

3.3.14. Consider any vectors a1 , a2 , . . . , an , and 0 in a vector space X. Then the equation s0 0 + s1 a1 + s2 a2 + · · · + sn an = 0 has the nontrivial solution s0 = 1, s1 = s2 = · · · = sn = 0, and so {0, a1 , a2 , . . . , an } is a dependent set.

3.3.16. Let a1 , a2 , a3 be independent vectors in R3 . Let A be the 3 × 3 matrix that has these vectors as columns. Then, by the definition of linear independence, the equation As = 0 has only the trivial solution. By the equivalence of Parts 4 and 6 of Theorem 2.5.5, this implies that As = b has a solution for every vector b in R3 , which means that a1 , a2 , a3 span R3 .

3.3.18.. Let A be an n × n matrix with columns a1 , a2 , . . . , an . By Part 6 of Theorem 2.5.5, A is singular if and only if As = 0 has nontrivial solutions. However, the latter condition holds if and only if the columns of A are dependent, 3.3.20.First, Rn was defined only for positive n, and so the problem makes sense only for 0 < n. Next, let A be the n × m matrix with columns a1 , a2 , . . . , am . If these vectors span Rn , then the equation As = b has a solution for every vector b in Rn . Thus , if A is row reduced to the echelon form U , then the corresponding equation Us = c must have a solution for 30

every vector c ∈ Rn , because otherwise, reversing the reduction steps for a c without a solution would result in an As = b without a solution. Thus U must not have any zero row, because if it had one, then U s = en would have no solution. Hence n ≤ m and r = n must hold.

Conversely, if 0 < n ≤ m and r = n, then reversing the foregoing argument, we find that the columns of A span Rn .

3.4.2. Row-reduce A as 

1  1 0

  1 1   3 → 0 4 0

2 3 2

2 1 2

  1 1   2 → 0 4 0

2 1 0

Thus a basis for Col(A) is the set      2   1  1 ,  3  ,  0 2 

 1 2 .

0

and a basis for Row(A) is

     0   1  2 ,  1  .  1 2 

For Null(A): x3 = s, x2 = −2s, x1 = 3s. Thus a basis for Null(A) is the set of one element,   3    −2  .   1 3.4.4. Row-reduce A as 

1  2 3

1 2 3

1 2 0

2 4 4

  −1 1   −2 → 0 4 0

1 0 0

1 −3 0

2 −2 0

 −1 7 .

0

Because the pivots are in the first and third columns of the reduced matrix, the corresponding columns of A constitute a basis for Col(A). Thus a basis 31

for Col(A) is the set

and a basis for Row(A) is

    1   1  2 , 2  ,  3 0 

     1 0      1   0         1  ,  −3  .         2   −2       −1 7 

To find a basis for Null(A), set x2 = s, x4 = t, and x5 = u. Then x3 = − 23 t + 73 u and x1 = −s − 43 t − 43 u. Hence       −1 −4 −4  1        t 0  u 0       x = s  0  +  −2  +  7    0  3 3  3 0  0 0 3 is the general form of a vector in Null(A), and so        −1 −4 −4              1 0 0        0  ,  −2  ,  7            0   3   0       0 0 3 

is a basis for Null(A).

3.4.6. Any echelon matrix E corresponding to A must have a zero last row. Thus no linear combination of the columns of E can equal any of the columns of A, which have nonzero third components. On the other hand, Theorem 3.4.1, coupled with the row reduction in Exercise 3.4.1, shows that the first two columns of A form a basis for Col(A).

3.4.8. First, assume AB = O. Now, this relation holds if and only if Abi = 0 for every column bi of B. Multiplying both sides of the latter 32

equation by parameters si and summing over i, we find that equivalently

A si bi = 0 for every vector of the form si bi . Since these vector constitute Col(B), we see that AB = O implies that Col(B) is a subset of Null(A). Since Col(B) is a vector space, it is then a subspace of Null(A) .

Conversely, if Col(B) is a subspace of Null(A), then A si bi = 0 for

every vector of the form si bi , since Col(B) is the set of such vectors, which must then lie in Null(A) as well. In particular, we may choose si = 1 for any given i, and sj = 0 for all j = i. Thus Abi = 0 for every i, and so AB = O. 3.4.10. Let B = (b1 , b2 , . . . , bn ) be the list of the given independent vectors and use the standard vectors for the spanning list, that is, let A = (e1 , e2 , . . . , en ) in the Exchange Theorem. Then by the theorem we can exchange all the standard vectors for the bi vectors and they will span Rn .

3.4.12. If the column vectors of the n × n matrix A = (a1 , a2 , . . . , an ) are independent, then Condition 6 of Theorem 2.5.5 holds, which is equivalent to Condition 4, and the latter shows that the column vectors of A = (a1 , a2 , . . . , an ) span Rn . Thus these vectors both span Rn and are independent, i.e. they form a basis for Rn .

Alternatively, if the n vectors that span Rn were dependent, then we could express one of them as a linear combination of the others, and eliminate that vector from any linear combination when spanning Rn . Thus we would have a set of n − 1 vectors spanning Rn , in contradiction to the result of Exercise 3.4.11.

3.4.14. Let a1 , a2 , a3 denote the given spanning vectors of U . It is easy to see that b1 = a2 + a3 − a1 and b2 = a1 + a3 , and so both b1 and b2 are in U . To find the required basis reduce the matrix A = (b1 , b2 , a1 , a2 , a3 ) to echelon form, and use the basic columns of A so found as the basis vectors:     0 1 1 1 0 1 1 0 0 1  1  1 0 0 1  1 1 1 0   → 0 →  1  1 1 0 0 1  1 0 0 1  1 0 0 1 0 1 0 0 1 0     1 0 0 1 0 1 0 0 1 0  0  1 1 1 0  1 0 −1 1   → 0 .  0  0 0 0 0 0  0 1 2 −1  0 −1 0 1 −1 0 0 0 0 0 Since in the last matrix the first three columns contain the pivots, the first 33

three columns of A are the basic columns and {b1 , b2 , a1 } is a basis for the subspace U .

3.5.2. a) The sole basis vector for Row(A) is b1 = AT = (1 1 1 2 −1)T . Thus dim(Row(A)) = dim(Col(A)) = 1, and, by Corollary 3.5.2 and Theorem 3.5.4, dim(Null(A)) = 4, and dim(Left-null(A)) = 0. b) To find a basis for Null(A), solve Ax = 0, that is, (1 1 1 2 −1)x = 0 by setting x2 = s, x3 = t, x4 = u, and x5 = v. Then we get x1 = −s − t − 2u + v, and so any vector of Null(A) can be written as x0 = sc1 + tc2 + uc3 + vc4 , where         −1 −1 −2 1  1   0   0   0                 c1 =   0  , c2 =  1  , c3 =  0  , c4 =  0  .  0   0   1   0  0 0 0 1 These are the basis vectors for Null(A). Thus, to get the desired decomposition of x = (1, 1, 1, 1, 1)T , we must solve x = sc1 + tc2 + uc3 + vc4 + wb1 .

By row reduction:   −1 − 1 − 2 1 1 1  1 0 0 0 1 1     0 1 0 0 1 1   →

 0 0 1 0 2 1  0 0 0 1 −1 1 

1  0   0   0 −1

0 1 0 0 −1

0 0 1 0 −2

1  0   0   0 0

0 1 0 0 0

0 0 1 0 0



0 0 0 1 1

0 0 0 1 0

 1 1 1 1   2 1  → − 1 1  1 1

 1 1 1 1   2 1  . − 1 1  8 4 34

Hence w = 1/2, v = 3/2, u = 0, t = 1/2, s = 1/2, and so 1 1 3 1 x0 = c1 + c2 + c4 = (1, 1, 1, 0, 3)T 2 2 2 2 and 1 xR = (1, 1, 1, 2, −1)T .

2 Alternatively, we may solve x = sc1 + tc2 + uc3 + vc4 + wb1 for w by left-multiplying it with bT1 and obtain, because of the orthogonality of the row space to the nullspace, bT1 x = wbT1 b1 , which becomes 4 = 8w. Thus again w = 1/2 and xR = 12 (1, 1, 1, 2, −1)T . From here x0 = x − xR = 1 (1, 1, 1, 0, 3)T , as before.

2 3.5.4. a) dim(Row (A)) = dim(Co1 (A)) = 3, and, by Corollary 3.5.2 and Theorem 3.5.4, dim(Null (A)) = 2, and dim(Left-null (A)) = 0. b) Since A is an echelon matrix, its transposed rows form a basis for Row(A).

To find a basis for Null(A), solve Ax = 0 by setting x3 = t and x5 = u.

Then we get x4 = −u, x2 = 0, and x1 = −2x4 + x5 = 3u. So any vector of Null(A) can be written as x0 = t1 c1 + t2 c2 , where     0 3  0   0       , c2 =  0  .

1 c1 =       0   −1  0 1 These are the basis vectors for Null(A). Thus, to get the desired decomposition of x = (6, 2, 1, 4, 8)T , we solve (6, 2, 1, 4, 8)T = t1 c1 + t2 c2 + s1 b1 + s2 b2 + s3 b3 , where the bi are the transposed rows of A. We solve by rowreduction:   0 3 1 0 0 6  0 0 0 2 0 2     1 1 → 0 0 0 0    0 −1 2 0 2 4  0 1 −1 0 2 8 35



1  0   0   0 0 

1  0   0   0 0 

1  0   0   0 0

0 −1 1 0 3

0 2 −1 0 1

0 0 0 2 0

0 1 0 0 0

0 −1 1 0 7

0 0 0 2 0

0 1 0 0 0

0 −1 1 0 0

 0 1 2 4   2 8  → 0 2  0 6

 0 1 2 8   4 12  →

0 2  6 18

 1 0 0 8  0 2  0 4 12  .

2  2 0 0 −22 −66

Hence we obtain s3 = 66/22 = 3, s2 = 1, s1 = 12 − 4 · 3 = 0, t2 = 8 − 2 · 3 = 2, and t1 = 1. Thus: 

       1 0 0 0  0   2   0   2          +1· 0 +3· 0 = 0  0 xR = 0 ·           2   0   2   6  −1 0 2 6 and 

     0 3 6  0   0   0        +2· 0  =  1 .

1 x0 = 1 ·         0   −1   −2  0 1 2 36

Indeed,



     0 6 6  2   0   2         0 + 1 =  1         6   −2   4  6 2 8

and

0

2

0

6



 6    0   6  1   = 0.  −2  2

3.5.6. This matrix is already in echelon form. Thus b1 = (1, 1, 1, 2, −1)T and b2 = (0, 2, 0, 0, 4)T form a basis for Row(A). Hence dim(Row (A)) = 2.

Consequently, by Theorem 3.5.2, dim(Col (A)) = 2, too, and, by Corollary 3.5.2 and Theorem 3.5.4, dim(Null (A)) = 3, and dim(Left-null (A)) = 0.

To find a basis for Null(A) we solve Ax = 0 by setting x3 = s, x4 = t, and x5 = u. Then 2x2 +4u = 0, and so x2 = −2u. Also, x1 = 2u−s−2t+u, and x1 = −s − 2t + 3u. Hence Null(A) consists of the vectors       −1 −2 3  0   0   −2         + t 0  + u 0  1 x = s        0   1   0  0 0 1 and



     −1 −2 3  0   0   −2         , c2 =  0  , c3 =  0  1 c1 =         0   1   0  0 0 1

form a basis for Null(A). To decompose x = (1, 3, 4, 2, 8)T into the sum of an x0 ∈ Null (A) and an xR ∈ Row (A), solve (1, 3, 4, 2, 8)T = s1 b1 + s2 b2 + t1 c1 + t2 c2 + t3 c3 by row reduction: 37



1  1   1   2 −1 

0 2 0 0 4

−1 0 1 0 0

−2 0 0 1 0

1  0   0   0 0

0 2 0 0 4

−1 1 2 2 −1

−2 2 2 5 −2

1  0   0   0 0

0 2 0 0 0

−1 1 2 0 −3

−2 2 2 3 −6

1  0   0   0 0

0 2 0 0 0

−1 1 2 0 0

1  0   0   0 0

0 2 0 0 0

−1 1 2 0 0







 3 1 − 2 3   0 4  → 0 2  1 8

 3 1 − 5 2   −3 3  → −6 0  4 9

 3 1 − 5 2   −3 3  →

−3 −3  14 5

 −2 3 1 2 − 5 2   2 −3 3  → 1 −1 −1  −3 19/2 19/2

 −2 3 1 2 − 5 2   2 −3 3  .

1 −1 −1  0 13/2 13/2

Hence t3 = 1, t2 = 0, t1 = 3, s2 = 2, s1 = 1,         −1 −2 3 0  0   0   −2   −2          +0· 0 +1· 0 = 3  1 x0 = 3 ·           0   1   0   0  0 0 1 1 38

and

Indeed,



     1 0 1  1   2   5             xR = 1 ·   1  + 2 ·  0  =  1 .  2   0   2  −1 4 7 

     0 1 1  −2   5   3         3 + 1 = 4         0   2   2  1 7 8

and

0

−2

3

0



 1    5   1   1  = 0.  2  7

3.5.8. a) 0 ∈ U + V , and so U + V is nonempty. b) U + V is closed under addition: Let x, y ∈ U + V . Then we can write x = u1 +v1 and y = u2 +v2 , where u1 , u2 ∈ U and v1 , v2 ∈ V . Since both U and V are vector spaces, they are closed under addition, and consequently u1 +u2 ∈ U and v1 +v2 ∈ V . Thus, x + y = (u1 +u2 )+(v1 +v2 ) ∈ U +V . c) U + V is closed under multiplication by scalars: Let x = u + v ∈ U + V and c ∈ R, with u ∈ U and v ∈ V . Since both U and V are vector spaces, they are closed under multiplication by scalars, that is, cu ∈ U and cv ∈ V .

Thus, cx = cu + cv ∈ U + V .

3.5.10. Write A = (a1 , a2 , . . . , an ) and B = (b1 , b2 , . . . , bp ). Then the members of Col(A) have the form ni=1 si ai , and those of Col(B) the p form bi . Hence any member of Col(A) + Col(B) can be written as i=1 ti n p s a + i=1 i i i=1 ti bi , which is exactly the same as the form of an arbitrary member of Col[A B].

3.5.12.

39

a) Col(A) has {a1 , a2 } for a basis, and since a1 = e1 and a2 − a1 = 2e2 , the set {e1 , e2 } for a basis as well. B can be reduced to the echelon form (0, e1 , e2 ), and so {b2 , b3 } is a basis for Col(B). Since   1 2 0 2 0 , A+B = 0 0 2 0

the vectors (1, 0, 0)T and (2, 2, 2)T form a basis for Col(A + B). b) Clearly, Col(A) + Col(B) = R3 , Col(A) ∩ Col(B) = {se1 }, and so they have {e1 , e2 , e3 } and {e1 }, respectively, for bases. c) No, because Col(A+B) is two-dimensional and Col(A)+Col(B) is threedimensional. Addition of matrices is different from addition of subspaces! d) 3 = 2 + 2 − 1.

3.5.14. If A is m × p and B is p × n, then AB is m × n. By the result of Exercise 3.4.9, we have nullity(B) = nullity(AB), and, by Corollary 3.5.1, rank(B) = n − nullity(B) and rank(AB) = n − nullity(AB). Thus rank(B) = rank(AB).

3.5.16. Assume that A is m × p and B is p × n.

The elements of Col(A) are of the form As, where s ∈ Rp , and the elements of Col(AB) are of the form ABt, where t ∈ Rn . Thus, writing s = Bt in ABt, we can see that every element of Col(AB) is also an element of Col(A).

The elements of Row(B) are of the form B T s, where s ∈ Rp , and the elements of Row(AB) are of the form B T AT t, where t ∈ Rm . Thus, writing s = AT t in B T AT t, we can see that every element of Row(AB) is also an element of Row(B).

Since the rank of a matrix equals the dimension of its column space, we have rank(AB) ≤ rank(A), and because it also equals the dimension of its row space, rank(AB) ≤ rank(B). Since, furthermore, the number of columns is the same n for both B and AB and nullity + rank = n, we have nullity(AB) ≥ nullity(B) 3.5.18. The first two columns are independent, and so they form a basis for U. Similarly, the last two columns form a basis for V .

To find a basis for U ∩ V , we must find all vectors x that can be written both as s1 a1 +s2 a2 and as t4 a4 +t5 a5 , that is, find all solutions of the equation 40

s1 a1 + s2 a2 = t4 a4 + t5 a5 or, equivalently, of           1 1 2 −1 0          4 = 0 . s1 0 + s2 2 − t4 0 − t5 0 0 2 2 0 Thus, in augmented matrix form we have   1 1 2 − 1 0  0 2 0 4 0  , 0 0 2 2 0

which is already in echelon form. Here t5 is free and setting t5 = t, from the third row we get t4 = −t. (We do not need to find s1 and s2 .) Thus, a parametric form of the line of intersection is         x1 2 −1 −3  x2  = −t  0  + t  4  = t  4  , x3 2 2 0    −3  and so  4  is a basis for U ∩ V.  0  U +V = R3 , and so any three independent vectors form a basis for U +V .

Furthermore,  

1 1 1 0 0 ⊥   2 = Null U = Left-null 0 1 2 0 0 0     0  has  0  for a basis, and  1   

2 −1 2 0 2 ⊥ 4  = Null V = Left-null  0 −1 4 2 2 2 41

  has  

 4  3  for a basis.  −4

3.5.20.The subspace S generated by U ∪ V is the set of all linear combinations of the elements of U ∪ V . Thus

m n

S = s s = ai ui + bj vj ; m, n ∈ N+ ; all ui ∈ U, vj ∈ V

i=1

j=1

Since

u=

n

i=1

ai ui ∈ U and v =

m

j=1

bj vj ∈ V,

we have S = U + V .

3.5.22.Assume that there is an x ∈ U + V with two different decompositions: x = u1 + v1 = u2 + v2 , with u1 , u2 ∈ U, u1 = u2 and v1 , v2 ∈ V , v1 = v2 . Then u1 − u2 = v2 − v1 = 0, where u1 − u2 ∈ U and v2 − v1 ∈ V .

Hence U ∩ V = {0}, and the sum is not direct.

Conversely, assume that U ∩ V = {0}; say, w ∈ U ∩ V and w = 0.

For any u ∈ U and v ∈ V , let x = u + w + v. Then x = (u + w) + v and x = u + (w + v) provide two different decompositions of x with one term from U and another from V . n 3.5.24.For n > 2 subspaces U of a vector space X, we define i i=1 Ui =

{ ni=1 ui |ui ∈ Ui for i = 1, . . . , n} .

3.5.26. If U ⊥ V , then, for all u ∈ U and v ∈ V, uT v = 0. Choose w ∈ U ∩ V for both u and v above. Then wT w = 0, which implies w = 0.

The converse is not true: In R2 let U = {u | u = ti} and V = {v | v = t(i + j)}. Then U ∩ V = {0}, but U is not orthogonal to V , since i · (i + j) = 0.

3.5.28.As in the solution of Exercise 3.5.27, let A be a basis for U ∩ V, B a basis for U ∩ V ⊥ , C a basis for U ⊥ ∩ V , and D a basis for U ⊥ ∩ V ⊥ . Then U + V has A ∪ B ∪ C for a basis, and so D is a basis not only for U ⊥ ∩ V ⊥ but also for (U + V )⊥ . Hence these two subspaces must be identical.

3.5.30.(U ⊥ )⊥ is the set of all vectors of Rn that are orthogonal to all vectors of U ⊥ . In other words, for all v ∈ (U ⊥ )⊥ we have vT u = 0 for all u ∈ U ⊥ . This identity is true, however, precisely for all v ∈ U. 42

3.5.32. Taking the orthogonal complement of both sides of the equation Null(A) = Row(B), we get Row(A) = Null(B). Hence the transposed column vectors of a basis for Null(B) may serve as the rows of A. Since B is an echelon matrix, no row reduction is needed for finding its nullspace. The variables x3 and x5 are free: Set x3 = s1 and x5 = s2 . Then x4 = −s2 , x2 = −2s2 and x1 − 2s2 + s1 − 2s2 − s2 = 0; and so finally x1 = −s1 + 5s2 . Hence we can write the solution vectors as   −1 5  0 −2    s 1 0 x=    0 −1  0 1 and obtain

A=

−1 5

0 −2

1 0

0 −1

0 1

.

3.5.34. According to the algorithm of Exercise 3.5.33, if we reduce [B|I] U L to the form with U an echelon matrix, then the transposed rows of O M M will form a basis for Left-null(B). Thus M can be taken as the matrix A such that Null(A) = Row(B). The reduction goes as follows:

  1 1 1 1 0 0 0  0 2 0 0 1 0 0   →  0 0 0 0 0 1 0  2 1 0 0 0 0 1 

1  2   0 0

1 1 2 0

1  0   0 0

1 −1 2 0



1 1 0 0 0 0 0 0

1 1 −2 −2 0 0 0 0

43

0 0 0 0

0 0 1 0

0 0 0 0

0 0 1 0

 0 1  → 0  1

 0 1  → 0  1



1  0   0 0

1 −1 0 0

1 1 −2 −2 −4 −4 0 0

0 0 1 0

0 0 0 0

Hence A = [0, 0, 1, 0], as before in Exercise 3.5.31.

 0 1  .

2  1

AT −bT

3.5.36. The equation Ax = b can be reformulated as [xT 1] =

x AT 0. This shows that is in the left nullspace of Thus the result of −bT 1

AT Exercise 3.5.33 is applicable. This would require that we augment −bT by the unit matrix of order n + 1. However, the last column may be omitted, since it does not change in the computation, and this just corresponds to the omission of the 1 in [xT 1]. The unit matrix of order n + 1 without its last AT I

By Exercise 3.5.33, the column is the second block column of 0 −bT reduction of this matrix yields both the particular solution x from the last row, and the general solution, by adding to x any linear combination of the transposed rows of L that correspond to the zero rows of U .

AT I 3.5.38.By the algorithm of Exercise 3.5.36, if we reduce −eTi 0 where ei is the ith column of the unit matrix I, we get the transposed solution

U L of Ax = ei in the last row of . The solutions of these equations, 0 xT for i = 1, 2, . . . , n, are the columns of A−1 . We can obtain these solutions AT I simultaneously for all i by reducing instead of reducing each −I 0

AT I separately. This computation would, however, result in the −eTi 0 transpose of A−1 in the lower right corner. So to obtain A−1 itself there, we just have to use A instead of AT in the block matrix that we reduce.

3.5.40.By the definition of dimension, any basis of an n-dimensional vector space X consists of n vectors. Let B = (b1 , b2 , . . . , bn ) be the list of the given independent vectors, and use the vectors ai of a basis for the spanning list, that is, let A = (a1 , a2 , . . . , an ) in the Exchange Theorem. Then, by the 44

theorem, we can exchange all the ai vectors for the bi vectors and they will span X, in addition to being independent. Alternatively, if the column vectors of the n × n matrix A = (a1 , a2 , . . . , an ) are independent, then Condition 6 of Theorem 2.5.5 holds, which is equivalent to Condition 4, and the latter shows that the column vectors of A = (a1 , a2 , . . . , an ) span Rn . Thus these vectors both span Rn and are independent, i.e. they form a basis for Rn .

3.5.42.If the n vectors that span X were dependent, then we could express one of them as a linear combination of the others, and eliminate that vector from any linear combination when spanning X. Thus we would have a set of n − 1 vectors spanning X, in contradiction to the result of Exercise 3.5.41.

3.6.2. a)   1 0 1 1 , S =A= 1 −2 2 1 0 b)



1  −1 −1 2 S = 4 5

c)

1 −2 −1

 2 0 , −2



 11 1 xA = S −1 x =  −2  .

4 1

3.6.4. a) We have 

Then

1  A= 2 0

−2 1 0

  0 3   0 and B = 2 1 0 

1 1 A−1 =  −2 5 0

45

2 1 0

 0 0  5

1 1 0

 0 1 .

1

and 

1 2 1 1 S = A−1 B =  −2 5 0 0   7 3 2 1 −4 − 1 1 . = 5 0 0 5

 0 3 0  2 5 0

1 1 0

 0 1  1

b) Hence 

and

 1 −3  1

−1 S −1 =  4 0

−3 7 0



    1 2 −11 − 3   4  =  27  .

1 3 3

−1 −1  4 xB = S xA = 0

−3 7 0

Thus x = −11b1 + 27b2 + 3b3 .

3.6.6. We have A = Rθ = and −1

A

= R−θ =

Thus, −1

MA = A M A =

cos θ sin θ

cos θ − sin θ

− sin θ cos θ

cos2 θ − sin θ cos θ

sin θ cos θ

.

− sin θ cos θ sin2 θ

.

3.6.8. The augmented matrix [A|B] corresponds to the matrix equation AS = B, and the equivalent, reduced form

I C O O 46

in Equation (3.138) to

I O

S=

C O

.

Hence IS = C, that is, the solution of AS = B is the S = C obtained by the reduction.

3.6.10. The vector c2 must be a linear combination x1 a1 + x2 a2 such that a1 · c2 = 0. Thus, we need to find a solution of a1 · (x1 a1 + x2 a2 ) = 0. This is equivalent to 14x1 + 5x2 = 0, which is solved by x1 = 5 and x2 = −14.

Hence c2 = 5a1 − 14a2 = 5(1, 2, 3)T − 14(0, 4, −1)T = (5, −46, 29)T , or any nonzero multiple of it, is a solution.

3.6.12. a) (L + M)A = A−1 (L + M )A = A−1 LA + A−1 M A = LA + MA . b) (LM )A = A−1 (LM)A = A−1 LAA−1 M A = LA MA .

0 t

3.6.14. For any nonzero t and t we can choose S = and then 0 t

1 −t SM(t)S −1 = M (t ). If t = 0, choose S = 0 1

0 1 is its own inverse, 3.6.16. Yes: The permutation matrix S = 1 0 and premultiplying the first given matrix by it switches the rows, and postmultiplying switches the columns.

3.6.18. Let X denote the set of n × n matrices. For matrices A, B of X, we say that B is similar to A if there exists an invertible matrix S in X such that B = S −1 AS. 1. This relation on X is reflexive: For any A we have A = I −1 AI. 2. It is symmetric: If B = S −1 AS, then A = T −1 BT with T = S −1 .

3. It is transitive: If B = S −1 AS and C = T −1 BT , then C = T −1 S −1 AST = (ST )−1 A(ST ).

3.6.20. n n n a) (AB)ik = aij bjk and so Tr(AB) = j=1 i=1 j=1 aij bji . Similarly, n n Tr(BA) = i=1 j=1 bij aji , and so Tr(AB) = Tr(BA). b) If A = SBS −1 , then Tr(A) = Tr(S(BS −1 )) = Tr((BS −1 )S) = Tr(B). c) They are not similar: the trace of the first matrix is 4, but the trace of the second matrix is 5.

47

4.1.2. Let x =

n

n

i=1 xAi ai and y =

x+y =

n

i=1 yAi ai . Then

(xAi + yAi ) ai

i=1

and ,by Equation 4.6 applied to x + y, T (x + y) = =

n

i=1 n

i=1

(xAi + yAi ) T (ai ) xAi T (ai ) +

n

yAi T (ai ) = T (x) + T (y).

i=1

Thus T satisfies Equation 4.1. Also, by Equation 4.6 applied to cx, T (cx) =

n

(cxAi ) T (ai ) = c

i=1

n

xAi T (ai ) = cT (x).

i=1

Thus T satisfies Equation 4.2. Therefore T is a linear transformation.

4.1.4. a. T : Rn → R such that T (x) = a · x, with a a fixed vector of Rn , is a linear transformation, since, by the properties of the dot product, T (x + y) = a · (x + y) = a · x + a · y =T (x) + T (y), and also T (cx) = a · (cx) = c (a · x) = cT (x). b. T : Rn → Rn such that T (x) = (A − λI)x, with A a fixed n × n matrix and λ any number, is a linear transformation, since, by the properties of matrix multiplication, T (x + y) = (A − λI)(x + y) = (A − λI)x + (A − λI)y = T (x) + T (y), and T (cx) = (A − λI) (cx) = c(A − λI)x = cT (x). c. T : Rn → Rm such that T (x) = Ax + b, with A a fixed m × n matrix and b a fixed non-zero vector of Rm , is not a linear transformation, since, for example, T (2x) = 2Ax + b but 2T (x) = 2Ax + 2b. d. T : Rn → R such that T (x) = |x| is not a linear transformation, since, for example, T ((−1) x) = | − x| = |x|, but (−1) T (x) = −|x|. e. T : Rn → R such that T (x) = xT a, with a a fixed vector of Rn , is a linear transformation by Part (a) above, since xT a = a · x. f. T : Rn → Rm such that T (x) = (aT x)b, with a a fixed vector of Rn and b a fixed vector of Rm , is a linear transformation, since, by the properties of matrix multiplication, T (x + y) = (aT (x + y))b = (aT x)b + (aT y)b = 48

T (x) + T (y) and T (cx) = (aT cx)b = c(aT x)b = cT (x).

4.1.6. We can write     x1 − x2 1x1 − 1x2 + 0x3 T (x) =  x2 − x3  =  0x1 + 1x2 − 1x3  x3 − x1 −1x1 + 0x2 + 1x3    1 −1 0 x1 −1 − 1   x2  . =  0 −1 0 1 x3 Thus



1 [T ] =  0 −1

−1 1 0

1 1

= [T ]

1 −1

 0 − 1 .

1

4.1.8. We have T

1 1

1 −1

and T

= [T ]

Hence we can write the matrix equation [T ]

1 1

1 −1

and multiplying from the right by

1 1

1 −1

−1

 1 = 1  1 

 1 =  −1  . −1

1  = 1 1

 1 −1  −1

1 −1

1 = 2 49





1 1

,

we get 

 1 1 −1  1 −1

1 1 [T ] =  1 2 1

Furthermore,

1 −1



1 = 0 0

 0 1 .

1

  

x1 0 x1 1  =  x2  . x2 1 x2



1  T (x) = [T ] x = 0 0

4.1.10. Writing B for the matrix (b1 , b2 , . . . , bn ), we have [T ]A = B and so [T ] = BA−1 .

4.1.12. The ax + by = 0 line has v = (b, −a)T as a direction vector. The projection of i onto v is

i·v b b v1 = v= 2 . a + b2 −a |v|2 y ax + by = 0

j v

t1 v1

v1 - i

1 t1 = i + 2 (v1 − i) = 2 a + b2

x

i

O Then, from the picture,

2b2 −2ab

−

1 0

1 = 2 a + b2

Similarly, j·v −a v2 = 2v = 2 a + b2 |v| 50

b −a

b2 − a2 −2ab

.

and 1 t2 = j + 2 (v2 − j) = 2 a + b2

−2ab 2a2

−

0 1

1 = 2 a + b2

−2ab a2 − b2

.

Thus, 1 [T ] = 2 a + b2

b2 − a2 −2ab −2ab a2 − b2

.

4.1.14. A linear transformation T : R2 → R2 maps a line p = p1 + tv1 , with v1 = 0, onto the line T (p) = T (p1 + tv1 ) = T (p1 ) + tT (v1 ) . Similarly, it maps the line p = p2 +sv2 , with v2 = 0, onto T (p) = T (p2 + sv2 ) = T (p2 ) + sT (v2 ) . Now p = p1 + tv1 is parallel to p = p2 + sv2 if and only if v2 = cv1 for some c = 0. But then also T (v2 ) = cT (v1 ) and the lines T (p) = T (p1 ) + tT (v1 ) and T (p) = T (p2 ) + sT (v2 ) = T (p2 ) + scT (v1 ) are also parallel.

4.1.16. It was shown in Example 4.1.12 that if T is the linear transformation T : R3 → R2 given by

x1 − x2 y = T (x) = , x1 + x3 then relative to the ordered bases given by the columns of   1 0 1 1 0  A= 1 0 1 1 and

B=

1 −1

1 1

the matrix TA,B that represents this transformation is

1 −1 − 2 − 1 TA,B = .

1 0 3 2 We want to verify Equation 4.27, that is, to show that yB = TAB xA , 51

where xA denotes the coordinate vector of x relative to A and yB that of y relative to B. Then    1 1 −1 x1 1 1 1   x2  xA = A−1 x =  −1 2 1 −1 1 x3   x1 + x2 − x3 1 −x1 + x2 + x3  , = 2 x −x +x 1

2

3

and

yB

1 = B y =B T (x) = 2

1 −x2 − x3 = .

2 2x1 − x2 + x3 −1

−1

1 −1

1 1

x1 − x2 x1 + x3

Thus, TAB xA =

1 2

1 = 2

−1 1

−2 0

−1 3

−x2 − x3 2x1 − x2 + x3

,



 x1 + x2 − x3 1 −x1 + x2 + x3  2 x1 − x2 + x3

the same as yB .

4.1.18. The matrix [A] that represents the transformation from the standard basis E = (1, x, . . . , xn ) to the new basis A = (1, 1 + x, 1 + x + x2 , . . . , 1 + x + · · · + xn ) is the (n + 1) × (n + 1) matrix   1 1 1 ··· 1  0 1 1 ··· 1    0 0 1 ··· 1 , [A] =   . .. .. ..   .. . . ··· .  0 0 0 ··· 1

that is, if p = (p0 , p1 , . . . , pn )T is the coefficient vector of the polynomial p0 +p1 x+· · ·+pn xn in the standard basis, then the coefficient vector relative to 52

the new basis is given implicitly by p = [A] pA or explicitly by pA = [A]−1 p.

The differentiation map from E to E is given by a slight modification of the matrix in Equation 4.41 as   0 0 0 ··· 0  0 1 0 ··· 0    0 0 2 ··· 0 .

[D] =   . .. .. ..   .. . . ··· .  0 0 0 ··· n

Thus the differentiation matrix relative to A = B can be defined by DA,B pA = [A]−1 [D] p = [A]−1 [D] [A] pA , where   1 −1 0 ··· 0  0 1 − 1 ··· 0    −1  , 0 0 1 · · · 0 [A] =   . . . .  .. .. .. · · · ..  0 0 0 ··· 1 and so



0  0  −1 0 DA,B = [A] [D] [A] =   .  ..

0

−1 1 0 .. .

0

−1 −1 2 .. .

0

··· ··· ··· ··· ···

 −1 −1   −1  . ..  .  n

4.1.20. Similarly to Example 4.1.13 in the text, consider the ordered bases A = (1, x, . . . , xn) and B = (1, x, . . . , xn+1 ). Then, using the notation aj = xj−1 for j = 1, 2, . . . , n + 1 and bi = xi−1 for i = 1, 2, . . . , n + 2, we have X(a1 ) = 0b1 + 1b2 + 0b3 + · · · + 0bn+2 , X(a2 ) = 0b1 + 0b2 + 1b3 + · · · + 0bn+2 , ··· X(an+1 ) = 0b1 + 0b2 + 0b3 + · · · + 1bn+2 .

According to Equation 4.29, the coefficients of the bi here form the transpose of the matrix that represents X relative to these bases. Thus XA,B is the 53

(n + 2) × (n + 1) matrix



0  1   0 XA,B =   0  .  ..

0

0 0 1 0 .. .

0

0 0 0 1 .. .

0

··· ··· ··· ···

··· ···

 0 0   0   0 . ..  . 

1

4.2.2. a. If a = 0, then Range(T ) = R, Ker(T ) is the (n − 1)-dimensional subspace {a}⊥ = {x : x = aT x = 0} of Rn and T is onto but not one-to-one for n > 1. If a = 0, then Range(T ) = {0} ⊂ R, Ker(T ) = Rn and T is neither onto nor one-to-one. b. Range(T ) = Col(A − λI), Ker(T ) = Null(A− λI). If rank(A − λI) = n, then T is onto and one-to-one. If, however, rank(A − λI) < n, as is the case when λ is an eigenvalue (see Chapter 7), then T is neither onto nor one-toone. c. Although this T is not linear, the concepts in question are still applicable and Range(T ) = {y + b : y ∈ Col(A)} ⊂ Rm and Ker(T ) is the solution set of Ax = −b. If m > n, then T is not onto and is one-to-one only if rank(A) = n. If m = n, then T is both onto and one-to-one if and only if rank(A) = n. If m < n, then T is not one-to-one and is onto only if rank(A) = m. T is not onto and is one-to-one if and only if n = 0. d. Again this T is not a linear transformation, but Range(T ) = {y : y ≥ 0} and Ker(T ) = {0} . e. Same as Part (a) above, since xT a = a · x. f. Range(T ) = Span ; thus T is onto if and only if m = 1 or a = 0 or {b} T b = 0. Ker(T ) = x : a x = 0 , the orthogonal complement of Span {a} ; hence T is one-to-one if and only if n = 1 and a = 0 and b = 0.

4.2.4. In Exercise 4.1.5, the matrix representing the linear transformation from R2 to R3 was determined to be   1 −1 3 .

[T ] =  2 3 2 This matrix has rank 2, and so Range(T ) = Col[T ], a two-dimensional sub54

space of R3 ; thus T is not onto. Also Ker(T ) = Null[T ] = {0} ; therefore T is one-to-one.

In Exercise 4.1.6, the matrix representing the linear transformation from R3 to R3 was determined to be   1 −1 0 1 − 1 .

[T ] =  0 −1 0 1

This matrix is row-eqivalent to the reduced echelon matrix   1 0 0 1 − 1 .

[T ] =  0 0 0 0 ! " Hence, Range(T ) = Span (1, 0, −1)T , (−1, 1, 0)T , a two-dimensional

3 subspace thus T is not onto. Also, we find that Ker(T ) = Null[T ] = ! of R ; " T Span (1, 1, 1) ; therefore T is not one-to-one.

In Exercise 4.1.7, the transformation is T : Rn → Rm such that T (x) = (a x)b, with a a fixed vector of Rn and b a fixed vector of Rm . Thus Range(T ) = {tb|t ∈ R} , a one-dimensional subspace of Rm if a = 0 and b = 0 and a zero-dimensional subspace of Rm if a = 0 or b = 0. Thus T is not onto unless m = 1.or a = 0 or b = 0. If a = 0 and b = 0, then

Ker(T ) = x ∈ Rn |aT x = 0 , an (n − 1)-dimensional subspace of Rn , and Ker(T ) = Rn if a = 0 or b = 0. Therefore T is not one-to-one unless n = 1 and a = 0 and b = 0.

In Exercise 4.1.8, the matrix representing the linear transformation from R2 to R3 was determined to be   1 0 1 .

[T ] =  0 0 1 T

This matrix has rank 2, and so Range(T ) = Col[T ], a two-dimensional subspace of R3 ; thus T is not onto. Also Ker(T ) = Null[T ] = {0} ; therefore T is one-to-one.

In Exercise 4.1.9, the matrix representing the linear transformation from 55

R3 to R3 was determined to be [T ] =

1 0

0 0

0 1

.

Thus Range(T ) = Col[T ]!= R2 , and " so T is onto. Also, we find that T Ker(T ) = Null[T ] = Span (0, 1, 0) ; therefore T is not one-to-one.

4.2.6. a) Range(T ) is not empty, since T (0) = 0 is always a member of

it. b) It is closed under addition, since, if u = T (x) and v = T (y) are two arbitrary vectors in Range(T ), then, by the linearity of T, we have u + v = T (x) + T (y) = T (x + y) and thus x + y ∈ Range(T ). c) It is closed under multiplication by scalars, since if u = T (x) is an arbitrary vector in Range(T ) and c any scalar, then, by the linearity of T, we have cu = cT (x) = T (cx) and so cu ∈ Range(T ).

4.2.8. Since T is onto, there exist ai vectors, for i = 1, 2, . . . , n, in U such that, for a given basis B = {b1 , b2 , . . . , bn }, we have T (ai ) = bi . Next, let us test the ai vectors for independence: Assume ni=1 ci ai = 0. Then, by the linearity of T and the fact that for every linear transformation T (0) = 0, we have n n n

T ci ai = ci T (ai ) = ci bi = T (0) = 0. i=1

i=1

i=1

The linear independence of the bi implies that ci = 0 for all i. Hence the ai vectors are independent, and consequently, by the result of Exercise 3.5.40, they form a basis for U. All that remains to show is that T is a one-to-one mapping of U onto V. But this proof is now easy: Let a ∈U such that T (a) = 0. Since the ai vectors form a basis for U, we can write a = ni=1 ci ai . But then 0 = T (a) = T

n

ci ai

=

i=1

n

i=1

ci T (ai ) =

n

ci bi .

i=1

The linear independence of the bi implies that ci = 0 for all i, and therefore a = 0. Thus the kernel of T contains just the zero vector, which, by Exercise 4.2.3, implies that T is one-to-one.

4.2.10. If TA,B represents an isomorphism of U onto V , then rank (TA,B ) = 56

−1 rank (T ) = dim (V ) = dim (U) . Thus TA,B is nonsingular. TA,B represents −1 the reverse isomorphism T .

4.2.12. One possible choice for K is the subspace of R3 generated by e2 and e3 , that is,

K 1 = {x : x = ae2 + be3 , for a, b ∈ R} , and another choice is the subspace of R3 generated by f = e2 + e3 and e3 , that is, K 2 = {x : x = af + be3 , for a, b ∈ R} .

4.3.2. First translate by (−1, 2) to move the first vertex to the origin, then shrink in the x-direction by a factor of 1/3, and in the y-direction by a factor of 1/4 (other solutions would be more involved): 

1 0  0 1/4 L = 0 0  1/3 0 1/4 =  0 0 0

 0 1/3   0 0 1 0  −1/3 1/2  .

1

 0 1   0 0 1 0

0 1 0

0 1 0

 −1 2  1

# 4.3.4. a) Let p = |p| and p12 = p21 + p22 , and let R1 denote the matrix of the rotation by θ about the z-axis, where θ stands for the angle from the y-axis to the vector (p1 , p2 , 0)T . Then     cos θ − sinθ 0 − p 0 p 2 1 1  cos θ 0 = p1 p2 0 .

R1 =  sinθ p 12 0 0 1 0 0 p 12

Similarly, the rotation by the angle φ about the x-axis is given by   p 0 0 1 p3 − p12  .

R2 =  0 p 0 p p 12

57

3

Hence, 

p 0 0 1  0 p3 − p12 R = R2 R1 = pp12 0 p p3 12   −pp1 0 pp2 1  p1 p3 p2 p3 −p212  . = pp12 p p pp pp 1 12

2 12



p2   p1 0

 0 0  p12

− p1 p2 0

3 12

# b) Let p = |p| and p23 = p22 + p23 , and let R1 denote the matrix of the rotation by ψ about the x-axis, where ψ stands for the angle from the z-axis to the vector (0, p2 , p3 )T . Then     0 1 0 0 p23 0 1  0 p3 − p2  . cos ψ − sinψ  = R1 =  0 p 23 0 p p 0 sinψ cos ψ 2

3

Similarly, the rotation by the angle φ about the y-axis is given by   p23 0 − p 1 1 p 0 .

R 2 =  0 p p 0 p 1

23

Thus,

R



 p23 0 p23 − p1 1  0 p 0  0 = R2 R1 = pp23 0 p23 0 p1  2  p −p1 p2 −p1 p3 1  23 0 pp3 −pp2  . = pp23 p p pp pp 1 23

4.3.6.

2 23

0 p3 p2

0 − p2 p3

 

3 23



1  0 T (t1 , t2 , t3 ) =   0 0

0 1 0 0

0 0 1 0

 t1 t2  . t3  1

4.3.8. First obtain the 4 × 4 matrix Rθ that represents in homogeneous 58

coordinates the rotation by an angle θ about the p = (1, 1, 1)T vector, by augmenting the matrix Rθ of Equation 4.98, as

Rθ 0 Rθ = .

0T 1 Then the required matrix is given by the product T (1, 0, 0) Rθ , where T (1, 0, 0) is the shift matrix from Exercise 4.3.6 with t1 = 1, t2 = t3 = 0.

4.3.10. By Corollary 3.6.1, the transition matrix from the standard basis to the new basis B is S = B = (u, v, n) , and, since S is orthogonal, the transition matrix in the reverse direction is S −1 = S T . (See Theorem 5.2.2.)

Also, for any vector x, the corresponding coordinate vector relative to the new basis is xB = S −1 x. If T denotes the matrix that deletes the n-component, then T xB = T S −1 x is the projection in the new basis, and xV = ST xB = ST S −1 x is the projection in the standard basis. Thus the projection matrix is P (u, v) = ST S −1   u1 1 0 v1 n1 1 v2 n2   0 =  u2 u3 v3 n3 0 0  2 2 u1 + v1 u1 u2 + v1 v2  u1 u2 + v1 v2 u22 + v22 = u1 u3 + v1 v3 u2 u3 + v2 v3

 0 u1 0   v1 0 n1

u2 v2 n2

 u1 u3 + v1 v3 u2 u3 + v2 v3  . u23 + v32

 u3 v3  n3

Another way of looking at the equation P (u, v) = ST S −1 is given by the result of Exercise 7.1.16.

5.1.2.   1 0 1 .

A= 0 0 0 5.1.4. We apply Theorem 5.1.1 with     1 2 1 A =  1 − 1  . and p =  −1  2 1 2

Hence, q = A(AT A)−1 AT p = 13 (5, −1, 4)T .

59

5.1.6. 

5.1.8.

1  P = 0 0

0 0 0 

a2 1  ab P = 2 a + b2 + c2 ac

 0 0 .

0 ab b2 bc

 ac bc  . c2

5.1.10. The orthogonal complement of the x = y plane is the line given by x = t, y = −t, z = 0. The projection matrix Q that represents the projection onto it is   1 −1 0 1 1 0 .

Q =  −1 2 0 0 0

This matrix can be obtained either from the solution of Exercise 5.1.8, or from Exercises 5.1.9 and 5.1.12.

5.1.12. I − P is a projection matrix, because it is symmetric and idempotent: (I − P )T = I T − P T = I − P and (I − P )2 = (I − P ) (I − P ) = I 2 − IP − P I + P 2 = I − 2P + P = I − P. I − P represents the projection onto Left-null (A) = Col (A)⊥ , because for any x ∈ Col (A) we have (I − P ) x = Ix−P x = x − x = 0, and for any y ∈ Col (A)⊥ we have (I − P ) y = Iy−P y = y − 0 = y.

5.1.14. a. This result follows from Lemma 5.1.3 by transposition. b. Clearly xR is a solution: AxR = AAT (AAT )−1 b = b. That it is in Row(A) can be seen by multiplying it by the transpose of any y ∈ Null (A) : then Ay = 0 implies that yT AT = 0T and yT xR = yT AT (AAT )−1 b = 0. Since Row(A) = Null (A)⊥ , it follows that xR ∈ Row(A). c. For the mapping A+ given by xR = AT (AAT )−1 b, we have Ker(A+ ) = {0} , because AT (AAT )−1 b = 0 implies AAT (AAT )−1 b = b = 0. Then the results of Exercises 4.2.3 and 4.2.9 show that A+ is an isomorphism.

5.1.16. 65x + 118y = −26.

60

5.1.18. 

x1 AT A =  y1 1

x2 y2 1

 2 xi =  xi yi xi

and



x1 AT z =  y1 1

x2 y2 1

··· ··· ···

xi y2i yi yi ··· ··· ···



x1 xm  x 2 ym    ...

1 xm  xi yi  m 

y1 y2 .. . ym

 1 1  ..  .  1

   z1 xm  z  x z i i

2    ym    ...  = yi zi .

1 zi zm 



5.1.20.

and



x21 2 2 2 x1 x2 ··· xm  x2 2 x2 ··· xm   AT A =  x1  ...

1 1 ··· 1 x2m  4 3 2  xi xi2 xi x xi  =  x3i i 2 xi xi m 



x21

AT y =  x1 1



x22 x2 1

··· ··· ···

x1 x2 .. . xm

 1 1  ..  .  1

  2  y1  y2  xi yi    xm   =  ...  xi yi .

1 yi ym x2m





5.2.2. From Equation 4.115, the viewplane V is the column space of the 61

matrix A = (u, v), that is,  

u1 v1 u1 T   v2 A = u2 and A = v1 u3 v3

u2 v2

u3 v3

.

Hence P = AAT = P (u, v) .

5.2.4.Orthogonal matrices preserve angles: Let Q be any n×n orthogonal matrix, x, y any vectors in Rn , θ the angle between Qx and Qy, and φ the angle between x and y. Then cos θ =

(Qx) · (Qy) (Qx)T Qy xT QT Qy xT y x·y = = = = = cos φ. |Qx| |Qy| |Qx| |Qy| |Qx| |Qy| |x| |y| |x| |y|

Since the angles are in [0, π] and the cosine function is one-to-one there, we obtain θ = φ.

5.2.6. (P Q)T = QT P T = Q−1 P −1 = (P Q)−1 .

5.2.8. 1. As in Example 5.2.2, let b1 = a1 , a2 − p2 = a2 − and

a2 · b1 1 b1 = (1, 0, 0, 1)T − (0, 0, −1, 1)T , b1 · b1 2

b2 = 2 (a2 − p2 ) = (2, 0, 0, 2)T − (0, 0, −1, 1)T = (2, 0, 1, 1)T .

Similarly, let a3 · b1 a3 · b2 b1 − b2 b1 · b1 b2 · b2 1 1 = (1, 0, −1, 0)T − (0, 0, −1, 1)T − (2, 0, 1, 1)T , 2 6

a3 − p3 = a3 −

and b3 = (6, 0, −6, 0)T − (0, 0, −3, 3)T − (2, 0, 1, 1)T = (4, 0, −4, −4)T .

Hence 1 1 1 q1 = √ b1 , q2 = √ b2 , q3 = b3 48 2 6 form an orthonormal basis for U. 62

2. On the other hand, it is clear that e2 is orthogonal to U, and so it extends the basis above to an orthonormal basis for R4 . But then e1 , e3 , e4 must also form an orthonormal basis for U. 3. The QR factorization of A is



 1 0   −1  0 √ √  √ √   √ 0 2/ 6 1/ 3 2 1/ 2 1/  √2 √ 0√ 0√ 0√   0  =  6/2 1/  −1/ 2 1/ 6 −1/ 3  √6 . √ √ √ 0 0 2/ 3 1/ 2 1/ 6 −1/ 3

0  0 A =   −1 1 

1 0 0 1

6.1.2.

0 0 0 1 1 2 1 2

4

0 0 3 = 3 4 0 3 = 4

1 3 3 3 3 1 2 1

1 1 1 1 1 1

0 3 = (−3) 0 − 4 − 1 = (−3) 4 0 0 1 2 1 2

1 1 1 1 1 1

1 2 = 3 0 1 2 = 21.

3 0 0 −4 −1 0 0 7 63

2 3 = 1

6.1.4.

−1 −2 2

0 0 − 1

0 0 1

1 −1 1

−1 −2 2

0 0 −1 (−1) 0 0 1

0 −3 3

−1 −2 2

0 −3 3 (−1) 0 0 1

0 0 0

−1

1 −2 2 1

0 0 0 −1 0

= = − 1 0 1 − 1 0 0 −3 −1 3 0

−1 1 −2 2 1

0 −3 0 3 0 = (−1) =

−1 0 1 − 1 0 0 0 0 −1 0

1 0 = 3. − 1 −1

6.1.6. By applying Axiom 1 repeatedly with s = c, t = 0, and i = 1, 2, . . . , n, we get det (cA) = |ca1 ca2 · · · can | = c|a1 ca2 · · · can | = c2 |a1 a2 ca3 · · · can | = · · · = cn det (A) .

6.1.8. Yes, a similar statement is true for any n. The fact that all products in Theorem 6.1.4 are zero, except the product of the secondary diagonal elements, can be proved much the same way as for Theorem 6.1.7; just replace the word "diagonal" by "secondary diagonal" everywhere in the proof and call the zero entries above the secondary diagonal the bad entries. Thus, Equation 6.1.17 becomes det(A) = :(P )an1 an−1,2 · · · a1n ,

where P = (n, n − 1, . . . , 1) , that is, P is the permutation of the first n natural numbers in descending order. This P requires [n/2] switches around the middle to be brought to the natural order (where [n/2] is the greatest integer ≤ n/2) and so :(P ) = (−1)[n/2] .

An alternative way of proving the statement is by exchanging columns (or rows) of the given matrix so as to change it into upper triangular form, counting the number of exchanges, and applying Theorem 6.1.1 to each exchange.

6.1.10. A and B are similar if there exists an invertible matrix S such that 64

B = S −1 AS. In that case det(B) = det(S −1 ) det(A) det(S) =

1 det(A) det(S) = det(A). det(S)

6.1.12. By Theorem 6.1.6, det(AT ) = det(A) and by the result in Exercise 6.1.6, with n = 3 and c = −1, det(−A) = (−1)3 det(A) = − det(A).

Thus, for a 3 × 3 skew-symmetric matrix AT = −A implies det(A) = − det(A), from which it follows that det(A) = 0. The same argument proves the same result for any odd positive integer n > 1, but no such relation holds for even n.

6.1.14.

2 1 a a

1 b b2

1 c c2

1 1 a a2 a a2

1 b + a = 0 b − a b2 − a2 = (b − a) (c − a) 0 2 2 0 c−a c −a 0 1 c+a

1 a a2

1 b + a = (b − a) (c − a) (c − b) . = (b − a) (c − a) 0 0 0 c−b

6.1.16. By the definition of linear independence, 1, x, x2 are linearly independent in the vector space P if 1s1 + xs2 + x2 s3 = 0 for all x implies that s1 = s2 = s3 = 0. It is sufficient to consider only three distinct values a, b, c for x. Then the equation above becomes the system 1s1 + as2 + a2 s3 = 0 1s1 + bs2 + b2 s3 = 0 1s1 + cs2 + c2 s3 = 0.

The determinant of this system is the Vandermonde determinant of Exercise 6.1.14. The right-hand side of that equation shows that it is not zero. Thus,

by Theorem 6.1.8, the corresponding matrix is nonsingular, and so the system has only the trivial solution.

65

6.2.2. Expansion along the first row gives

0

1 2

0 4 4 3 3 0 4

0 3 = 0 −1 + 2

2 1 3 1 3 2 3 2 1 = 0 (0 − 6) − 1 (4 − 9) + 2 (8 − 0) = 21.

6.2.4. Expansion along the second row and then along the second row of the resulting determinant gives

−1

−2 2 1

−1 −2 1 0

0 −1 0

0 − 1 = 0 = − (−1) 0 0 1 − 1

1 −1 −1 1 −1 1 −1

−2 2 −1 (−1) = 3.

1 −1 6.2.6.

and

−1 x1 = 3

1 x2 = 2

3 1 ÷ 1 2

−1 1 ÷ 3 2

3 = 2, 1

3 = −1.

1

6.2.8. |A| = 4, |b a2 a3 a4 | = −15, |a1 b a3 a4 | = 1, |a1 a2 b a4 | = 7, |a1 a2 a3 b| = 2, and so x1 = 15/4, x2 = 1/4, x3 = 7/4, x4 = 2/4.

6.2.10. |A| = −2, A11 = −12, A12 = −2, A13 = 6, A21 = 5, A22 = 1, A23 = −3, A31 = 4, A32 = 0, A33 = −2. Thus,   −12 5 4 1 0  Adj (A) =  −2 6 −3 −2 and



1 −12 A−1 = −  −2 2 6

5 1 −3

  4 6 −5/2 0  =  1 −1/2 −2 −3 3/2 66

 −2 0 .

1

6.2.12. If we apply Theorem 6.2.3 to A−1 in place of A, we get −1 −1 Adj (A−1 ) Adj (A−1 ) A = = , |A−1 | 1/ |A|

and so

−1

Adj A

(A−1 ) = |A|

−1

−1 = |A| A−1 = (Adj (A))−1 .

6.2.14. Expanding the determinant along the first row, we we can see that this is a linear equation; and substituting the coordinates of either of the given points for x, y in the determinant, we get two equal rows, which makes the determinant vanish. Thus this is an equation of the line containing the given points.

6.2.16. Expanding the determinant along the first row, we obtain a linear combination of the elements of that row. That combination can be rearranged into the given form, because the the coefficient of x2 +y 2 is a nonzero determinant by the result in Exercise 6.2.13, since the given points are noncollinear.

Substituting the coordinates of any of the given points for x, y in the determinant, we get two equal rows, which makes the determinant vanish, and so these points lie on the circle.

6.2.18.

a1

a1

a a 1 a a 1 2 3 2 3

1 b1 b2 b3 1 1 b1 − a1 b2 − a2 b3 − a3 0 = c2 c3 1 6 c1 − a1 c2 − a2 c3 − a3 0 6 c1 d1 d1 − a1 d2 − a2 d2 d3 1 d3 − a3 0

b − a1 1 1 = − c1 − a1 6 d −a 1 1

b2 − a2 c2 − a2 d2 − a2

b3 − a3 1 c3 − a3 = − |b − a c − a d − a| .

6 d3 − a3

By the remarks in the last paragraph of Section 6.2, the absolute value of the determinant above gives the volume of the parallelepiped with the given edges. (A geometric proof, involving the cross product, is suggested in Exercise 6.3.5.) The volume of the tetrahedron is one sixth of that of the corresponding parallelepiped.

6.2.20. Write T (e1 ) = t1 and T (e2 ) = t2 . Then, by Theorem 4.1.2, 67

[T ] = [t1 t2 ] . Thus the unit square is mapped to the parallelogram spanned by t1 and t2 , which has area |det ([T ])| .

6.3.2. We may choose

i j k

−→ −→ 0 1 = −i + 2j − k n = AB × AC = −1 2 1 0 as the normal vector of the plane, and so the plane’s equation can be written as n · (r − a) = 0, that is, as − (x − 1) + 2 (y + 1) − (z − 2) = 0.

6.3.4.

w · (u × v) = w · (u2 v3 − u3 v2 )i − w · (u1 v3 − u3 v1 )j + w · (u1 v2 − u2 v1 )k = w1 (u2 v3 − u3 v2 ) − w2 (u1 v3 − u3 v1 ) + w3 (u1 v2 − u2 v1 )

w1 w2 w3 w1 u v 1 1

u2 u3 = w2 u2 v2 = u1 v1 v2 v3 w3 u3 v3 = det(w, u, v) = det(u, v, w).

Similarly,

u · (v × w) = det(u, v, w) and v · (w × u) = det(v, w, u) = det(u, v, w).

6.3.6. Let the edge vectors be directed as shown in the diagram, and let n1 = 12 a1 × a2 , n2 = 12 a2 × a3 , n3 = 12 a3 × a4 , and n4 = 12 a4 × a5 .

a3

a2 a4

a5

a1

Then a4 = a1 − a3 and a5 = a1 − a2 . Hence 2 (n1 + n2 + n3 + n4 ) = 68

a1 × a2 + a2 × a3 + a3 × (a1 − a3 ) + (a1 − a3 ) × (a1 − a2 ) = a1 × a2 + a2 × a3 + a3 × a1 − a3 × a1 − a1 × a2 + a3 × a2 = 0.

6.3.8. To prove the first statement, note that (a × b) × (c × d) = [a · (c × d)] b − [b · (c × d) ]a by Thm. 6.3.1, Part 12, = [ det (a, c, d)]b − [ det (b, c, d)]a by Thm. 6.3.1, Part 11.

For the second statement, we have (a × b) × (c × d) = [(a × b) · d)] c − [(a × b) · c]d by Thm. 6.3.1, Part 13, = [ det (a, b, d)]c − [ det (a, b, c)]d by Thm. 6.3.1, Part 11.

6.3.10. When air flows toward the center of a low pressure area with velocity v, then in the northern hemisphere the Coriolis force F points counterclockwise. For example, if v points from south to north as in the diagram, then, by the right-hand rule, F = v × ω points east, which corresponds to a counterclockwise direction around the center of the low pressure area to which v points. v F

On the other hand, in the southern hemisphere the circulation of hurricanes must be clockwise. There, if v points from south to north, then v × ω points west, which corresponds to a clockwise direction.

7.1.2. λ1 = λ2 = 2, and every s ∈ R2 is an eigenvector.

7.1.4. The characteristic equation is

0−λ 0

|A − λI| = = 0, 1 0−λ or equivalently, (−λ)2 = 0. Thus λ = 0 is the only eigenvalue and it has 69

algebraic multiplicity two.

The eigenvectors are the solutions of

0 0 s1 0 = , (A − 0I)s = 0 1 0 s2 or in echelon form if

0 (A − 0I)s = = .

0

0 Thus, s2 is free and s1 = 0 or, equivalently, s = s , which shows that 1 λ = 0 has geometric multiplicity 1, and so A is defective.

7.1.6. The characteristic equation is

2−λ 0 1

2−λ 0 = 0, |A − λI| = 0 0 0 2−λ 1 0

0 0

s1 s2

or equivalently, (2 − λ)3 = 0. Thus λ = 2 is the only eigenvalue and it has algebraic multiplicity three. The eigenvectors are the solutions of      0 0 1 0 s1      0 0 s2 = 0  .

(A − 2I)s = 0 0 0 0 s3 0

The variables s1 and s2 are free and s3 = 0. Thus the geometric multiplicity of the eigenvalue λ = 2 is two, the corresponding eigenspace is spanned by s1 = e1 and s2 = e2 , and A is defective.

7.1.8. The characteristic equation is

1−λ

0 0 1

0 1−λ 1 1

|A − λI| = = 0, 0 2−λ 0 0 0 0 0 2−λ or equivalently, (1−λ)2 (2−λ)2 = 0. Thus λ1 = 1 and λ2 = 2 are eigenvalues of algebraic multiplicity two.

70

The eigenvectors corresponding to λ1 = 1 are the solutions of     1−1 0 0 1 0 0 0 1  0  1−1 1 1  0 1 1   s =  0  s = 0,  0  0 0 2−1 0  0 1 0  0 0 0 2−1 0 0 0 1

or in reduced echelon form, of  0 0  0 0   0 0 0 0

1 0 0 0

 0 1   s = 0.

0  0

Hence s1 and s2 are free, s3 = s4 = 0, and so the vectors s = se1 + te2 form the two-dimensional eigenspace belonging to λ1 = 1.

The eigenvectors corresponding to λ2 = 2 are the solutions of     1−2 0 0 1 −1 0 0 1  0  1−2 1 1  1 1   s =  0 − 1  s = 0,  0   0 2−2 0 0 0 0 0  0 0 0 2−2 0 0 0 0 This system is in echelon form, s3 and s4 are free, and so, setting s3 = s and s4 = t, we get s1 = t and s2 = s + t. Hence the vectors     0 1  1   1     s = s  1  + t 0 .

0 1

form the two-dimensional eigenspace belonging to λ2 = 2.

7.1.10. If B = A + cI, then |B − λB I| = |A − (λB − c) I| and so the characteristic equation of B is equivalent to the characteristic equation of A with λB − c = λA or λB = λA + c. Thus the eigenvalues are paired by this equation, and As = λA s is equivalent to (A + cI) s = (λA + c) s, that is, to Bs = λB s with the same s.

7.1.12. Multiplying As = λs by A, we get A2 s = λAs = λ2 s. Hence s is an eigenvector of A2 as well, belonging to the eigenvalue λ2 .

71

7.1.14. The characteristic equations of A and AT are equivalent: From the definition of transposes we have (A − λI)T = AT −λI and by Theorem 6.1.6, T T

| (A − λI) | = |A − λI|. Thus |A − λI| = 0 if and only if A − λI = 0.

7.1.16. By definition, P is a projection matrix if and only if P = P 2 and P = P T . Thus, P s = λs implies P s = P 2 s = λP s = λ2 s and so, for any nonzero eigenvector s, λ = λ2 . Hence, the eigenvalues are 1 and, unless P = I, 0. The eigenvectors belonging to the eigenvalue λ = 1 are the solutions of P s = s. Since P s ∈ Col (P ) , the eigenvectors are the vectors s of Col (P ) . Since P = P T , we also have Row (P ) = Col (P ) , and so this is the eigenspace corresponding to λ = 1. Similarly, the eigenspace corresponding to λ = 0 is Null (P ) = Left-null (P ) .

7.1.18. Let u and v be eigenvectors of A and AT belonging to the distinct eigenvalues λ1 and λ2 respectively, that is, let Au = λ1 u and AT v = λ2 v.

Left-multiplying these equations by vT and uT , we get vT Au = λ1 vT u and uT AT v = λ2 uT v.

Taking the transpose of the last equation, we can change it to vT Au = λ2 vT u.

Thus (λ2 − λ1 ) vT u = 0,

and since λ2 − λ1 = 0, we must have vT u = 0.

7.2.2. Ax = SΛx S −1 , where  x  λ1 0 ··· 0  0 λx2 · · · 0  , Λx =    ··· x 0 0 ··· λn

whenever λxi exists for every i.

7.2.4. The characteristic equation is

3−λ 2

|A − λI| = = 0, 2 3−λ 72

or equivalently, (3 − λ)2 − 22 = 0. The solutions are λ1 = 1 and λ2 = 5, 1 and corresponding normalized eigenvectors are s1 = √12 and s2 = −1

1 √1 . Hence 2 1

1 1 1 S=√ 1 2 −1 is an orthogonal matrix, and so S

−1

1 =S = √ 2 T

1 1

−1 1

.

Also, Λ=

1 0

0 5

.

Thus, 100

A

1 = SΛ S = 2

100 1 1+5 = 2 −1 + 5100 100

−1

1 −1

1 1

1 0

100

.

1 0

−1+5 1 + 5100

0 5100

1 1

−1 1

1 1

−1 1

Similarly, 1/2

A

1 = SΛ S = 2

1 1 + 51/2 = 2 −1 + 51/2 1/2

−1

1 −1

1 1

1/2

−1+5 1 + 51/2

7.2.6. From the solution of Exercise 7.1.8,  1 0 0  0 1 1 S=  0 0 1 0 0 0 73

0 51/2

.

 1 1   0  1

and



Hence

1  0 Λ=  0 0

and

1  0 S −1 =   0 0



0 1 0 0

0 0 2 0

 0 0  .

0  2

0 1 0 0

0 −1 1 0

 −1 −1   0  1

A4 = SΛ4 S −1 = 

1  0   0 0

0 1 0 0

0 1 1 0

 1 1  0 1   0  0 1 0 

1  0 =  0 0

0 1 0 0

0 0 16 0 0 1 0 0

0 15 16 0

 0 1  0 0   0  0 16 0

0 1 0 0

 15 15  .

0  16

0 −1 1 0

 −1 −1   0  1

7.2.8. If s is an eigenvector of A belonging to the eigenvalue λ, then As = λs implies B (T s) = T AT −1 (T s) = T As = λ (T s) .

Conversely, if T s is an eigenvector of B belonging to the eigenvalue λ, then B (T s) = λ (T s) implies As = T −1 BT s =T −1 λT s = λT −1 T s =λs.

7.2.10. Substituting the expressions (7.91) and (7.92) into the initial conditions (7.93) and (7.94) , we obtain c11 + c12 = 0 and c21 + c22 = Q. Now, substituting the expressions (7.91) and (7.92) into the differential equations 74

(7.83) and (7.84) , and considering the above relations, we get

R λ1 t 1 Q λ2 t c11 λ1 eλ1 t − λ2 eλ2 t = − e − eλ2 t − c21 eλ1 t − eλ2 t − e L LC LC and

c21 λ1 eλ1 t − λ2 eλ2 t + Qλ2 eλ2 t = c11 eλ1 t − eλ2 t .

Setting here t = 0 results in

c11 (λ1 − λ2 ) = −

Q LC

and c21 (λ1 − λ2 ) = −Qλ2 .

These equations in conjunction with Equations (7.89) and (7.90) yield the desired formulas for the cij .

7.2.12. a. From

(At)2 (At)3 At + + · · · u0 u(t) = e u0 = I + At + 2!

3! we get

1 2 1 3 2 O + A + A 2t + A 3t + · · · u0 2!

3!

2 (At) (At)3 + + · · · u0 = Au. = A I + At + 2!

3!

du(t) = dt

Similarly,

(A0)2 (A0)3 u(0) = e u0 = I + A0 + + + · · · u0 = Iu0 = u0 .

2!

3!

A0

b.

(SΛS −1 t)2 (SΛS −1 t)3 −1 eAt = eSΛS t = I + SΛS −1 t + + + ··· 2!

3!

(Λt)2 (Λt)3 = S I + Λt + + + · · · S −1 = SeΛt S −1 , 2!

3!

75

and, by Equation (7.51),

Λt (Λt)2 (Λt)3 e ii = I + Λt + + + ··· = eλi t 2!

3! ii Λt and e ij = 0 if i = j. Thus 

eλ1 t  0 eΛt =   0

0 eλ2 t

··· ··· ··· ···

0

0 0

eλn t



 . 

Writing skl for the kl component of S −1 and u0l for the l component of u0 , we obtain

SeΛt S −1 u0 j = eλk t sjk skl u0l k

.

l

Letting ck denote the vector that has the sum in parentheses on the right above as its jth component results in the desired expression Λt

−1

u (t) = Se S u0 =

n

eλk t ck .

k=1

c. From the given data and Equation (7.85) , the matrix of the system is

−5 − 4 A= .

1 0 The characteristic equation yields λ1 = −4 and λ2 = −1, and a corresponding diagonalizing matrix is

−4 1 S= , 1 −1 with inverse S

−1

1 =− 3

76

1 1

1 4

.

Thus, u (t) = SeΛt S −1 u0

−4t

1 −4 1 e 0 1 = − −t 1 − 1 0 e 1 3

1 40 (e−4t − e−t ) = , 3 10 (4e−t − e−4t )

1 4

with the first component giving i (t) and the second one q (t) .

Plots: 6e+09 5e+09 4e+09 3e+09

i

2e+09 1e+09

-5

-4

-3

t -2

-1

0

1

i = 40 (e−4t − e−t ) 3 -5

-4

-3

t -2

-1

1 0 -2e+08 -4e+08 -6e+08 -8e+08 -1e+09 -1.2e+09 -1.4e+09

q = 10 (4e−t − e−4t ) 3 77

q

0 10

d. From the given data and Equation (7.85) , the matrix of this system is

−2 − 1 A= .

1 0 This matrix is defective, and therefore undiagonalizable. The power series approach can, however, be used as follows: We compute

3 2 −4 − 3 5 4 2 3 4 A = , A = , A = , −2 − 1 3 2 −4 − 3 and so on. Hence the components of eAt = I + At +

(At)2 (At)3 + + ··· 2!

3!

are At 3t2 4t3 − + · · · = e−t (1 − t) , e 11 = 1 − 2t + 2!

3!

At

t3 t4 e 21 = − eAt 12 = t − t2 + − + · · · = −te−t , 2! 3!

Thus

At t2 2t3 3t4 e 22 = 1 − − + − · · · = e−t (1 + t) .

2!

3!

4!

At

e

−t

=e

1 − t −t −t 1 + t

and At

−t

u(t) = e u0 = 10e Plots: 78

−t 1+t

.

7000 6000 5000 4000

i

3000 2000 1000 -5

-4

-3

t -2

0

-1

1

−t

i = −10te -5

-4

-3

t -2

-1

1 0 -1000 -2000 -3000

q

-4000 -5000

q = 10 (1 + t) e−t 7.3.2. The matrix of the given quadratic form is 1 A= 25

7 24

24 −7

.

The eigenvalues and corresponding unit eigenvectors are λ1 = 1, λ2 = −1, s1 = 15 (4.3)T , s2 = 15 (−3.4)T . Hence the given equation represents a hyperbola centered at the origin, whose axis has half-length 1, and points in the direction of s1 , with vertices at x = ± 15 (4.3)T . The asymptotes have the equation 7x21 + 48x1 x2 −7x22 = 0, or equivalently (7x2 + x1 ) (7x1 − x2 ) = 0, that is, x2 = 7x1 and x1 = −7x2 .

79

3 2 1

-3

-2

-1

0

1

2

3

-1 -2 -3

7.3.4. The matrix of the given quadratic form is   3 −1 −1 1 3 −1  .

A =  −1 4 −1 −1 3

The eigenvalues are λ1 = λ2 = 1, and λ3 = 1/4 and an orthogonal matrix of corresponding eigenvectors is √   √ 3 1 √ √2 1   S=√ − 3 1 √2 .

6 0 −2 2 In the basis S the equation of the ellipsoid takes on the standard form y12 + y22 +

y32 = 1.

4

This is an ellipsoid of revolution, with major axis of half-length 2 pointing in the s3 = √13 (1, 1, 1)T direction, and a circle of radius 1 as its cross section in the plane of the minor axes.

80

7.3.6. The matrix of the given quadratic form is   0 1 1 0 1 .

A= 1 1 1 0

The eigenvalues are λ1 = 2 and λ2 = λ3 = −1 and an orthogonal matrix of corresponding eigenvectors is √   √ 2√3 0√ 2 √6 1 S =  2√3 −3√ 2 −√6  .

6 2 3 3 2 − 6 In the basis S the equation of the surface takes on the standard form 2y12 − y22 − y32 = 1.

This is a hyperboloid of revolution of two sheets, with its axis pointing in the s1 = √13 (1, 1, 1)T direction.

7.4.2. a) xH = (2 − 4i, 1 + 2i) and yH = (1 + 5i, 4 − 2i), b)

2 + 4i 2 H |x| = x x = (2 − 4i, 1 + 2i) = 25 and |x| = 5, 1 − 2i

√ 1 − 5i 2 H |y| = y y = (1 + 5i, 4 − 2i) = 46 and |y| = 46, 4 + 2i c)

1 − 5i xH y = (2 − 4i, 1 + 2i) = −18 − 14i + 10i = −18 − 4i, 4 + 2i

2 + 4i H y x = (1 + 5i, 4 − 2i) = −18 + 14i − 10i = −18 + 4i.

1 − 2i 7.4.4. a) xH = (2, −2i, 1 − i) and yH = (−5i, 4 − i, 4 + i), b) 81



 2 √ |x|2 = xH x = (2, −2i, 1 − i)  2i  = 10 and |x| = 10, 1+i 

 5i √ |y|2 = yH y = (−5i, 4 − i, 4 + i)  4 + i  = 59 and |y| = 59, 4−i c)   5i xH y = (2, −2i, 1 − i)  4 + i  = 5 − 3i, 4−i 

 2 yH x = (−5i, 4 − i, 4 + i)  2i  = 5 + 3i.

1+i

7.4.6. The characteristic equation is (cos θ − λ)2 + sin2 θ = 0. Thus, the eigenvalues are λ = cos θ ± i sin θ = e±iθ . The corresponding eigenvectors i 1 are s1 = s and s2 = t .

1 i 7.4.8. (AH A)H = AH AHH = AH A. 7.4.10. |Ux|2 = (Ux)H U x = xH U H U x = |x|2 .

7.4.12. First verify that Equations 7.169 and 7.170 are solutions of Equations 7.83 and 7.84. Indeed, using Equation 7.164, we get

−at

di(t) 2Q # = αe sin ωt − ωe−at cos ωt dt LC |D| QR Q −at # e−at sin ωt − = e cos ωt, LC L2 C |D| 1 and this expression clearly equals − R i − LC q.

L

82

2

4 Similarly, using also |D| = LC −R , we find that L2

dq(t) = αe−at dt

QR # sin ωt + Q cos ωt L |D|

QR −Qω sin ωt + # ω cos ωt L |D|

+e−at =

−2Q −at # e sin ωt = i(t).

LC |D|

The initial conditions i(0) = 0 and q(0) = Q are clearly satisfied.

H T T H 7.4.14. If A = A, then |A| = A = A = A = |A|.

7.4.16. The characteristic equation is

1−λ 1 1

1−λ 1 = 0, |A − λI| = 0 0 −1 1 − λ

or equivalently, (1 − λ) [(1 − λ)2 + 1] = 0. The eigenvalues are λ1 = 1, λ2 = 1 + i and λ3 = 1 − i, and the corresponding eigenvectors are s1 = s(1, 0, 0)T , s2 = t(1 − i, 1, i)T , and s3 = t(1 + i, i, 1)T .

8.1.2.

1 2

2 4

=

1 2

0 1

1 0

0 1

1 0

2 0

.

8.1.4. Approximately (109 /3)/105 ≈ 3333 seconds, that is, 55 minutes.

8.1.6. By Theorem 8.1.1, L is m × m and lower triangular. Thus, solving Lc = b is just back substitution and needs m2 /2 long operations.

Substituting the free variables into Ux = c, if there are any, takes anywhere from 0 to (n − r)2 /2 operations, depending on the locations of the pivots. Here r denotes the common rank of both A and U. To solve for the r basic variables takes about r2 /2 long operations.

8.1.8. The forward elimination phase requires n3 /3 operations for A and only n3 /6 operations for the I on the right side, because the zeroes above the 83

diagonal do not change. The back substitution uses n2 /2 operations for each of the n columns of I, for a total of n(n2 /2) = n3 /2 These numbers add up to n3 /3 + n3 /6 + n3 /2 = n3 .

8.2.2. The first step of Gaussian Elimination would produce

2 1000 4000 , 0 −3001 −11998 and our machine would round the second row to give

2 1000 4000 , 0 −3000 −12000

The machine would then solve this system by back substitution and obtain the wrong solution x2 = 4 and x1 = 0. The correct solution is x2 = 11998 = 3001 3.9980 . . . and x1 = 6000 = 0.9996 . . ., the same as in Exercise 8.2.1.

6002 The reason for the discrepancy is that in the first step of the back substitution the machine rounded x2 = 3.998 . . . to 4, and in the next step the machine had to multiply x2 by 1000 in solving for x1 . Here the small roundoff error, hidden in taking x2 as 4, got magnified a thousandfold.

8.2.4. The scale factors are s1 = 4, s2 = 4 and s3 = 5, and the ratios r1 = 1/2, r2 = 1/4 and r3 = 1. Since r3 > r1 > r2 , we put the third row on top, and then proceed with the row reduction:

    5 2 1 2 5 2 1 2  2 4 − 2 6  →  0 16 − 12 26 

1 3 4 −1 0 13 19 −7 The new scale factors are s2 = 16 and s3 = 19, and the ratios r2 = 1 and r3 = 13/19. Thus we leave the rows in place, and reduce further to obtain

  5 2 1 2 16 − 12 26  → 0 0 0 46 −45 Hence x3 = −45/46, x2 = 41/46, and x1 = 11/46.

8.2.6. a. Since for such matrices, c = |λ2 |/|λ1 | and |λ2 | ≥ |λ1 |, the relation c ≥ 1 follows at once.

84

b. By the given formula, c = 1 implies λ2 = ±λ1 for the two eigenvalues. c. The characteristic equation is

0.001 − λ

1 = 0, |A − λI| = 1 1−λ

or equivalently, √ (0.001−λ)(1−λ)−1 √ = 0. The eigenvalues are approximately λ1 ≈ (1 + 5)/2 and λ2 ≈ (1 − 5)/2. Thus c ≈ 2.62. d. The given formula does not apply in this case, because # the matrix is not symmetric. In this case, c is given by the formula c = |µn |/|µ1 | with µi denoting eigenvalues of the matrix AT A. Hence c ≈ 2.62 again. e When c = 1, the solution of Ax = b is 100% accurate. For c = 2.62 it can be in serious error on a two-digit machine, but the condition number does not show the advantage resulting from partial pivoting.

8.3.2. The eigenvalues of B = A − cI are 1 − c, 2 − c, and 3 − c. For a given c, the eigenvalue 2 − c would be dominant if and only if |2 − c| > |1 − c| and |2 − c| > |3 − c| were true simultaneously. But this is impossible: iy c

|2 - c| |1 - c|

|3 - c| 1

2

3

x

Write c = c1 + ic2 , with c1 , c2 real. If c1 ≥ 2, then |2 − c| < |1 − c|, because |2 − c|2 = (c1 − 2)2 + c22 , |1 − c|2 = (c1 − 1)2 + c22 , and |1 − c|2 − |2 − c|2 = (c1 − 1)2 − (c1 − 2)2 = 2c1 − 3 ≥ 4 − 3 > 0.

Similarly, if c1 < 2, then |2 − c| < |3 − c|.

85

A.2.2. The second part of inequality A.23 is just the triangle inequality for vectors in the plane. A direct algebraic proof can be obtained starting from Equation A.21, expanding |z1 + z2 |2 with the aid of Equations A.12 and A.9, and noting that Rz ≤ |z| : |z1 + z2 |2 = (z1 + z2 ) (z1 + z2 ) = (z1 + z2 ) (z1 + z2 ) = |z1 |2 + (z1 z2 + z1 z2 ) + |z2 |2 = |z1 |2 + 2R (z1 z2 ) + |z2 |2 ≤ |z1 |2 + 2 |z1 z2 | + |z2 |2 = (|z1 | + |z2 |)2 .

Taking square roots results in the second part of inequality A.22. The first part follows from the second one by observing that |z1 | = |(z1 − z2 ) + z2 | ≤ |z1 − z2 | + |z2 | , and thus |z1 | − |z2 | ≤ |z1 − z2 | . Similarly, |z2 | − |z1 | ≤ |z2 − z1 | . Hence ||z1 | − |z2 || ≤ |z1 − z2 | . k k k n=0 xn + i n=0 yn . Suppose that ∞A.2.4. Note first that n=0 zn = n=0 zn converges to z = x + iy. Then, since

k k k

xn − x = R zn − z ≤ zn − z ,

n=0

n=0

n=0

∞ the real part ∞ n=0 xn converges to x. Similarly, the imaginary part n=0 yn converges to y.

∞ Conversely, if ∞ n=0 xn converges to x and n=0 yn converges to y, then the inequality k

k

k

zn − z = xn − x + i yn − y

n=0

n=0

k k n=0

≤ xn − x + yn − y n=0 n=0

implies that ∞ n=0 zn converges to z.

A.2.6. If

z = ei(φ+2kπ) , 86

with z = 0, and if wn = z, where w = ReiΦ , then wn = Rn einΦ .

Thus, we must have Rn = r and einΦ = eiΦ , from which it folows that R = r1/n and Φ = (φ + 2kπ)/n. Therefore, z 1/n = r1/n ei(φ+2kπ)/n .

Substituting k = 0, 1, 2, . . . , n − 1 into the last expression results in n distinct roots of z. (Note that k = n leads to the angle Φ = (φ+2nπ)/n = (φ/n)+2π, which is equivalent to the angle φ/n corresponding to k = 0, and k = n + 1 leads to an angle equivalent to the earlier angle corresponding to k = 1, etc.)

87

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