7.2 • Solutions 425
⎢
⎢
⎢
⎣
19/2 0 6
9/2 1 6 0 .
−
⎡
⎤
⎢
⎥
⎢
⎥
17/2 and –13/2, so the quadratic form is indefinite. The corresponding eigenvectors may be
computed:
43 4 3
50 5 0
⎧⎫ ⎧ ⎫−
⎡⎤⎡⎤ ⎡⎤⎡⎤
⎢⎥⎢⎥ ⎢⎥⎢⎥
⎪⎪ ⎪ ⎪
⎢⎥⎢⎥ ⎢⎥⎢⎥
⎣⎦⎣⎦ ⎣⎦⎣⎦
⎩⎭ ⎩ ⎭
Each set of eigenvectors above is already an orthogonal set, so they may be normalized to form the
columns of P, and
1
APDP
−
=
, where
⎢
⎢
⎢
⎢
⎢
⎣
3/ 50 4/ 50 3/ 50 4/ 50 17 / 2 0 0 0
⎡⎤
−
⎡
⎤
The desired change of variable is x = Py, and the new quadratic form is
18. [M] The matrix of the quadratic form is
11 6 6 6
6100
.
6001
6010
A
−−−
⎡
⎤
⎢
⎥
−−
⎢
⎥
=
⎢
⎥
−−
⎢
⎥
−−
⎢
⎥
⎣
⎦
The eigenvalues of A are 17, 1, –
1, and –7, so the quadratic form is indefinite. The corresponding eigenvectors may be computed:
30 0 1
11 11
−
⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤
⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥
⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥
⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦
These eigenvectors may be normalized to form the columns of P, and
1
APDP
−
=
, where
⎢
⎢
⎢
3/ 12 0 0 1/2 17 0 0 0
00 0 7
1/ 12 1/ 2 1/ 6 1/2
⎡⎤
−
⎡
⎤
⎢
⎥
⎢⎥
−
⎣
⎦
⎣⎦
The desired change of variable is x = Py, and the new quadratic form is
426 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
19. Since 8 is larger than 5, the
2
2
x
term should be as large as possible. Since
22
12
1xx+=
, the largest
20. Since 5 is larger in absolute value than –3, the
2
1
x
term should be as large as possible. Since
21. a. True. See the definition before Example 1, even though a nonsymmetric matrix could be used to
compute values of a quadratic form.
22. a. True. See the paragraph before Example 1.
b . False. The matrix P must be orthogonal and make
T
PAP
diagonal. See the paragraph before
23. The characteristic polynomial of A may be written in two ways:
22
det( )det ()
ab
AI ad adb
bd
−
⎡⎤
−= =−+ +−
⎢⎥
−
⎣⎦
24. If det A > 0, then by Exercise 23,
12
0>
, so that
1
and
2
have the same sign; also,
2
det 0ad A b=+>
.
a . If det A > 0 and a > 0, then d > 0 also, since ad > 0. By Exercise 23,
12
0ad+=+>
. Since
1
7.3 • Solutions 427
25. Exercise 27 in Section 7.1 showed that
T
BB
is symmetric. Also
( ) || || 0
TT T
BB B B B==≥xxxxx
, so
0. Thus if x ≠ 0,
0
TT
BB >xx
and
T
BB
is positive definite.
26. Let
,
T
APDP=
where
1
.
T
PP
−
=
The eigenvalues of A are all positive: denote them
1
,,.
n
…
Let C
27. Since the eigenvalues of A and B are all positive, the quadratic forms
T
Axx
and
T
Bxx
are positive
28. The eigenvalues of A are all positive by Theorem 5. Since the eigenvalues of
1
A
−
are the reciprocals
7.3 SOLUTIONS
Notes:
Theorem 6 is the main result needed in the next two sections. Theorem 7 is mentioned in Example
1. The matrix of the quadratic form on the left is
520
262.
027
A
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
−
⎣
⎦
The equality of the quadratic
forms implies that the eigenvalues of A are 9, 6, and 3. An eigenvector may be calculated for each
eigenvalue and normalized:
⎡
⎢
⎢
⎢
⎣
1/3 2/3 2/3
−
⎡⎤ ⎡⎤ ⎡⎤
2. The matrix of the quadratic form on the left is
122.
122
A
⎡
⎤
⎢
⎥
=
⎢
⎥
⎢
⎥
⎣
⎦
The equality of the quadratic forms
implies that the eigenvalues of A are 5, 2, and 0. An eigenvector may be calculated for each
eigenvalue and normalized:
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎢
⎣
1/ 3 2/ 6 0
⎡⎤ ⎡ ⎤
−
⎡
⎤
3. (a) By Theorem 6, the maximum value of
T
Axx
subject to the constraint
1
T
=xx
is the greatest
eigenvalue
1
of A. By Exercise 1,
1
9.=
4. (a) By Theorem 6, the maximum value of
Axx
subject to the constraint
1
=xx
is the greatest
eigenvalue
1
of A. By Exercise 2,
1
5.=
(b) By Theorem 6, the maximum value of
T
Axx
subject to the constraint
1
T
=xx
occurs at a unit
⎡⎤
5. The matrix of the quadratic form is
52
.
25
A
−
⎡
⎤
=
⎢
⎥
−
⎣
⎦
The eigenvalues of A are
1
7=
and
2
3.=
7.3 • Solutions 429
(b) By Theorem 6, the maximum value of
T
Axx
subject to the constraint
1
T
=xx
occurs at a unit
(c) By Theorem 7, the maximum value of
T
Axx
subject to the constraints
1
T
=xx
and
0
T
=xu
is
the second greatest eigenvalue
2
of A, which is 3.
(a) By Theorem 6, the maximum value of
T
Axx
subject to the constraint
1
T
=xx
is the greatest
eigenvalue
1
of A, which is 15/2.
(b) By Theorem 6, the maximum value of
T
Axx
subject to the constraint
1
T
=xx
occurs at a unit
(c) By Theorem 7, the maximum value of
T
Axx
subject to the constraints
1
T
=xx
and
0
T
=xu
is
7. The eigenvalues of the matrix of the quadratic form are
1
2,=
2
1,=−
and
3
4.=−
By Theorem
6, the maximum value of
T
Axx
subject to the constraint
1
T
=xx
occurs at a unit eigenvector u
⎡
⎢
⎢
⎢
⎣
8. The eigenvalues of the matrix of the quadratic form are
1
9,=
and
2
3.=−
By Theorem 6, the
maximum value of
T
Axx
subject to the constraint
1
T
=xx
occurs at a unit eigenvector u
⎢
⎢
⎢
⎣
1
−
⎡
⎤
2
−
⎡⎤
430 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
9. This is equivalent to finding the maximum value of
T
Axx
subject to the constraint
1.
T
=xx
By
Theorem 6, this value is the greatest eigenvalue
of the matrix of the quadratic form. The matrix of
10. This is equivalent to finding the maximum value of
T
Axx
subject to the constraint
1
T
=xx
. By
Theorem 6, this value is the greatest eigenvalue
1
of the matrix of the quadratic form. The matrix of
11. Since x is an eigenvector of A corresponding to the eigenvalue 3, Ax = 3x, and
(3 )
TT
A==xxx x
12. Let x be a unit eigenvector for the eigenvalue λ. Then
( ) ()
TT T
A===xxx x xx
since
1
T
=xx
.
So λ must satisfy m ≤ λ ≤ M.
13. If m = M, then let t = (1 – 0)m + 0M = m and
.
n
=xu
Theorem 6 shows that
.
T
nn
Am=uu
Now
suppose that m < M, and let t be between m and M. Then 0 ≤ t – m ≤ M – m and 0 ≤ (t – m)/(M – m)
≤ 1. Let
7.3 • Solutions 431
14. [M] The matrix of the quadratic form is
⎡
⎢
3/2 15 0 1/2
15 3/ 2 1/ 2 0
A
⎢
⎥
=
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
The eigenvalues of A are
1
17,=
2
13,=
3
14,=−
and
4
16.=−
(a) By Theorem 6, the maximum value of
T
Axx
subject to the constraint
1
T
=xx
is the greatest
eigenvalue
1
of A, which is 17.
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎦
(c) By Theorem 7, the maximum value of
T
Axx
subject to the constraints
1
T
=xx
and
0
T
=xu
is
the second greatest eigenvalue
2
of A, which is 13.
⎢
⎢
⎢
0 3/2 5/2 7/2
7/2 5/2 3/2 0
⎡
⎤
⎢
⎥
⎢
⎥
⎣
⎦
1
15/ 2,=
2
1/2,=−
3
5/2,=−
and
4
9/2.=−
(b) By Theorem 6, the maximum value of
T
Axx
subject to the constraint
1
T
=xx
occurs at a unit
1
⎡⎤
⎢
⎢
⎢
1/2
1/2
⎡
⎤
⎢
⎥
⎢
⎥
⎣
⎦
432 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
16. [M] The matrix of the quadratic form is
4355
3033
.
5301
5310
A
−−−
⎡
⎤
⎢
⎥
−−−
⎢
⎥
=
⎢
⎥
−− −
⎢
⎥
−−−
⎢
⎥
⎣
⎦
The eigenvalues of A are
1
9,=
2
3,=
3
1,=
and
4
9.=−
(a) By Theorem 6, the maximum value of
T
Axx
subject to the constraint
1
T
=xx
is the greatest
1
⎢⎥
⎢⎥
⎣⎦
⎢
⎢
⎢
⎢
⎣
2/ 6
0.
⎡
⎤
−
⎢
⎥
17. [M] The matrix of the quadratic form is
6222
21000
.
20133
20313
A
−−−−
⎡
⎤
⎢
⎥
−−
⎢
⎥
=
⎢
⎥
−−
⎢
⎥
−−
⎢
⎥
⎣
⎦
The eigenvalues of A are
1
4,=−
2
10,=−
3
12,=−
and
4
16.=−
eigenvector u corresponding to the greatest eigenvalue
1
of A. One may compute that
3
1
1
1
−
⎡⎤
⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎣⎦
is
⎢
⎢
⎢
⎢
3/ 12
1/ 12
⎡
⎤
−
⎢
⎥
⎣
⎦
(c) By Theorem 7, the maximum value of
T
Axx
subject to the constraints
1
T
=xx
and
0
T
=xu
is
the second greatest eigenvalue
2
of A, which is –10.
7.4 • Solutions 433
7.4 SOLUTIONS
Notes:
The section presents a modern topic of great importance in applications, particularly in computer
calculations. An understanding of the singular value decomposition is essential for advanced work in
science and engineering that requires matrix computations. Moreover, the singular value decomposition
explains much about the structure of matrix transformations. The SVD does for an arbitrary matrix almost
what an orthogonal decomposition does for a symmetric matrix.
1. Let
10
.
03
A⎡⎤
=⎢⎥
−
Then
10
,
09
T
AA ⎡⎤
=⎢⎥
and the eigenvalues of
T
AA
are seen to be (in decreasing
2. Let
50
.
00
A
−
⎡⎤
=⎢⎥
Then
25 0 ,
00
T
AA ⎡⎤
=⎢⎥
and the eigenvalues of
T
AA
are seen to be (in decreasing
3. Let
61
.
06
A
⎡⎤
=⎢⎥
⎢⎥
⎣⎦
Then
66
,
67
T
AA
⎡⎤
=⎢⎥
⎢⎥
⎣⎦
and the characteristic polynomial of
T
AA
is
4. Let
32
.
03
A
⎡⎤
=⎢⎥
⎢⎥
⎣⎦
Then
323
,
23 7
T
AA
⎡⎤
=⎢⎥
⎢⎥
⎣⎦
and the characteristic polynomial of
T
AA
is
2
1.=
Thus the singular values of A are
1
93
σ
==
and
2
11.
σ
==
5. Let
30
.
00
A
−
⎡⎤
=⎢⎥
⎣⎦
Then
90
,
00
T
AA ⎡⎤
=⎢⎥
⎣⎦
and the eigenvalues of
T
AA
are seen to be (in decreasing
order)
1
9=
and
2
0.=
Associated unit eigenvectors may be computed:
434 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
6. Let
20
.
01
A
−
⎡⎤
=⎢⎥
−
⎣⎦
Then
40
,
01
T
AA ⎡⎤
=⎢⎥
⎣⎦
and the eigenvalues of
T
AA
are seen to be (in decreasing
order)
1
4=
and
2
1.=
Associated unit eigenvectors may be computed:
Thus the matrix Σ is
20
.
01
⎡⎤
Σ=⎢⎥
⎣⎦
Next compute
0 10101
T
AU V
=Σ =
⎢⎥⎢⎥⎢⎥
−
⎣⎦⎣⎦⎣⎦
7. Let
21
.
22
A
−
⎡⎤
=⎢⎥
⎣⎦
Then
82
,
25
T
AA ⎡⎤
=⎢⎥
⎣⎦
and the characteristic polynomial of
T
AA
is
2/ 5 1/ 5
9: , 4:
1/ 5 2/ 5
⎡⎤ ⎡ ⎤
−
==
⎢⎥ ⎢ ⎥
⎢⎥ ⎢ ⎥
⎣⎦ ⎣ ⎦
7.4 • Solutions 435
8. Let
23
.
02
A⎡⎤
=⎢⎥
⎣⎦
Then
46
,
613
T
AA ⎡⎤
=⎢⎥
⎣⎦
and the characteristic polynomial of
T
AA
is
2
17 16 ( 16)( 1),−+=− −
and the eigenvalues of
T
AA
are (in decreasing order)
1
16=
and
2
1.=
Associated unit eigenvectors may be computed:
1/ 5 2/ 5
16 : , 1:
2/ 5 1/ 5
⎡⎤ ⎡ ⎤
−
==
⎢⎥ ⎢ ⎥
⎢⎥ ⎢ ⎥
⎣⎦ ⎣ ⎦
2
11.
σ
==
Thus the matrix Σ is
40
.
01
⎡
⎤
Σ=
⎢
⎥
⎣
⎦
Next compute
Since
12
{, }uu
is a basis for
2
, let
2/ 5 1/ 5 .
1/ 5 2/ 5
U
⎡
⎤
−
=
⎢
⎥
⎢
⎥
⎣
⎦
Thus
9. Let
71
00.
55
A
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
Then 74 32 ,
32 26
T
AA
⎡⎤
=⎢⎥
⎣⎦
and the characteristic polynomial of
T
AA
is
2/ 5 1/ 5
90 : , 10 :
1/ 5 2/ 5
⎡⎤ ⎡ ⎤
−
==
⎢⎥ ⎢ ⎥
⎢⎥ ⎢ ⎥
⎣⎦ ⎣ ⎦
436 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
11 2 2
12
1/ 2 1/ 2
11
0, 0
1/ 2 1/ 2
AA
σσ
⎡⎤ ⎡ ⎤
−
⎢⎥ ⎢ ⎥
== = =
⎢⎥ ⎢ ⎥
⎢⎥ ⎢ ⎥
⎣⎦ ⎣ ⎦
uv u v
Since
12
{, }uu
is not a basis for
3
, we need a unit vector
3
u
that is orthogonal to both
1
u
and
2
.u
The vector
3
u
must satisfy the set of equations
1
0
T
=ux
and
2
0.
T
=ux
These are equivalent to the
linear equations
123
3
123
00
00
,so 1 ,and 1
00
00
xxx
xxx
⎡⎤ ⎡⎤
++ = ⎢⎥ ⎢⎥
==
⎢⎥ ⎢⎥
−+ + = ⎢⎥ ⎢⎥
⎣⎦ ⎣⎦
xu
10. Let
42
21.
00
A
−
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
⎣⎦
Then
20 10 ,
10 5
T
AA
−
⎡⎤
=⎢⎥
−
⎣⎦
and the characteristic polynomial of
T
AA
is
2
25 (25)−=−
, and the eigenvalues of
T
AA
are (in decreasing order)
1
25=
and
2
0.=
Associated unit eigenvectors may be computed:
11
1
2/ 5
11/ 5
0
A
σ
⎡⎤
⎢⎥
==
⎢⎥
⎢⎥
uv
7.4 • Solutions 437
Therefore, let
2/ 5 1/ 5 0
1/ 5 2/ 5 0 .
001
U
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
⎢⎥
Thus
11. Let
31
62.
62
A
−
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
−
⎣⎦
Then
81 27 ,
27 9
T
AA
−
⎡⎤
=⎢⎥
−
⎣⎦
and the characteristic polynomial of
T
AA
is
2
90 (90),−=−
and the eigenvalues of
T
AA
are (in decreasing order)
1
90=
and
2
0.=
Associated unit eigenvectors may be computed:
3/ 10 1/ 10
90 : , 0: .
1/ 10 3/ 10
⎡⎤⎡⎤
==
⎢⎥⎢⎥
−
⎢⎥⎢⎥
⎣⎦⎣⎦
438 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
12. Let
11
01.
11
A
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
−
⎣⎦
Then
20
,
03
T
AA ⎡⎤
=⎢⎥
⎣⎦
and the eigenvalues of
T
AA
are seen to be (in decreasing
order)
1
3=
and
2
2.=
Associated unit eigenvectors may be computed:
matrix Σ is
30
02.
00
⎡⎤
⎢⎥
Σ=⎢⎥
⎢⎥
⎢⎥
⎣⎦
Next compute
⎡
⎢
⎢
⎢
⎣
123
3
123
1/ 6
1
0,so 2 ,and 2/ 6
00
11/ 6
xx x
xxx
⎡
⎤
⎡⎤
⎢
⎥
++ = ⎢⎥
=− =−
⎢
⎥
⎢⎥
+− =
⎢
⎥
⎢⎥
⎣⎦
⎢
⎥
⎣
⎦
xu
. Therefore let
1/ 3 1/ 2 1/ 6
1/ 3 0 2/ 6 .
1/ 3 1/ 2 1/ 6
U
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
−
⎢
⎥
⎣
⎦
⎡
⎢
⎣
13. Let
32 2
.
23 2
A⎡⎤
=⎢⎥
−
⎣⎦
Then
32
23,
22
T
A
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
−
17 8 ,
817
TT T T
AA AA
⎡
⎤
==
⎢
⎥
⎣
⎦ and the eigenvalues of