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 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
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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 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 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 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, so that aii + ajj > 2aij . Thus the larger of the diagonal entries aii and ajj must exceed aij .
433