Section 8.

Chapter 8
Section 8.1
8.1.1 ⃗e1 , ⃗e2 is an orthonormal eigenbasis.
! " ! "
√1
1 √1 1
8.1.2 , 2 is an orthonormal eigenbasis.
2 1 −1

! " ! "
√1
2 √1 −1
8.1.3 , 5 is an orthonormal eigenbasis.
5 1 2

⎤
⎡ ⎡ ⎤ ⎡ ⎤
1 1 1
8.1.4 √13 ⎣ 1 ⎦ , √12 ⎣ −1 ⎦ , √16 ⎣ 1 ⎦ is an orthonormal eigenbasis.
−1 0 2

8.1.5 Eigenvalues −1, −1, 2

⎡ ⎤ ⎡ ⎤ ⎡ ⎤
−1 1 1
1 √1 √1
Choose ⃗v1 = √2 ⎣ 1 ⎦ in E−1 and ⃗v2 = 3
⎣ 1 ⎦ in E2 and let ⃗v3 = ⃗v1 × ⃗v2 =
6
⎣ 1 ⎦.
0 1 −2

⎡ ⎤ ⎡ ⎤ ⎡ ⎤
2 2 1
1 1 1
8.1.6 3
⎣ 2 ⎦,
3
⎣ −1 ⎦,
3
⎣ −2 ⎦ is an orthonormal eigenbasis.
1 −2 2

! " ! " ! " ! "

√1
1 √1
1 √1
1 1 5 0
8.1.7 , is an orthonormal eigenbasis, so S = and D = .
2 1 2 −1 2 1 −1 0 1

! " ! " ! "

√1
3 −1 3 −1
8.1.8 , √110 is an orthonormal eigenbasis, with λ1 = 4 and λ2 = −6, so S = √1 and
! 1 3 1 3
10 10
"
4 0
D= .
0 −6

⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 −1 0
8.1.9 √12 ⎣ 0 ⎦, √12 ⎣ 0 ⎦, ⎣ 1 ⎦ is an orthonormal eigenbasis, with λ1 = 3, λ2 = −3, and λ3 = 2, so
1 1 0
⎡ ⎤ ⎡ ⎤
1 −1 √0 3 0 0
1
S = √2 ⎣ 0 0 2 ⎦ and D = ⎣ 0 −3 0 ⎦.
1 1 0 0 0 2

8.1.10 λ1 = λ2 = 0 and λ3 = 9.
⎡ ⎤ ⎡ ⎤
2 1
⃗v1 = √15 ⎣ 1 ⎦ is in E0 and ⃗v2 = 1 ⎣
3 −2 ⎦ is in E9 .
0 2

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Chapter 8

⎤ ⎡
2
1 ⎣
Let ⃗v3 = ⃗v1 × ⃗v2 = 3√ 5
−4 ⎦; then ⃗v1 , ⃗v2 , ⃗v3 is an orthonormal eigenbasis.
−5
⎡ ⎤
√2 1 2
√ ⎡ ⎤
5 3 3 5 0 0 0
⎢ ⎥
S=⎢ √1 − 32 4
− 3√ ⎥ and D = ⎣ 0 9 0 ⎦.
⎣ 5 5 ⎦
√ 0 0 0
2 5
0 3 − 3

⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 −1 0
8.1.11 √1 1
⎣ 0 ⎦, √ ⎣ 0 ⎦, ⎣ 1 ⎦ is an orthonormal eigenbasis, with λ1 = 2, λ2 = 0, and λ3 = 1, so S =
2 2
⎡ 1 ⎤ 1 0
⎡ ⎤
1 −1 0 2 0 0
1 ⎢ √ ⎥
√ ⎣0 0 2 ⎦ and D = ⎣ 0 0 0 ⎦.
2
1 1 0 0 0 1

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 1 0 2
8.1.12 a E1 = span ⎣ 0 ⎦ and E−1 = (E1 )⊥ . An orthonormal eigenbasis is √1 ⎣ 0 ⎦, ⎣ 1 ⎦, √1 ⎣ 0 ⎦.
5 5
2 2 0 −1
⎡ ⎤
1 0 0
b Use Theorem 7.4.1: B = ⎣ 0 −1 0 ⎦.
0 0 −1
⎡ 1 ⎤
⎡ ⎤ √ 0 √2
−0.6 0 0.8 5 5
c A = SBS −1
⎢ 0 1 0 ⎥
=⎣ 0 −1 0 ⎦ , where S = ⎣ ⎦.
0.8 0 0.6 √2 0 − √15
5

8.1.13 Yes; if ⃗v is an eigenvector of A with eigenvalue λ, then ⃗v = I3⃗v = A2⃗v = λ2⃗v , 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
8.1.14 Let S be as in Example 3. Then S −1 AS = ⎣ 0 0 0 ⎦.
0 0 3
⎡ ⎤
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 ⎤
−2 0 0
1
)
−1 1
* 1 −1
AS − I3 ) = ⎣ 0 − 12
⎢ ⎥
c. This is 2 (A − I3 ), so that S 2 (A − I3 ) S = 2 (S 0 ⎦.
0 0 1

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 Section 8.1

8.1.15 Yes, if A⃗v = λ⃗v , then A−1⃗v = λ1 ⃗v , so that an orthonormal eigenbasis for A is also an orthonormal eigenbasis
for A−1 (with reciprocal eigenvalues).

8.1.16 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)

8.1.17 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).
+
8.1.18 By Theorem 6.3.6, the volume is |det A| = det(AT A). Now ⃗vi · ⃗vj = ∥⃗vi ∥∥⃗vj ∥cos(θ) = 12 , so that AT A has
) * ) *n−1 ) 1 *
all 1’s on the diagonal and 21 ’s outside. By Exercise 17 with p = 1 and q = 12 , det(AT A) = 21 1
2n + 2 =
) 1 *n + ) *n/2 √
2 (n + 1), so that the volume is det(AT A) = 12 n + 1.

8.1.19 Let L(⃗x) = A⃗x. 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 A⃗v1 , . . . , A⃗vm are orthogonal, since A⃗vi · A⃗vj = (A⃗vi )T A⃗vj =
⃗viT AT A⃗vj = ⃗vi · (AT A⃗vj ) = ⃗vi · (λj ⃗vj ) = λj (⃗vi · ⃗vj ) = 0 if i ̸= j.

8.1.20 By Exercise 19, there is an orthonormal basis ⃗v1 , . . . , ⃗vm of Rm such that T (⃗v1 ), . . . , T (⃗vm ) are orthogonal.
1
Suppose that T (⃗v1 ), . . . , T (⃗vr ) are nonzero and T (⃗vr+1 ), . . . , T (⃗vm ) are zero. Then let w ⃗ i = ∥T (⃗ vi )∥ T (⃗
vi ) for
⊥
i = 1, . . . , r and choose an orthonormal basis w ⃗ r+1 , . . . , w
⃗ n of [span(w
⃗ 1, . . . , w
⃗ r )] . Then w
⃗ 1, . . . , w
⃗ n does the job.

8.1.21 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.

8.1.22 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.

8.1.23 The eigenvalues are real (by Theorem 8.1.3), so that the only possible eigenvalues are ±1. Since A is
symmetric, E1 and E−1 are orthogonal complements. Thus A represents a reflection about E1 .

8.1.24 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 0 1 0
1 ⎢ 0 ⎥ 1 ⎢ 1 ⎥
√ ⎣ ⎦, √ ⎣ ⎦, √ ⎣ 1 ⎢ 0 ⎥ 1 ⎢ 1⎥
⎦ , √2 ⎣ ⎦ is an orthonormal eigenbasis.
2 0 2 1 2 0 −1
1 0 −1 0

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Chapter 8

8.1.25 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
⎡ ⎤
1 0 0 1 0
⎢ 0 1 √0 0 1 ⎥
The columns of S must form an eigenbasis for A : S = √12 ⎢ 0 0
⎢ ⎥
2 0 0 ⎥ is one possible choice.
⎣ ⎦
0 1 0 0 −1
1 0 0 −1 0

8.1.26 Since Jn is both orthogonal and symmetric, the eigenvalues are 1 and −1. If n is even, then both have
multiplicity n2 (as in Exercise 24). If n is odd, then the multiplicities are n+1 2 for 1 and n−1
2 for −1 (as in
Exercise 25). One way to see this is to observe that tr(Jn ) is 0 for even n, and 1 for odd n (recall that the trace
is the sum of the eigenvalues).

8.1.27 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 n2 each. E2 is the span of all ⃗ei + ⃗en+1−i , for i = 1, . . . , n2 , and E0 is spanned by all
⃗ei −⃗en+1−i . If n is odd, then E2 is spanned by all ⃗ei +⃗en+1−i , for i = 1, . . . , n−1
2 ; E0 is spanned by all ⃗
ei −⃗en+1−i ,
for i = 1, . . . , n−1
2 , and E 1 is spanned by ⃗
e n+1 .
2

8.1.28 For λ ̸= 0
⎡ −λ 0 | 1 ⎤ ⎡ −λ 0 | 1 ⎤
| |
⎢ −λ | 1 ⎥ ⎢ −λ | 1 ⎥
⎢ .. | .. ⎥ ⎢ .. | .. ⎥
⎢
fA (λ) = det ⎢ . | . ⎥
⎥= 1
det
⎢
⎢ . | . ⎥
⎥
⎢ | ⎥ λ ⎢ | ⎥
⎢ 0 −λ | 1 ⎥ ⎢ 0 −λ | 1 ⎥
⎣ | ⎦ ⎣ | ⎦
| |
1 1 ··· 1 | 1−λ λ λ ··· λ | λ − λ2
⎡ −λ 0 | 1 ⎤
|
⎢ −λ | 1 ⎥
⎢ .. | .. ⎥
1
⎢
= λ det ⎢ . | . ⎥
⎥
⎢ | ⎥
⎢ 0 −λ | 1 ⎥
⎣ | ⎦
|
0 0 ··· 0 | −λ2 + λ + 12

= −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⎥
⎢.⎥ ⎢ .⎥
.
E4 = span ⎢ . ⎥ (12 ones), E−3 = span ⎢ .⎥
⎢ . ⎥ (12 ones)
⎢ ⎥
⎣1⎦ ⎣ 1⎦
4 −3

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 Section 8.1

so
⎡ ⎤
1 1 1 1 1 1 1 1 1 1 1 1 1
⎢ −1 0 0 0 0 0 0 0 0 0 0 1 1⎥
⎢ ⎥
⎢ 0 −1 0 0 0 0 0 0 0 0 0 1 1⎥
⎢ ⎥
⎢ 0 0 −1 0 0 0 0 0 0 0 0 1 1⎥
⎢ ⎥
⎢ 0 0 0 −1 0 0 0 0 0 0 0 1 1⎥
⎢ ⎥
⎢ 0 0 0 0 −1 0 0 0 0 0 0 1 1⎥
⎢ ⎥
S=⎢ 0 0 0 0 0 −1 0 0 0 0 0 1 1⎥
⎢ ⎥
⎢ 0 0 0 0 0 0 −1 0 0 0 0 1 1⎥
⎢ ⎥
⎢ 0 0 0 0 0 0 0 −1 0 0 0 1 1⎥
⎢ ⎥
⎢ 0 0 0 0 0 0 0 0 −1 0 0 1 1⎥
⎢ ⎥
⎢ 0 0 0 0 0 0 0 0 0 −1 0 1 1⎥
⎣ ⎦
0 0 0 0 0 0 0 0 0 0 −1 1 1
0 0 0 0 0 0 0 0 0 0 0 4 −3

diagonalizes A, and D = S −1 AS will have all zeros as entries except d12, 12 = 4 and d13, 13 = −3.

8.1.29 By Theorem 5.4.1 (im A)⊥ = ker(AT ) = ker(A), so that ⃗v is orthogonal to w.

⃗

8.1.30 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

A⃗v = ⃗v v T ⃗v = ⃗v (⃗v · ⃗v ) = ⃗v , (since ⃗v · ⃗v = 1) and A⃗vi = ⃗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 = ⎢ ⎣ ... .. .. ⎥
. .⎦
0 0 ... 0

8.1.31 True; A is diagonalizable, that is, A is similar to a diagonal matrix D; then A2 is similar to D2 . Now
rank(D) = rank(D2 ) is the number of nonzero entries on the diagonal of D (and D2 ). Since similar matrices
have the same rank (by Theorem 7.3.6b) we can conclude that rank(A) = rank(D) = rank(D2 ) = rank(A2 ).

1
8.1.32 By Exercise 17, det(A) = (1 − q)n−1 (qn + 1 − q). A is invertible if det(A) ̸= 0, that is, if q ̸= 1 and q ̸= 1−n .

2π
8.1.33 The angles must add up to 2π, so θ = 3 = 120◦ . (See Figure 8.1.)

v1

v2 θ
unit
θ circle
θ

v3

Figure 8.1: for Problem 8.1.33.

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Chapter 8

Algebraically, we can see this as follows: let A = [ ⃗v1 ⃗v2 ⃗v3 ], a 2 × 3 matrix.
⎡ ⎤
1 cos θ cos θ
1
Then AT A = ⎣ cos θ 1 cos θ ⎦ is a noninvertible 3 × 3 matrix, so that cos θ = 1−3 = − 21 , by Exercise 32,
cos θ cos θ 1
and θ = 2π
3 = 120 ◦
.

8.1.34 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
⎢ cos θ 1 cos θ cos θ ⎥ 1
Then AT A = ⎣ ⎦ is noninvertible, so that cos θ = 1−4 = − 13 , by Exercise 32, and
cos θ cos θ 1 cos θ
) * cos θ cos θ cos θ 1
θ = arccos − 31 ≈ 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.

8.1.35 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 ⎥
Then AT A = ⎢ ⎣ ... .. ⎥ is a noninvertible (n + 1) × (n + 1) matrix with 1’s on the diagonal and
. ⎦
cos θ ··· 1 2 3
1 1
cos θ outside, so that cos θ = 1−n , by Exercise 32, and θ = arccos 1−n .

8.1.36 If ⃗v is an eigenvector with eigenvalue λ, then λ⃗v = A⃗v = A2⃗v = λ2⃗v , so that λ = λ2 and 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 .

8.1.37 In Example 4 we see that the image of the unit circle is an ellipse with semi-axes 2 and 3. Thus ||A⃗u|| takes
all values in the interval [2, 3].

8.1.38 The spectral theorem tells us that there exists an orthonormal eigenbasis ⃗v1 , ⃗v2 for A, with associated
eigenvalues -2 and 3. Consider a unit vector ⃗u = c1⃗v1 + c2⃗v2 in R2 , with c21 + c22 = 1. Then ⃗u · A⃗u =
(c1⃗v1 + c2⃗v2 )·(−2c1⃗v1 + 3c2⃗v2 ) = −2c21 +3c22 , which takes all values on the interval [−2, 3] since −2 = −2c21 −2c22 ≤
−2c21 + 3c22 ≤ 3c21 + 3c22 = 3.

8.1.39 The spectral theorem tells us that there exists an orthonormal eigenbasis ⃗v1 , ⃗v2 , ⃗v3 for A, with associated
eigenvalues -2, 3 and 4. Consider a unit vector ⃗u = c1⃗v1 + c2⃗v2 + c3⃗v3 in R3 , with c21 + c22 + c23 = 1. Then

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 Section 8.1

⃗u · A⃗u = (c1⃗v1 + c2⃗v2 + c3⃗v3 ) · (−2c1⃗v1 + 3c2⃗v2 + 4c3⃗v3 ) = −2c21 + 3c22 + 4c23 , which takes all values on the interval
[−2, 4] since −2 = −2c21 − 2c22 − 2c23 ≤ −2c21 + 3c22 + 4c23 ≤ 4c21 + 4c22 + 4c23 = 4.

8.1.40 Using the terminology introduced+ in Exercise 8.1.39, we have

∥A⃗u∥ = ∥−2c1⃗v1 + 3c2⃗v2 + 4c3⃗v3 ∥ = 4c21 + 9c22 + 16c22 , which takes all values on the interval [2, 4]. Geometri-
cally, the image of the unit sphere under A is the ellipsoid with semi-axes 2, 3, and 4.

8.1.41 The spectral theorem tells us that there exists an orthogonal matrix S such that S −1 AS = D is diagonal.
Let D1 be the diagonal matrix such that D13 = D; the diagonal entries of D1 are the cube roots of those of D.
) *3
Now B = SD1 S −1 does the job, since B 3 = SD1 S −1 = SD13 S −1 = SDS −1 = A.
! "
12 14
8.1.42 We will use the strategy outlined in Exercise 8.1.41. An orthogonal matrix that diagonalizes A = 15
! " ! " ! " ! 14 " 33
1 1 −2 −1 8 0 2 0 −1 1 6 2
is S = 5
√ , with S AS = D = . Now D1 = and B = SD1 S = 5 .
2 1 0 1 0 1 2 9
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 1 1
8.1.43 There is an orthonormal eigenbasis ⃗v1 = √12 ⎣ −1 ⎦ , ⃗v2 = √16 ⎣ 1 ⎦ , ⃗v3 = √13 ⎣ 1 ⎦ with associated
0 −2 1
eigenvalues -9, -9, 24. We are looking for a nonzero vector ⃗v = c1⃗v1 + c2⃗v2 + c3⃗v3 such that ⃗v · A⃗v =
(c1⃗v1 + c2⃗v2 + c3⃗v3 ) · (−9c1⃗v1 − 9c2⃗v2 + 24c3⃗v3 ) = −9c21 −⎡9c22 + ⎤
24c23 = 0 or −3c21 − 3c22 + 8c23 = 0. One possible
√ √ √ 3
solution is c1 = 8 = 2 2, c2 = 0, c3 = 3, so that ⃗v = ⎣ −1 ⎦.
1

8.1.44 Use Exercise 8.1.43 as a guide. Consider an orthonormal eigenbasis ⃗v1 , ...., ⃗vn for A, with associated eigen-
values λ1 ≤ λ2 ≤ .... ≤ λn , listed in ascending order. If ⃗v = c1⃗v1 + ... + cn⃗vn is any nonzero vector in Rn , then
⃗v · A⃗v = (c1⃗v1 + ... + cn⃗vn ) · (λ1 c1⃗v1 + ... + λn cn⃗vn ) = λ1 c21 + ... + λn c2n . If all the eigenvalues are positive, then
⃗v · A⃗v will be positive. Likewise, if all the eigenvalues are negative, then ⃗v · A⃗v will be negative. However, if A
has positive as well as negative eigenvalues, meaning√that λ1 < √ 0 < λn (as in Example 8.1.43) , then there exist
nonzero vectors ⃗v with ⃗v · A⃗v = 0, for example, ⃗v = λn⃗v1 + −λ1⃗vn .

8.1.45 a If S −1 AS is upper triangular then the first column of S is an eigenvector

! " of A. Therefore, any matrix
0 −1
without real eigenvectors fails to be triangulizable over R, for example, .
1 0

b Proof by induction on n: For an n × n matrix A ! we can "choose a complex invertible n × n matrix P whose first
−1 λ ⃗v
column is an eigenvector for A. Then P AP = . B is triangulizable, by induction hypothesis, that is,
0 B
there is an invertible
! "(n − 1) × (n −!1) matrix
" Q such
! that "Q!−1 BQ =" !T is upper
" triangular.
! "
1 0 λ ⃗
v 1 0 λ ⃗
v 1 0 λ ⃗v Q
Now let R = . Then R−1 R= = is upper triangular.
! " 0 Q 0 B 0 Q−1 0 B 0 Q 0 T
λ ⃗v
R−1 R = R−1 P −1 AP R = S −1 AS, where S = P R, proving our claim.
0 B

8.1.46 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.

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