ISM: Linear Algebra
Section 8.1
Chapter 8 8.1
1. e1 , e2 is an orthonormal eigenbasis. 2. 3.
1 √ 2
1 √ 1 , 12 1 −1
is an orthonormal eigenbasis.
2 √ −1 , 15 is an orthonormal eigenbasis. 1 2 1 1 1 1 1 1 4. √3 1 , √2 −1 , √6 1 is an orthonormal eigenbasis. −1 0 2
1 √ 5
5. Eigenvalues −1, −1, 2 −1 1 1 1 Choose v1 = √2 1 in E−1 and v2 = √3 1 in E2 and let v3 = v1 × v2 = 0 1 2 2 1 1 6. 1 2 , 3 −1 , 1 −2 is an orthonormal eigenbasis. 3 3 1 −2 2 7.
1 √ 2
1 1 √ 1 . 6 −2
1 , 1
1 √ 2
1 −1
is an orthonormal eigenbasis, so S =
1 √ 2
1 1 1 −1
and D =
5 0 . 0 1
3 −1 , √1 is an orthonormal eigenbasis, with λ1 = 4 and λ2 = −6, so S = 10 1 3 3 −1 4 0 1 √ and D = . 10 1 3 0 −6 1 −1 0 1 1 9. √2 0 , √2 0 , 1 is an orthonormal eigenbasis, with λ1 = 3, λ2 = −3, and 1 1 0 λ3 = 2, so 1 −1 √0 3 0 0 1 S = √2 0 2 and D = 0 −3 0 . 0 1 1 0 0 0 2 8.
1 √ 10
10. λ1 = λ2 = 0 and λ3 = 9. 401
Chapter 8
ISM: Linear Algebra
1 2 1 v1 = √5 1 is in E0 and v2 = 1 −2 is in E9 . 3 2 0 2 1 Let v3 = v1 × v2 = 3√5 −4 ; then v1 , v2 , v3 is an orthonormal eigenbasis. −5 2 2 1 √ √ 3 5 3 5 0 0 0 1 2 4 S = √5 − 3 − 3√5 and D = 0 9 0 . √ 0 0 0 2 0 − 35 3 0 1 −1 1 1 11. √2 0 , √2 0 , 1 is an orthonormal eigenbasis, with λ1 = 2, λ2 = 0, and λ3 = 1, 0 1 1 1 −1 0 2 0 0 √ 1 so S = √2 0 0 2 and D = 0 0 0 . 0 0 1 1 1 0 1 12. a. E1 = span 0 and E−1 = (E1 )⊥ . An orthonormal eigenbasis is 2 2 1 √ 0 . 5 −1 1 0 0 0 . b. Use Fact 7.4.1: B = 0 −1 0 0 −1 1 2 √ √ 0 −0.6 0 0.8 5 5 0 . c. A = SBS −1 = 0 −1 0 , where S = 0 1 2 1 0.8 0 0.6 √ 0 − √5 5 0 1 1 √ 0 , 1 , 5 0 2
13. Yes; if v is an eigenvector of A with eigenvalue λ, then v = I3 v = A2 v = λ2 v, so that λ2 = 1 and λ = 1 or λ = −1. Since A is symmetric, E1 and E−1 will be orthogonal complements, so that A represents the reflection about E1 . 0 0 0 14. Let S be as in Example 3. Then S −1 AS = 0 0 0 . 0 0 3 402
ISM: Linear Algebra
Section 8.1
0 0 0 a. This matrix is 2A so that S −1 (2A)S = 0 0 0 . 0 0 6 −3 0 0 b. This is A − 3I3 , so that S −1 (A − 3I3 )S = S −1 AS − 3I3 = 0 −3 0 . 0 0 0
1 c. This is 2 (A − I3 ), so that S −1 1 −2 1 1 −1 AS − I3 ) = 0 2 (A − I3 ) S = 2 (S
0 −1 2 0
0
1
0 0 .
1 15. Yes, if Av = λv, then A−1 v = λ v, so that an orthonormal eigenbasis for A is also an −1 orthonormal eigenbasis for A (with reciprocal eigenvalues).
16. a. ker(A) is four-dimensional, so that the eigenvalue 0 has multiplicity 4, and the remaining eigenvalue is tr(A) = 5. b. B = A + 2I5 , so that the eigenvalues are 2, 2, 2, 2, 7. c. det(B) = 24 · 7 = 112 (product of eigenvalues)
17. If A is the n × n matrix with all 1’s, then the eigenvalues of A are 0 (with multiplicity n − 1) and n. Now B = qA + (p − q)In , so that the eigenvalues of B are p − q (with multiplicity n − 1) and qn + p − q. Thus det(B) = (p − q)n−1 (qn + p − q).
1 18. By Fact 6.3.7, the volume is |det A| = det(AT A). Now vi · vj = vi vj cos(θ) = 2 , so 1 T that A A has all 1’s on the diagonal and 2 ’s outside. By Exercise 17 with p = 1 and q = 1 , 2 1 det(AT A) = 2 √ 1 n/2 n + 1. 2 n−1 1 2n
+
1 2
=
1 n 2
(n + 1), so that the volume is
det(AT A) =
19. Let L(x) = Ax. Then AT A is symmetric, since (AT A)T = AT (AT )T = AT A, so that there is an orthonormal eigenbasis v1 , . . . , vm for AT A. Then the vectors Av1 , . . . , Avm T are orthogonal, since Avi · Avj = (Avi )T Avj = vi AT Avj = vi · (AT Avj ) = vi · (λj vj ) = λj (vi · vj ) = 0 if i = j. 403
Chapter 8
ISM: Linear Algebra
20. By Exercise 19, there is an orthonormal basis v1 , . . . , vm of Rm such that T (v1 ), . . . , T (vm ) are orthogonal. Suppose that T (v1 ), . . . , T (vr ) are nonzero and T (vr+1 ), . . . , T (vm ) are 1 zero. Then let wi = T (vi ) T (vi ) for i = 1, . . . , r and choose an orthonormal basis wr+1 , . . . , wn of [span(w1 , . . . , wr )]⊥ . Then w1 , . . . , wn does the job. 21. For each eigenvalue there are two unit eigenvectors: ±v1 , ±v2 , and ±v3 . We have 6 choices for the first column of S, 4 choices remaining for the second column, and 2 for the third. Answer: 6 · 4 · 2 = 48.
22. a. If we let k = 2 then A is symmetric and therefore (orthogonally) diagonalizable. b. If we let k = 0 then 0 is the only eigenvalue (but A = 0), so that A fails to be diagonalizable.
23. The eigenvalues are real (by Fact 8.1.3), so that the only possible eigenvalues are ±1 (by Fact 7.1.2). Since A is symmetric, E1 and E−1 are orthogonal complements. Thus A represents a reflection about E1 . 24. Note that A is symmetric and orthogonal, so that the eigenvalues are 1 and −1 (see Exercise 23). 1 0 1 0 0 1 0 1 E1 = span , and E−1 = span , , so that 0 1 0 −1 1 0 −1 0 1 1 0 √ , 2 0 1 0 1 1 √ , 2 1 0 1 1 0 √ , 2 0 −1 0 1 1 √ is an orthonormal eigenbasis. 2 −1 0
25. Note that A is symmetric an orthogonal, so that the eigenvalues of A are 1 and −1. 1 0 0 1 0 0 1 0 0 1 E1 = span 0 , 0 , 1 , E−1 = span 0 , 0 0 1 0 0 −1 1 0 0 −1 0 404
ISM: Linear Algebra 1 0 1 The columns of S must form an eigenbasis for A : S = √2 0 0 1 possible choice.
Section 8.1 0 0 1 0 1 √ 0 0 1 0 2 0 0 is one 1 0 0 −1 0 0 −1 0
26. Since Jn is both orthogonal and symmetric, the eigenvalues are 1 and −1. If n is even, then both have multiplicity n (as in Exercise 24). If n is odd, then the multiplicities are 2 n+1 for 1 and n−1 for −1 (as in Exercise 25). One way to see this is to observe that 2 2 tr(Jn ) is 0 for even n, and 1 for odd n (recall that the trace is the sum of the eigenvalues). 27. If n is even, then this matrix is Jn + In , for the Jn introduced in Exercise 26, so that the eigenvalues are 0 and 2, with multiplicity n each. E2 is the span of all ei + en+1−i , for 2 i = 1, . . . , n , and E0 is spanned by all ei − en+1−i . If n is odd, then E2 is spanned by all 2 ei + en+1−i , for i = 1, . . . , n−1 ; E0 is spanned by all ei − en+1−i , for i = 1, . . . , n−1 , and 2 2 E1 is spanned by e n+1 .
2
28. For λ = 0 fA (λ) = det 1 = λ det −λ −λ −λ | 1 | −λ | 1 | . .. 1 . . . = det | | λ 0 −λ | 1 | | 1 1 ··· 1 | 1−λ λ 0 | 1 | −λ | 1 | . .. . . . | | 0 −λ | 1 | | 2 0 · · · 0 | −λ + λ + 12 0 −λ 0 λ | 1 | | 1 | . .. . . | . | −λ | 1 | | 2 ··· λ | λ −λ 0
0
= −λ11 (λ2 − λ − 12) = −λ11 (λ − 4)(λ + 3) Eigenvalues are 0 (with multiplicity 11), 4 and −3. Eigenvalues for 0 are e1 − ei (i = 2, . . . , 12), 1 1 1 1 . . (12 ones), E−3 = span . (12 ones) . E4 = span . . 1 1 4 −3 405
Chapter 8 so 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 −1 0 −1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 S= 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0
ISM: Linear Algebra
diagonalizes A, and D = S −1 AS will have all zeros as entries except d12, d13, 13 = −3. 29. By Fact 5.4.1 (im A)⊥ = ker(AT ) = ker(A), so that v is orthogonal to w.
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 −3
12
= 4 and
30. The columns v, v2 , . . . , vn of R form an orthogonal eigenbasis for A = v v T , with eigenvalues 1, 0, 0, . . . , 0(n − 1 zeros), since Av = vv T v = v(v·v) = v, (since v·v = 1) and Avi = v v T vi = v(v·vi ) = 0 (since v·vi = 0). 1 0 ··· 0 0 0 ··· 0 . Therefore we can let S = R, and D = . . . 31. True; A is diagonalizable, that is, A is similar to a diagonal matrix D; then A2 is similar to D2 . Now rank(D) = rank(D 2 ) is the number of nonzero entries on the diagonal of D (and D2 ). Since similar matrices have the same rank (by Fact 7.3.6b) we can conclude that rank(A) = rank(D) = rank(D 2 ) = rank(A2 ). 32. By Exercise 17, det(A) = (1 − q)n−1 (qn + 1 − q). A is invertible if det(A) = 0, that is, if 1 q = 1 and q = 1−n . 33. The angles must add up to 2π, so θ =
2π 3
0 0 ··· 0
= 120◦ . (See Figure 8.1.)
Algebraically, we can see this as follows: let A = [ v1 v2 v3 ], a 2 × 3 matrix. 1 cos θ cos θ Then AT A = cos θ 1 cos θ is a noninvertible 3×3 matrix, so that cos θ = cos θ cos θ 1 2π 1 − 2 , by Exercise 32, and θ = 3 = 120◦ . 406
1 1−3
=
ISM: Linear Algebra
v1 v2 θ θ θ unit circle
Section 8.1
v3
Figure 8.1: for Problem 8.1.33. 34. Let v1 , v2 , v3 , v4 be such vectors. Form A = [ v1 v2 v3 v4 ], a 3 × 4 matrix. 1 cos θ cos θ cos θ 1 cos θ cos θ cos θ 1 1 Then AT A = is noninvertible, so that cos θ = 1−4 = − 3 , cos θ cos θ 1 cos θ cos θ cos θ cos θ 1 1 by Exercise 32, and θ = arccos − 3 ≈ 109.5◦ . See Figure 8.2.
Figure 8.2: for Problem 8.1.34. The tips of v1 , v2 , v3 , v4 form a regular tetrahedron. 35. Let v1 , . . . , vn+1 be these vectors. Form A = [ v1 · · · vn+1 ], an n × (n + 1) matrix. 1 cos θ · · · cos θ cos θ 1 · · · cos θ is a noninvertible (n + 1) × (n + 1) matrix Then AT A = . .. . . . cos θ ··· 1 with 1’s on the diagonal and cos θ outside, so that cos θ = θ = arccos
1 1−n 1 1−n ,
by Exercise 32, and
.
36. If v is an eigenvector with eigenvalue λ, then λv = Av = A2 v = λ2 v, so that λ = λ2 and 407
Chapter 8
ISM: Linear Algebra
therefore λ = 0 or λ = 1. Since A is symmetric, E0 and E1 are orthogonal complements, so that A represents the orthogonal projection onto E1 .
37. a. If S −1 AS is upper triangular then the first column of S is an eigenvector of A. Therefore, any matrix without real eigenvectors fails to be triangulizable over R, for example, 0 −1 . 1 0 b. Proof by induction on n: For an n × n matrix A we can choose a complex invertible λ v n × n matrix P whose first column is an eigenvector for A. Then P −1 AP = . 0 B B is triangulizable, by induction hypothesis, that is, there is an invertible (n−1)×(n−1) matrix Q such that Q−1 BQ = T is upper triangular. Now let R = 1 0 . 0 Q Then R−1 1 0 λ v 1 0 λ v = R = 0 Q 0 B 0 Q−1 0 B λ v R = R−1 P −1 AP R = S −1 AS, where 0 B
λ vQ is upper triangular. R−1 0 T S = P R, proving our claim.
38. a. By definition of an upper triangular matrix, e1 is in ker U , e2 is in ker(U 2 ), . . . , en is in ker(U n ), so that all x in Cn are in ker(U n ), that is, U n = 0. b. By Exercise 37b, there is an invertible S such that S −1 AS = U is upper triangular. The diagonal entries of U are all zero, since A and U have the same eigenvalues; therefore U n = 0 by part a. Now A = SU S −1 and An = SU n S −1 = 0, as claimed.
n n n
39. a. For all i, j,
k=1
aik bkj ≤ ↑
k=1
|aik bkj | =
k=1
|aik ||bkj |
triangle inequality b. By induction on t: |At | = |At−1 A| ≤ |At−1 ||A| ≤ |A|t−1 |A| = |A|t ↑ part a ↑ by induction hypothesis
408
ISM: Linear Algebra 40. If t ≥ n − 1 then (In + U )t = In + Now t 1 t 2 U2 +· · ·+ t n−1
Section 8.2 U n−1 , since U n = 0.
U+
t ≤ tn for k = 1, . . . , n − 1, so that (In + U )t ≤ tn (In + U + · · · + U n−1 ), as k claimed. Check that the formula holds for t < n − 1 as well. |r11 | ∗ λ ∗ .. .. = 41. Let λ be the largest |rii |; note that λ < 1. Then |R| = ≤ . . 0 λ 0 |rnn | 1 ∗ .. = λ(In + U ), and |Rt | ≤ |R|t ≤ λt (In + U )t ≤ λt tn (In + U + · · · + U n−1 ). λ . 0 1 We learn in Calculus that lim (λt tn ) = 0, so that lim (Rt ) = 0.
t→∞ t→∞
42. a. From Exercise 37b we know that there is an invertible S and an upper triangular R such that S −1 AS = R, and |rii | < 1 for all i, since the diagonal entries of R are the eigenvalues of A. Now lim Rt = 0 by Exercise 41. Note that A = SRS −1 and At = SRt S −1 , so that lim At = 0, as claimed.
t→∞ t→∞
b. See the remark after Definition 7.6.1
8.2
1 1. We have a11 = coefficient of x2 = 6, a22 = coefficient of x2 = 8, a12 = a21 = 2 ( coefficient of x1 x2 ) = 1 2 7 6 −2 7 − 2 . So, A = 7 −2 8
2. A =
0
1 2
1 2
0 0 4
7 2
3 3. A = 0 4. A = 5. A =
3
7 2
3
5
6 2 , positive definite 2 3 1 2 , indefinite (since det(A) < 0) 2 1 409
Chapter 8 2 3 , indefinite (since det(A) < 0) 3 4 0 0 2 7. A = 0 3 0 , indefinite (eigenvalues 2, −2, 3) 2 0 0
ISM: Linear Algebra
6. A =
8. If S −1 AS = D is diagonal, then S −1 A2 S = D2 , so that all eigenvalues of A2 are ≥0. So A2 is positive semi-definite; it is positive definite if and only if A is invertible.
9. a. (A2 )T = (AT )2 = (−A)2 = A2 , so that A2 is symmetric. b. q(x) = x T A2 x = x T AAx = −x T AT Ax = −(Ax) · (Ax) = − Ax 2 ≤ 0 for all x, so that A2 is negative semi-definite. The eigenvalues of A2 will be ≥ 0. c. If v is a complex eigenvector of A with eigenvalue λ, then A2 v = λ2 v, and λ2 ≤ 0, by part b. Therefore, λ is imaginary, that is, λ = bi for a real b. Thus, the zero matrix is the only skew-symmetric matrix that is diagonalizable over R.
10. L(x) = (x+v)T A(x+v)−x T Ax−v T Av = x T Ax+x T Av +v T Ax+v T Av −x T Ax−v T Av = x T Av + v T Ax = v T Ax + v T Ax = (2v T A)x, ↑ note that x T Av is a scalar so that x T Av = (x T Av)T = v T AT x = v T Ax if A is symmetric. So L is linear with matrix 2v T A. 11. The eigenvalues of A−1 are the reciprocals of those of A, so that A and A−1 have the same definiteness. 12. det(A) is the product of the two (real) eigenvalues. q is indefinite if an only if those have different signs, that is, their product is negative. 13. q(ei ) = ei · Aei = aii > 0 14. If det(A) is positive then both eigenvalues have the same sign, so that A is positive definite or negative definite. Since e1 · Ae1 = a > 0, A is in fact positive definite. 15. A = 6 2 ; eigenvalues λ1 = 7 and λ2 = 2 2 3
1 √ 5
orthonormal eigenbasis v1 =
2 , v2 = 1
1 √ 5
−1 2
410
ISM: Linear Algebra
Section 8.2
v2
√2
E7
√7
1 v 2
1 v 1
v1 1
– 1 v1
√7
E2 – 1 v2
√2
Figure 8.3: for Problem 8.2.15. λ1 c2 + λ2 c2 = 1 or 7c2 + 2c2 = 1. (See Figure 8.3.) 1 2 1 2 16. A = 0
1 2
0
1 2
1 ; eigenvalues λ1 = 1 , and λ2 = − 2 2 1 √ 2
orthonormal eigenbasis v1 =
1 2 2 c1
1 1
and v2 =
1 √ 2
1 −1
1 − 2 c2 = 1. (See Figure 8.4.) 2
17. A =
3 2 , eigenvalues λ1 = 4, λ2 = −1 2 0
1 √ 5
orthonormal eigenbasis v1 =
2 , v2 = 1
1 √ 5
−1 2
4c2 − c2 = 1 (hyperbola) (See Figure 8.5.) 1 2 18. A = 9 −2 , eigenvalues λ1 = 10, λ2 = 5 −2 6
1 √ 5
orthonormal eigenbasis v1 =
2 , v2 = 1
1 √ 5
−1 2
10c2 + 5c2 = 1. This is an ellipse, as shown in Figure 8.6. 1 2 411
Chapter 8
ISM: Linear Algebra
E1
2
√2 v1 v1
1 v2 –√2 v1 E– 1
2
Figure 8.4: for Problem 8.2.16. 1 2 ; eigenvalues λ1 = 5, λ2 = 0 2 4
1 √ 5
19. A =
eigenvectors v1 =
1 , v2 = 2
1 √ 5
−2 1
5c2 = 1 (a pair of lines) (See Figure 8.7.) 1 Note that (x2 + 4x1 x2 + 4x2 ) = (x1 + 2x2 )2 = 1, so that x1 + 2x2 = ±1, and the two lines 1 2 are x2 = 20. A =
1−x1 2
and x2 =
−1−x1 . 2
−3 3 ; eigenvalues λ1 = 6 and λ2 = −4 3 5
√1 10
orthonormal eigenbasis v1 =
1 , v2 = 3 412
1 √ 10
−3 1
ISM: Linear Algebra
Section 8.2
v2
1 2 v1
E4
v1 1
– 1 v1 2
E–1
Figure 8.5: for Problem 8.2.17.
v2
1 v √5 2 1 v √10 1
E10
v1 1
– 1 v1
√10
– 1 v2
√5
E5
Figure 8.6: for Problem 8.2.18.
6c2 − 4c2 = 1. This is a hyperbola, as shown in Figure 8.8. 1 2 413
Chapter 8
ISM: Linear Algebra
v2
1 v √5 1
v1
– 1 v1
√5
1
E5
E0
Figure 8.7: for Problem 8.2.19.
v1 v2
1 v √6 1
– 1 v1
√6
E–4
E6
Figure 8.8: for Problem 8.2.20.
414
ISM: Linear Algebra
Section 8.2
21. a. In each case, it is informative to think about the intersections with the three coordinate planes: x1 − x2 , x1 − x3 , and x2 − x3 . • For the surface x2 + 4x2 + 9x2 = 1, all these intersections are ellipses, and the surface 1 2 3 itself is an ellipsoid . 0 This surface is connected and bounded; the points closest to the origin are ± 0 ,
1 3
1 and those farthest ± 0 . (See Figure 8.9.) 0
x3
x2 x1
Figure 8.9: for Problem 8.2.21a: x2 + 4x2 + 9x2 = 1, an ellipsoid (not to scale). 1 2 3 • In the case of x2 + 4x2 − 9x2 = 1, the intersection with the x1 − x2 plane is an ellipse, 1 2 3 and the two other intersections are hyperbolas. The surface is connected and not 0 bounded; the points closest to the origin are ± 1 . (See Figure 8.10.) 2 0 • In the case −x2 − 4x2 + 9x2 = 1, the intersection with the x1 − x2 plane is empty, and 1 2 3 the two other intersections are hyperbolas. The surface consists of two pieces and is 415
Chapter 8
x3
ISM: Linear Algebra
x2
x1
Figure 8.10: for Problem 8.2.21a: x2 +4x2 −9x2 = 1, a hyperboloid of one sheet (not to scale). 1 2 3 0 unbounded. The points closest to the origin are ± 0 . (See Figure 8.11.)
1 3
b. A = 1 2 1
1
1 2
1
3 2
2
3 2
3
is positive definite, with three positive eigenvalues λ1 , λ2 , λ3 .
The surface is given by λ1 c2 + λ2 c2 + λ3 c2 = 1 with respect to the principal axes, an 1 2 3 ellipsoid. To find the points closest to and farthest from the origin, use technology to find the eigenvalues and eigenvectors: eigenvalues: λ1 ≈ 0.56, λ2 ≈ 4.44, λ3 = 1 0.31 0.86 unit eigenvectors: v1 ≈ 0.19 , v2 ≈ 0.54 , v3 = 0.78 −0.47 Equation: 0.56c2 + 4.44c2 + c2 = 1 1 2 3
1 Farthest points when c1 = ± √0.56 and c2 = c3 = 0 1 Closest points when c2 = ± √4.44 and c1 = c3 = 0
1 1 √ −2 6 1
416
ISM: Linear Algebra
x3
Section 8.2
x2
x1
Figure 8.11: for Problem 8.2.21a: −x2 − 4x2 + 9x2 = 1, a hyperboloid of two sheets (not to 1 2 3 scale). 0.86 1.15 1 Farthest points ≈ ± √0.56 0.19 ≈ ± 0.26 −0.47 −0.63 0.31 0.15 1 Closest points ≈± √4.44 0.54 ≈ ± 0.26 0.78 0.37 −1 0 5 22. A = 0 1 0 ; eigenvalues λ1 = 4, λ2 = −6, λ3 = 1 5 0 −1 Equation with respect to principal axes: 4c2 − 6c2 + c2 = 1, a hyperboloid of one sheet 1 2 3 (see Figure 8.10). Closest to origin when c1 = ± 1 , c2 = c3 = 0. 2 1 1 A unit eigenvector for eigenvalue 4 is v = √2 0 , so that the desired points are 1 417
Chapter 8 1 0.35 1 ± 1 √2 0 ≈ ± 0 . 2 1 0.35 23. Yes; M = 1 (A + AT ) is symmetric, and 2
1 x T M x = 2 x T Ax + 1 x T AT x = 1 x T Ax + 1 x T Ax = x T Ax 2 2 2
ISM: Linear Algebra
Note that x T Ax is a 1 × 1 matrix, so that x T Ax = (x T Ax)T = x T AT x. 24. q(e1 ) = e1 · Ae1 = e1 · (first column of A) = a11 25. q(v) = v · Av = v · (λv) = λ(v · v) = λ ↑ v is a unit vector 26. False; If A = 0 1 1 0 then q 1 1 0 1 = · 0 0 1 0 1 1 0 = · = 0. 0 0 1
27. Let v1 , . . . , vn be an orthonormal eigenbasis for A with Avi = λi vi . We know that q(vi ) = λi (see Exercise 25), so that q(v1 ) = λ1 and q(vn ) = λn are in the image. We claim that all numbers between λn and λ1 are in the image as well. To see this, apply the Intermediate Value Theorem to the continuous function f (t) = q((cos t)vn + (sin t)v1 ) on 0, π (note that f (0) = q(vn ) = λn and f π = q(v1 ) = λ1 ). (See Figure 8.12.) 2 2
λ1 λn 0 vn f(t)
π 2
v1
(cos t)vn + (sin t)v1, a unit vector
t
Figure 8.12: for Problem 8.2.27. The Intermediate Value Theorem tells us that for any c between λn and λ1 , there is a t0 such that f (t0 ) = q((cos t0 )vn + (sin t0 )v1 ) = c. Note that (cos t0 )vn + (sin t0 )v1 is a unit vector. Now we will show that, conversely, q(v) is on [λn , λ1 ] for all unit vectors v. Write v = c1 v1 + · · · + cn vn and note that v 2 = c2 + · · · + c2 = 1. Then q(v) = 1 n λ1 c2 + λ2 c2 + · · · + λn c2 ≤ λ1 c2 + λ1 c2 + · · · + λ1 c2 = λ1 . Likewise, q(v) ≥ λn . We have 1 2 n 1 2 n shown that the image of S n−1 under q is the closed interval [λn , λ1 ]. 418
ISM: Linear Algebra
Section 8.2
28. The hint almost gives it away. Since D is a diagonal matrix with positive diagonal entries, 2 we can write D = D1 , where D1 is diagonal with positive diagonal entries (the square roots of the entries of D). Now A = SDS T = SD1 D1 S T = SD1 (SD1 )T = BB T where B = SD1 . The columns of B are scalar multiples of the corresponding columns of S, so that they are orthogonal. 29. From Example 1 we have S = B = SD1 =
1 √ 5 1 √ 5
6 2 . −3 4
2 1 −1 2
and D =
9 0 . Let D1 = 0 4
3 0 0 2
and
30. Define D1 as in Exercise 28. Then A = SDS −1 = SD1 D1 S −1 = (SD1 S −1 )(SD1 S −1 ) = B 2 , where B = SD1 S −1 . B is positive definite, since S −1 BS = D1 is diagonal with positive diagonal entries. 31. S =
1 √ 5
2.8 −0.4 3 0 2 1 . (see Exercise 29), so that B = SD1 S −1 = and D1 = −0.4 2.2 0 2 −1 2 x2 = a xy = b y2 + z 2 = c
32. Recall that a = q(e1 ) > 0 and det A = ac − b2 = λ1 λ2 > 0. x a b = y b c 0 z x y 0 z = x2 xy xy y2 + z 2 means that
It is required that x and z be positive. This system has the unique solution √ x= a y= z=
b x
=
b √ a
c − y2 =
c−
b2 a
=
ac−b2 a
33. Use the formulas for x, y, z derived in Exercise 32. √ √ √ x= a= 8=2 2 y= z= L=
b √ a 2 1 = − 2√2 = − √2
ac−b2 a
= 0
3 √ 2
36 8
=
3 √ , 2
so that
2 2 1 − √2
√
.
34. • (i) implies (ii): See the hint at the end of the exercise. 419
Chapter 8
ISM: Linear Algebra
• (ii) implies (iii): det A(m) is the product of the (positive) eigenvalues. • (iii) implies (iv): A= A(n−1) vT v k = B xT 0 1 BT 0 BB T x = x T BT t Bx xT x + t
The system x = B −1 v
Bx = v has the unique solution xT x + t = k
t = k − x T x = k − B −1 v 2 . Note that t is positive since 0 < det(A(n) ) = det(A) = det (det B)2 · t. • (iv) ⇒ (i) x T Ax = x T LLT x = (LT x)T (LT x) = LT x
2
B xT
BT 0 det 0 1
x t
=
> 0 if x = 0, since L is invertible. z t s
4 −4 8 x 0 0 x 35. Solve the system −4 13 1 = y w 00 8 1 26 z t s 0 x2 = 4, so x = 2 2y = −4, so y = −2 2 0 0 2z = 8, so z = 4 L = −2 3 0 4 + w2 = 13, so w = 3 4 3 1 −8 + 3t = 1, so t = 3 2 16 + 9 + s = 26, so s = 1 37.
∂q ∂x1
y w 0
36. If A = QR, then AT A = (QR)T QR = RT QT QR = RT R = LLT , L = RT . = 2ax1 + bx2 and
∂2q ∂x1 ∂x2 ∂q ∂x2
= bx1 + 2cx2 , so that q11 = q11 q12 q12 q22 = det
∂2q ∂x2 1
= 2a, q22 =
∂2q ∂x2 2
= 2c, and
q12 =
= b, and D = det a
b 2 b 2
2a b = 4ac − b2 > 0. b 2c
The matrix A =
of q is positive definite, since a > 0 and det(A) = 1 D > 0. This 4 c means, by definition, that q has a minimum at 0, since q(x) > 0 = q(0) for all x = 0.
38. The eigenvalues of B are p − q and nq + p − q = p + (n − 1)q, so that B is positive definite if p − q > 0 and p + (n − 1)q > 0. 420
ISM: Linear Algebra 39. If v1 , . . . , vn is such a 1 cos θ 1 cos θ AT A = . .. . . . cos θ cos θ
Section 8.2 basis consisting of unit vectors, and we let A = [v1 · · · vn ], then · · · cos θ .. . cos θ . is positive definite, so that, by Exercise 38, 1−cos θ > 0 .. . . . ··· 1
1 1−n
and 1 + (n − 1) cos θ > 0 or 1 > cos θ >
or 0 < θ < arccos 1
1 1−n
.
· · · cos θ .. . cos θ 1 cos θ Conversely, if θ is in this range, then the matrix . . is positive .. .. . . . . . . cos θ cos θ · · · 1 definite, so that it has a Cholesky factorization LLT . The columns of LT give us a basis with the desired property. cos θ 40. Let λ be the smallest eigenvalue of A. If we let k = 1 − λ, then the smallest eigenvalue of the matrix A + kIn will be λ + k = 1, so that all the eigenvalues of A + kIn will be positive. Thus matrix A + kIn will be positive definite, by Fact 8.2.4. 41. The functions x2 , x1 x2 , x2 form a basis of Q2 , so that dim(Q2 ) = 3. 1 2 42. The functions xi xj form a basis of Qn , where 1 ≤ i ≤ j ≤ n. A little combinatorics shows that there are 1+2+3+· · ·+n = n(n+1)/2 such functions, so that dim(Qn ) = n(n+1)/2 43. Note that T (ax2 +bx1 x2 +cx2 ) = ax2 (we let x2 = 0). Thus im(T ) = span(x2 ), rank(T ) = 1 2 1 1 1, ker(T ) = span(x1 x2 , x2 ), nullity(T ) = 2. 2 44. Note that T (ax2 + bx1 x2 + cx2 ) = ax2 + bx1 + c (we let x2 = 1). Thus im(T ) = P2 , 1 2 1 rank(T ) = 3, ker(T ) = {0}, nullity(T ) = 0 (T is an isomorphism). 45. Note that T (ax2 + bx2 + cx2 + dx1 x2 + ex1 x3 + f x2 x3 ) = ax2 + b + c + dx1 + ex1 + f 1 2 3 1 (we let x2 = x3 = 1). Thus im(T ) = P2 and rank(T ) = 3. The kernel of T consists of the quadratic forms with a = 0, d + e = 0, and b + c + f = 0 (consider the coefficients of x2 , x1 , and 1). The general element of the kernel is q(x1 , x2 , x3 ) = (−c − f )x2 + 1 2 cx2 − ex1 x2 + ex1 x3 + f x2 x3 = c(x2 − x2 ) + e(x1 x3 − x1 x2 ) + f (x2 x3 − x2 ). Thus 3 3 2 2 ker(T ) = span(x2 − x2 , x1 x3 − x1 x2 , x2 x3 − x2 ) and nullity(T ) = 3. 3 2 2 46. Note that T (ax2 +bx2 +cx2 +dx1 x2 +ex1 x3 +f x2 x3 ) = ax2 +bx2 +cx2 +dx1 x2 +ex2 +f x1 x2 1 2 3 1 2 1 1 (we let x3 = x1 ). Thus im(T ) = Q2 and rank(T ) = 3. The kernel of T consists of the quadratic forms with a + c + e = 0, b = 0, and d + f = 0 (consider the coefficients of x2 , x2 , and x1 x2 ). The general element of the kernel is q(x1 , x2 , x3 ) = (−c − e)x2 + 1 2 1 cx2 − f x1 x2 + ex1 x3 + f x2 x3 = c(x2 − x2 ) + e(x1 x3 − x2 ) + f (x2 x3 − x1 x2 ). Thus 3 3 1 1 ker(T ) = span(x2 − x2 , x1 x3 − x2 , x2 x3 − x1 x2 ) and nullity(T ) = 3. 3 1 1 421
Chapter 8
ISM: Linear Algebra
47. T (A + B)(x) = x T (A + B)x = x T Ax + x T Bx equals (T (A) + T (B))(x) = T (A)(x) + T (B)(x) = x T Ax + x T Bx. The verification of the second axiom of linearity is analogous. By definition of a quadratic form, im(T ) = Qn : For every quadratic form q in Qn there is a symmetric n × n matrix A such that q = T (A). Thus, the rank of T is dim(Qn ) = n(n + 1)/2 (see Exercise 42). By the rank nullity theorem, nullity(T ) = dim(Rn×n ) − rank(T ) = n2 − n(n − 1) n(n + 1) = 2 2
Next, let’s think about the kernel of T . In our solution to Exercise 23 we observed that 1 1 1 T (A) = T ( 2 (A + AT )); note that matrix 2 (A + AT ) is symmetric. Now 2 (A + AT ) = 0 T if (and only if) A = −A, that is, if A is skew-symmetric. Thus the skew-symmetric matrices are in the kernel of T . Since the space of skew-symmetric matrices has the same dimension as ker(T ), namely, n(n − 1)/2, we can conclude that ker(T ) consists of all skew-symmetric n × n matrices. 0 0 1 48. The matrix of T with respect to the basis x2 , x1 x2 , x2 is A = 0 1 0 , with the 2 1 1 0 0 0 1 1 eigenvalues 1,1, −1 and corresponding eigenvectors 1 , 0 , 0 . Thus x1 x2 and 0 1 −1 x2 + x2 are eigenfunctions with eigenvalue 1, and x2 − x2 has eigenvalue −1. Yes, T is 1 2 1 2 diagonalizable, since there is an eigenbasis. 1 0 0 49. The matrix of T with respect to the basis x2 , x1 x2 , x2 is A = 0 2 0 , with the 1 2 0 0 4 0 0 1 eigenvalues 1,2,4 and corresponding eigenvectors 0 , 1 , 0 . Thus x2 , x1 x2 , x2 1 2 1 0 0 are eigenfunctions with eigenvalues 1, 2, and 4, respectively. Yes, T is diagonalizable, since there is an eigenbasis. 0 1 0 50. The matrix of T with respect to the basis x2 , x1 x2 , x2 is A = 2 0 2 , with the 1 2 1 0 0 1 1 1 eigenvalues 0,2, −2 and corresponding eigenvectors 0 , 2 , −2 . −1 1 1 422
ISM: Linear Algebra
Section 8.3
Thus x2 − x2 , x2 + 2x1 x2 + x2 , x2 − 2x1 x2 + x2 are eigenfunctions with eigenvalues 0, 2, 1 2 1 2 1 2 and −2, respectively. Yes, T is diagonalizable, since there is an eigenbasis. 51. If B is negative definite, then A = −B is positive definite, so that the determinants of all principal submatrices A(m) are positive. Thus det(B (m) ) = det(−A(m) ) = (−1)m det(A(m) ) is positive for even m and negative for odd m.
8.3
1. σ1 = 2, σ2 = 1 2. The image of the unit circle is the unit circle, since the transformation defined by A preserves length. Thus σ1 = σ2 = 1 by Fact 8.3.2. 3. AT A = In ; the eigenvalues of AT A are all 1, so that the singular values of A are all 1. 4. AT A =
√ 3+ 5 2
1 1 , with eigenvalues λ1,2 = 1 2 =
√ 1+ 5 2
√ 3± 5 2 . √ −1+ 5 2
The singular values of A are σ1 = ≈ 0.62.
≈ 1.62 and σ2 =
√ 3− 5 2
=
5. AT A =
p2 + q 2 0 , with eigenvalues λ1 = λ2 = p2 +q 2 . The singular values of A 0 p2 + q 2 are σ1 = σ2 = p2 + q 2 . A represents a rotation combined with a scaling, with a scaling factor of p2 + q 2 , so that the image of the unit circle is a circle with radius p2 + q 2 .
6. The eigenvalues of AT A are λ1 = 25 and λ2 = 0, so that the singular values of A are σ1 = 5 and σ2 = 0 (these are also the eigenvalues of A; compare with Exercise 24). 1 1 1 , so that v1 = √5 works. The image of the unit circle is the line 2 2 segment connecting the tips of Av1 = 5v1 and A(−v1 ) = −5v1 . See Figure 8.13. E5 = span 7. AT A = 1 0 0 4
λ1 = 4, λ2 = 1; σ1 = 2, σ2 = 1 eigenvectors of AT A : v1 = 1 0 1 , u1 = σ1 (Av1 ) = , v2 = 0 1 1 0 1 2 0 0 1 , so that U = ,Σ= ,V = . 0 −1 0 0 1 1 0 p2 + q 2 0 0 ; λ1 = λ 2 = p 2 + q 2 ; σ1 = σ 2 = p2 + q 2 423 0 , u2 = −1
1 σ2 (Av2 )
=
8. AT A =
p2 + q 2
Chapter 8
5 v1
ISM: Linear Algebra
v1
A
– v1 –5 v1
Figure 8.13: for Problem 8.3.6. eigenvectors of AT A : v1 = √
1 p2 +q 2
p 1 0 1 1 , u2 = , v2 = , u1 = σ1 Av1 = √ 2 2 p +q q 0 1 −q p −q 1 , so that U = √ 2 2 , Σ = ( p2 + q 2 )I2 , V = I2 . p +q p q p 5 10 10 20 (See Exercise 6)
1 σ2 Av2
=
9. AT A =
λ1 = 25, λ2 = 0; σ1 = 5, σ2 = 0; eigenvectors of AT A : v1 = u1 =
1 √ 5 1 √ 5
1 , v2 = 2 −2 1
1 √ 5
−2 , u1 = 1
1 σ1 Av1
=
1 √ 5
1 , u2 = a unit vector orthogonal to 2
so that U = V =
1 √ 5
1 −2 5 0 ,Σ= . 2 1 0 0 1 2 −2 1 1 −2 2 1 ↑ VT 10 0 0 5
1 √ 5
10. In Example 4 we found transpose of both sides: 6 −7 = 2 6
1 √ 5
6 2 = −7 6 10 0 0 5 ↑ Σ
1 √ 5
2 −1 1 2
; now take the
2 1 −1 2 ↑ U
1 √ 5
.
11. AT A =
1 0 ; λ1 = 4, λ2 = 1; σ1 = 2, σ2 = 1 eigenvectors of AT A : 0 4 0 1 0 0 1 1 1 v1 = , v2 = , u1 = σ1 Av1 = 1 , u2 = σ2 Av2 = 0 , u3 = 0 , 1 0 0 0 1 424
ISM: Linear Algebra 0 1 0 2 0 0 1 U = 1 0 0 , Σ = ,V = . 0 1 1 0 0 0 1 12. In Example 5 we see that 0 1 1 = 1 1 0
1 √ 2
Section 8.3
1 −1 1 1
√
3 0 0 0 1 0
1 √ 6 1 √ 2 1 √ 3
2 √ 6
0
1 − √3
1 √ 6 1 − √2 1 √ 3
.
Now take the transpose of both sides. 1 1 1 √ √ √ √ 6 2 3 0 1 3 0 2 1 1 1 = √ 0 − √3 0 1 6 1 0 0 0 1 1 1 √ √ −√
6 2 3
1 √ 2
1 1 −1 1 ↑ VT
↑ U 13. AT A = v1 =
1 √ 5
↑ Σ
√ √ 37 16 ; λ1 = 45, λ2 = 5; σ1 = 3 5, σ2 = 5 eigenvectors of AT A : 16 13 2 , v2 = 1
1 √ 5
−1 , u1 = 2
1 σ1 Av1
=
1 , v2 = 0
1 σ2 Av2
=
0 , so that 1
√ 1 0 3 5 U= ,Σ= 0 0 1 14. AT A =
0 √ ,V = 5
1 √ 5
2 −1 . 1 2
4 6 ; λ1 = 16, λ2 = 1; σ1 = 4, σ2 = 1 6 13
1 √ 5
eigenvectors of AT A : v1 = =
1 √ 5
1 , v2 = 2
1 √ 5
−2 , u1 = 1
1 σ1 Av1
=
1 √ 5
2 , u2 = 1
1 σ2 Av2
−1 , so that 2
1 √ 5
U=
4 0 2 −1 ,V = ,Σ= 0 1 1 2
1 √ 5
1 −2 . 2 1
1 σ2 v 2 ,
1 15. If Av1 = σ1 u1 and Av2 = σ2 u2 , then A−1 u1 = σ1 v1 and A−1 u2 = singular values of A−1 are the reciprocals of the singular values of A.
so that the
16. If A = U ΣV T then A−1 = V Σ−1 U T and (A−1 )T A−1 = U Σ−1
−1 2
2
U −1 . Thus (A−1 )T A−1
is similar to Σ , so that the eigenvalues of (A−1 )T A−1 are the squares of the reciprocals of the singular values of A. It follows that the singular values of A−1 are the reciprocals of those of A. 425
Chapter 8 17. We need to check that A But A
b·u1 σ1 v 1 b·u1 σ1 v 1
ISM: Linear Algebra +··· +
b·um σm v m
= projimA b (see page 221 of the text).
+ ···+
b·um σm v m
=
b·u1 σ1 Av1
+ · · · + b·um Avm = (b · u1 )u1 + · · · + (b · um )um σm
= projimA b, since u1 , . . . , um is an orthonormal basis of im(A) (see Fact 5.1.5). 1 1 1 3 4 2 1 1 1 1 1 18. b = , u1 = 2 , u2 = 2 , v2 = 1 , v1 = 5 5 3 , σ1 = 2, σ2 = 1, so that 3 1 −1 −4 4 1 −1 x∗ =
b·u1 σ1 v 1
+
b·u2 σ2 v 2
=
−0.1 . −3.2
19. x = c1 v1 + · · · + cm vm is a least-squares solution if Ax = c1 Av1 + · · · + cm Avm = c1 σ1 u1 + · · · + cr σr ur = projimA b. But projimA b = (b · u1 )u1 + · · · + (b · ur )ur , since u1 , . . . , ur is an orthonormal basis of im(A). Comparing the coefficients of ui above we find that it is required that ci σi = b · ui or ci = b·ui , for i = 1, . . . , r, while no σi condition is imposed on cr+1 , . . . , cm . The least-squares solutions are of the form x ∗ = b·ur b·u1 σ1 v1 +· · ·+ σr vr +cr+1 vr+1 +· · · cm vm , where cr+1 , . . . , cm are arbitrary (see Exercise 17 for a special case).
20. a. A = U ΣV T = U V T V ΣV T = QS, where Q = U V T and S = V ΣV T . Note that Q is orthogonal, being the product of orthogonal matrices; S is symmetric as S T = (V T )T ΣT V T = V ΣV T = S; and S is similar to Σ, so that the eigenvalues of S are the (non-negative) diagonal entries of Σ. b. Yes, write A = U ΣV T = U ΣU T U V T = S1 Q1 where S1 = U ΣU T and Q1 = U V T .
1 1 2 21. A = √ 5 −2 1 U 1 1 2 =√ 5 −2 1 U = 1 4 3 −3 4 5 Q
10 0 0 5
Σ
1 2 −1 √ 2 5 1 VT 1 2 1 √ 5 −1 2 V 10 0 0 5
Σ
1 2 −1 √ 2 5 1 VT 9 −2 −2 6 S
1 2 −1 √ 2 5 1 VT
426
ISM: Linear Algebra
Section 8.3
22. a. Note that the transformation S preserves length, as shown in Figure 8.14.
SA unit circle
σ2
σ1
A σ2 σ1
S
Figure 8.14: for Problem 8.3.22a. b. (SA)T SA = AT S T SA = AT A since S is orthogonal. Thus, SA and A have the same singular values.
23. AAT U = U ΣV T V ΣT U T U = U ΣSigmaT , since V T V = Im and U T U = In , so that AAT ui =
2 σi u i
for i = 1, . . . , r for i = r + 1, . . . , n
0
The nonzero eigenvalues of AT A and AAT are the same. 24. The eigenvalues of AT A = A2 are the squares of the eigenvalues of A, so that the singular values of A are the absolute values of the eigenvalues of A. 25. See Figure 8.15. 427
Chapter 8
ISM: Linear Algebra
A unit circle
→
Au σ2
→
u
σ1
Figure 8.15: for Problem 8.3.25. Algebraically: Write u = c1 v1 + c2 v2 and note that u Then Au = c1 σ1 u1 + c2 σ2 u2 , so that Au Au ≥ σ2 . Likewise Au ≤ σ1 .
2 2
= c2 + c2 = 1. 1 2
2 2 2 2 2 = c2 σ1 + c2 σ2 ≥ c2 σ2 + c2 σ2 = σ2 and 1 2 1 2
26. Write v = c1 v1 + · · · + cm vm and note that v 2 = c2 + · · · + c2 . Then Av = c1 σ1 u1 + m 1 2 2 2 2 2 2 2 · · · + cr σr ur and Av 2 = c2 σ1 + c2 σ2 + · · · + c2 σr ≤ c2 σ1 + c2 σ1 + · · · + c2 σ1 ≤ σ1 v 2 1 2 r 1 2 r so that Av ≤ σ1 v . Likewise, Av ≥ σm v . 27. Let v be a unit eigenvector with eigenvalue λ and use Exercise 26. 28. If λ1 , . . . , λn are the eigenvalues of AT A, then (det A)2 = det(AT A) = λ1 · · · λn = 2 2 σ1 · · · σn , so that | det A| = σ1 · · · σn . For a 2 × 2 matrix:
A unit circle Ω σ2 σ1 A(Ω)
Figure 8.16: for Problem 8.3.28. area of ellipse A(Ω) = area of unit circle Ω 428
πσ1 σ2 π
| det(A)| = expansion factor =
= σ1 σ2 . See Figure 8.16.
ISM: Linear Algebra σ1 0 .. . σr 0 .. . 0
T v1 . . . T vr . . .
True or False
29. A = U ΣV T
= [u1 · · · ur · · ·] 1 2 −2 1 2] =
σ vT 1 1 . . = [u1 · · · ur · · ·] . T σr v r 0
T T = σ1 u1 v 1 + · · · + σ r ur v r
30.
6 2 = −7 6
1 +5 √5
1 √ 5 1 √ [1 5
10 0 0 5
1 √ 5
2 −1 1 1 = 10 √5 1 2 −2
1 √ [2 5
− 1]
2 1
4 −2 2 4 + −8 4 1 2
T T 31. The formula A = σ1 u1 v1 + · · · + σr ur vr gives such a representation.
32. (SAR)T SAR = RT AT S T SAR = RT AT AR is similar to AT A, so that the matrices AT A and (SAR)T SAR have the same eigenvalues. Thus A and SAR have the same singular values. 33. Yes; since AT A is diagonalizable and has only 1 as an eigenvalue, we must have AT A = In . 34. A = U ΣU T means that U T AU = U −1 AU = Σ, i.e., A is orthogonally diagonalizable, with eigenvalues ≥ 0. This is the case if and only if A is symmetric and positive semidefinite. 35. We will freely use the diagram on Page 389 (with r = m). We have AT Avi = AT (σi ui ) = 1 2 σi vi and therefore (AT A)−1 vi = σ2 vi for i = 1, . . . , m. Then (AT A)−1 AT ui = (AT A)−1 (σi vi ) =
i
for i = 1, . . . , m and (AT A)−1 AT ui = 0 for i = m + 1, . . . , n since ui is in ker(AT ) in this case. Note that (AT A)−1 AT ui is the least-squares solution of the equation Ax = ui ; for i = 1, . . . , m this is the exact solution since ui is in im(A). 36. We will freely use the diagram on page 389. By construction of the vi as eigenvectors 1 2 of AT A we have AT Avi = λi vi = σi vi , or (AT A)−1 vi = σ2 vi . Then A(AT A)−1 AT ui =
i
1 σi v i
A(AT A)−1 (σi vi ) = A
1 σi v i
=
1 σi Avi
= ui for i = 1, . . . , m and A(AT A)−1 AT ui = 0 for
i = m + 1, . . . , n since ui is in ker(AT ) in this case. The fact that A(AT A)−1 AT ui = ui if i = 1, . . . , m means that the matrix A(AT A)−1 AT represents the orthogonal 0 if i = m + 1, . . . , n projection onto im(A) = span(u1 , . . . , um ).
True or False
1. T, by the spectral theorem (Fact 8.1.1) 429
Chapter 8 a b b c 1 = a > 0, by Definition 8.2.3. 0 0 −1 1 0
ISM: Linear Algebra
2. T. Note that [ 1 0 ]
3. F. The orthogonal matrix A = 4. T. If A =
fails to be diagonalizable (over R). = [25] is λ = 25, so that the
3 3 , then the eigenvalue of AT A = [ 3 4 ] 4 4 √ singular value of A is σ = λ = 5.
5. F. The last term, 5x2 , does not have the form required in Definition 8.2.1 6. F. The singular values of A are the square roots of the eigenvalues of A T A, by Definition 8.3.1. 7. T, by Fact 8.2.4. 8. T, by Definition 8.2.1 2 λ1 . 0 λ1 . 0 9. T. If D = . . . , then DT D = D2 = . . . . The eigenvalues of D T D are 0 . λ2 0 . λn n λ2 , . . . , λ2 , and the singular values of D are λ2 = |λ1 |, . . . , λ2 = |λn |. 1 n n 1 10. F, since det 2
5 2
3
5 2
1 = − 4 < 0 (see Fact 8.2.7).
11. F. Consider A =
0 −1 . 1 0
12. T, since AAT is symmetric (use the spectral theorem) 13. T, by Fact 8.2.4: all the eigenvalues are positive. 14. T, since the matrix is symmetric. 15. F. Consider the shear matrix A = 1 1 . The unit circle isn’t mapped into itself, so 0 1 that the singular values fail to be 1, 1.
16. F. In general, (AT A)T = AT A = AAT 17. T, by Fact 8.3.2 18. T. All four eigenvalues are negative, so that their product, the determinant, is positive. 19. T, by Fact 8.1.2 430
ISM: Linear Algebra 20. F, since the determinant is 0, so that 0 is an eigenvalue. 21. F. As a counterexample, consider A = S = 2In . 22. T, since eiT Aei = aii < 0.
True or False
23. T. By Fact 7.3.6, matrices A and B have the same eigenvalues. Now use Fact 8.2.4. 24. T. The spectral theorem guarantees that there is an orthogonal R such that R T AR is diagonal. Now let S = RT . 25. F. Let A = I2 , v = 1 1 . ,w= 1 0
26. T. Consider the singular value decomposition A = U V T , or AV = U , where V is orthogonal (see Fact 8.3.5). We can let S = V , since the columns of AS = AV = U are orthogonal, by construction. 27. T. By the spectral theorem, A is diagonalizable: S −1 AS = D for some invertible S and a diagonal D. Now D n = S −1 An S = S −1 0S = 0, so that D = 0 (since D is diagonal). Finally, A = SDS −1 = S0S −1 = 0, as claimed. 28. F. If k is negative, then kq(x) will be negative definite. 29. T. The eigenvalues λ1 , . . . , λn of A are nonzero, since A is invertible, so that the eigenvalues λ2 , . . . , λ2 of A2 are positive. Now use Fact 8.2.4. 1 n 30. T, by Fact 8.3.2, since v = Ae1 and w = Ae2 are the principal semi-axes of the image of the unit circle. 31. F. For example, (x2 )(x2 x3 ) fails to be a quadratic form. 1 32. T. We can write q(x) = x T 1
5 2 5 2
4
x.
−1 0 0 33. F. Consider A = 0 1 0 , which is indefinite. 0 0 1 34. T, by Definition 8.2.3: x T (A + B)x = x T Ax + x T Bx > 0 for all nonzero x. 35. T, since x · Ax is positive, so that cos θ is positive, where θ is the angle enclosed by x and Ax. 36. T. Preliminary remark: If σ is the largest singular value of an n × m matrix M , then M v ≤ σ v for all v in Rm (see Exercise 8.3.26). Now let σ1 , σ2 be the singular values of matrix AB , with σ1 ≥ σ2 , and let v1 be a unit vector in R2 such that ABv1 = σ1 (see Fact 8.3.3). Now σ2 ≤ σ1 = A(Bv1 ) ≤ 3 Bv1 ≤ 3 · 5 v1 = 15, proving our claim; note that we have used the preliminary remark twice. 431
Chapter 8 0 1 . 0 0
ISM: Linear Algebra
37. F. Consider A =
38. T. If λ is the smallest eigenvalue of A, let k = 1 − λ. Then the smallest eigenvalue of A + kIn is λ + k = 1, so that all the eigenvalues of A + kIn are positive. Now use Fact 8.2.4. 39. F. Let A = 0 1 0 0 and B = 1 0 . Then 1 is a singular value of BA but not of AB . 0 0
40. T, since A + A−1 = A + AT is symmetric. b c x1 41. T. The quadratic form q(x1 , x2 ) = [ x1 d e 0 = ax2 + 2cx1 x2 + f x2 1 2 x2 e f a c , and det(A) = is positive definite. The matrix of this quadratic form is A = c f af − c2 > 0 since A is positive definite. Thus af > c2 , as claimed. a 0 x2 ] b c 42. F. Consider the positive definite matrix A = 43. F. Consider the indefinite matrix A = 1 −1 . −1 2
1 0 . 0 −1
44. T. By Fact 8.3.2., the continuous function f (x) = A
cos x has the global maximum 5 sin x and the global minimum 3. Note that the image of the unit circle consists of all vectors cos x of the form A . By the intermediate value theorem, f (c) = 4 for some c. Let sin x cos c (draw a sketch!). u= sin c
2
45. T, since x T A2 x = −x T AT Ax = −(Ax)T Ax = − Ax
≤ 0 for all x.
46. T. If λ1 , . . . , λn are the eigenvalues of AT A, then λ1 λ2 . . . λn = det(AT A) = (det A)2 . If √ √ σ1 = λ1 , . . . , σn = λn are the singular values of A, then √ σ1 σ2 . . . σn = λ1 λ2 . . . λn = | det A|, as claimed. 47. F. Note that the columns of S must be unit eigenvectors of A. There are two distinct real eigenvalues, λ1 , λ2 , and for each of them there are two unit eigenvectors, ±v1 (for λ1 ) and ±v2 (for λ2 ). (Draw a sketch!) Thus there are 8 matrices S, namely S = [ ±v1 ±v2 ] and S = [ ±v2 ±v1 ] 48. T. See the remark following Definition 8.2.1. 432
ISM: Linear Algebra 49. F. Some eigenvalues of A may be negative. 50. F. Consider the similar matrices A = 0 0 and B = 0 3 singular values 0 and 3, while those of B are 0 and 5.
True or False
0 4 . Matrix A has the 0 3
51. T. Let v1 , v2 be an orthonormal eigenbasis, with Av1 = v1 and Av2 = 2v2 . Consider a nonzero vector x = c1 v1 + c2 v2 ; then Ax = c1 v1 + 2c2 v2 . If c1 = 0, then x = c2 v2 and Ax = 2c2 v2 are parallel, and we are all set. Now consider the case when c1 = 0. Then the angle between x and Ax is arctan(2c2 /c1 ) − arctan(c2 /c1 ); to see this, subtract the angle between v1 and x from the angle between v1 and Ax (draw a sketch). Let m = c2 /c1 and use calculus to see that the function f (m) = arctan(2m) − arctan(m) 1 assumes its global maximum at m = √2 . The maximal angle between x and Ax is √ √ arctan( 2) − arctan(1/ 2) < 0.34 < π/6. 52. T. Let A = √ a b 1 a . By Fact 8.3.2, A = = a2 + c2 < 5 (since the length of c d 0 c the semi-major axis of the image of the unit circle is less than 5). Thus a < 5 and c < 5. Likewise, b < 5 and d < 5.
53. T. We need to show that each entry aij = aji off the diagonal is smaller than some entry on the diagonal. Now (ei − ej )T A(ei − ej ) = aii + ajj − 2aij > 0, so that aii + ajj > 2aij . Thus the larger of the diagonal entries aii and ajj must exceed aij .
433