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r
7.1 SOLUTIONS
Notes:
Students can profit by reviewin
working on this section. Theorems 1 an
d
sections that follow. Note that symmetri
c
text have real entries, as mentioned at
been constructed so that mastery of the
G
Theorem 2 is easily proved for the
2
1. Since
35 ,
57
T
AA
⎡⎤
==
⎢⎥
−
⎣⎦
the matri
3. Since
22 ,
44
T
AA
⎡⎤
=≠
⎢⎥
⎣⎦
the matri
x
r
n
g Section 5.3 (focusing on the Diagonalization The
o
d
2 and the calculations in Examples 2 and 3 are imp
o
c matri
x
means real symmetric matrix, because all m
a
the beginning of this chapter. The exercises in this
s
G
ra
m
-Schmidt process is not needed.
2
× 2 case:
i
x is symmetric.
x
is not symmetric.
o
rem) before
o
rtant for the
a
trices in the
s
ection have
406 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
6. Since A is not a square matrix
T
AA≠
and the matrix is not symmetric.
7. Let .6 .8 ,
.8 .6
P⎡⎤
=⎢⎥
−
⎣⎦
and compute that
8. Let
1/ 2 1/ 2 ,
1/ 2 1/ 2
P
⎡⎤
−
=⎢⎥
⎢⎥
⎣⎦
and compute that
9. Let 52
,
25
P−
⎡⎤
=⎢⎥
⎣⎦
and compute that
Thus P is not orthogonal.
10. Let
122
212,
221
P
−
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
−
⎣⎦
and compute that
Thus P is not orthogonal.
7.1 • Solutions 407
12. Let
.5 .5 .5 .5
.5 .5 .5 .5 ,
.5 .5 .5 .5
.5 .5 .5 .5
P
−−
⎡⎤
⎢⎥
−−
⎢⎥
=⎢⎥
⎢⎥
−−
⎢⎥
⎣⎦
and compute that
.5 .5 .5 .5
.5 .5 .5 .5 .
−−
⎡
⎤
⎢
⎥
13. Let 31
.
13
A⎡⎤
=⎢⎥
⎣⎦
Then the characteristic polynomial of A is
22
(3 ) 1 6 8 ( 4)( 2),−λ −=λ−λ+ =λ− λ−
so the eigenvalues of A are 4 and 2. For λ = 4, one computes that a basis for the eigenspace is 1,
1
⎡
⎤
⎢
⎥
⎣
⎦
which can be normalized to get
1
1/ 2 .
1/ 2
⎡⎤
=⎢⎥
⎢⎥
u
For λ = 2, one computes that a basis for the eigenspace
14. Let 15
.
51
A⎡⎤
=⎢⎥
⎣⎦
Then the characteristic polynomial of A is
22
(1 ) 25 2 24 ( 6)( 4),−λ − =λ − λ− = λ− λ+
so the eigenvalues of A are 6 and –4. For λ = 6, one
2
1/ 2 .
1/ 2
⎡⎤
−
=⎢⎥
⎢⎥
⎣⎦
u
Let
15. Let 16 4 .
41
A−
⎡⎤
=⎢⎥
−
⎣⎦
Then the characteristic polynomial of A is
2
(16 )(1 ) 16 17 ( 17)−λ −λ − =λ − λ= λ− λ
, so the eigenvalues of A are 17 and 0. For λ = 17, one
⎣
2
1/ 17 .
4/ 17
⎡⎤
=⎢⎥
⎢⎥
⎣⎦
u
Let
16. Let 724
.
24 7
A−
⎡⎤
=⎢⎥
⎣⎦
Then the characteristic polynomial of A is
2
(7 )(7 ) 576 625−−λ −λ− =λ− =
(25)(25)λ− λ+
, so the eigenvalues of A are 25 and –25. For λ = 25, one computes that a basis for
7.1 • Solutions 409
17. Let
113
131.
311
A
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
The eigenvalues of A are 5, 2, and –2. For λ = 5, one computes that a basis for
basis for the eigenspace is
1
2,
1
⎡⎤
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
which can be normalized to get
2
1/ 6
2/ 6 .
1/ 6
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
u
For λ = –2, one
1
−
1/ 2
⎡⎤
−
1/ 3 1/ 6 1/ 2 50 0
⎡⎤
−
⎡
⎤
18. Let
2360
36 23 0 .
003
A
−−
⎡⎤
⎢⎥
=− −
⎢⎥
⎢⎥
⎣⎦
The eigenvalues of A are 25, 3, and –50. For λ = 25, one computes that a
computes that a basis for the eigenspace is
0
0,
1
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
which is of length 1, so
2
0
0.
1
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
u For λ = –50, one
410 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
19. Let
324
262.
423
A
−
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
⎣⎦
The eigenvalues of A are 7 and –2. For λ = 7, one computes that a basis for
⎣
⎦⎣⎦
⎩⎭
1
1/ 5
2/ 5 ,
0
⎡⎤
−
⎢⎥
=⎢⎥
⎢⎥
⎢⎥
⎣⎦
u
2
4/ 45
2/ 45 .
5/ 45
⎡
⎤
⎢
⎥
=
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
u
For λ = –2, one computes that a basis for the eigenspace is
2
1,
2
−
⎡⎤
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
Then P orthogonally diagonalizes A, and
1
APDP
−
=
.
20. Let
744
450.
409
A
−
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
⎣⎦
The eigenvalues of A are 13, 7, and 1. For λ = 13, one computes that a basis
7.1 • Solutions 411
Then P orthogonally diagonalizes A, and
1
APDP
−
=
.
21. Let
4131
1413
.
3141
1314
A
⎡⎤
⎢⎥
⎢⎥
=⎢⎥
⎢⎥
⎢⎥
The eigenvalues of A are 9, 5, and 1. For λ = 9, one computes that a basis
basis for the eigenspace is
1
1,
1
1
−
⎡⎤
⎢⎥
⎢⎥
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
which can be normalized to get
2
1/2
1/2 .
1/2
1/2
−
⎡
⎤
⎢
⎥
⎢
⎥
=
⎢
⎥
−
⎢
⎥
⎢
⎥
⎣
⎦
u
For λ = 1, one
1/ 2
⎡
⎤
−
0
⎡⎤
[]
1234
1/2 1/ 2 1/ 2 0 9000
1/2 1/2 0 1/2 0500
and 0010
1/2 1/ 2 1/ 2 0
0001
1/2 1/ 2 0 1/ 2
PD
⎡⎤
−−
⎡
⎤
⎢⎥
⎢
⎥
−
⎢⎥
⎢
⎥
== =
⎢⎥
⎢
⎥
−
⎢⎥
⎢
⎥
⎢⎥
⎢
⎥
⎣
⎦
⎣⎦
uu uu
412 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
22. Let
2000
0101
.
0020
0101
A
⎡⎤
⎢⎥
⎢⎥
=⎢⎥
⎢⎥
⎢⎥
⎣⎦
The eigenvalues of A are 2 and 0. For λ = 2, one computes that a basis for
1
⎡
⎤
0
⎡
⎤
⎣
⎦
0
⎡
⎤
that a basis for the eigenspace is
0
1,
0
1
⎡⎤
⎢⎥
−
⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎣⎦
which can be normalized to get
4
0
1/ 2 .
0
1/ 2
⎡⎤
⎢⎥
−
⎢⎥
=⎢⎥
⎢⎥
⎢⎥
⎣⎦
u
Let
100 0 2000
⎡⎤
Then P orthogonally diagonalizes A, and
1
APDP
−
=
.
23. Let
311
131
113
A
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
. Since each row of A sums to 5,
131115 1
11311551
111315 1
A
⎡
⎤⎡ ⎤⎡⎤⎡⎤ ⎡⎤
⎢
⎥⎢ ⎥⎢⎥⎢⎥ ⎢⎥
===
⎢
⎥⎢ ⎥⎢⎥⎢⎥ ⎢⎥
⎢
⎥⎢ ⎥⎢⎥⎢⎥ ⎢⎥
⎣
⎦⎣ ⎦⎣⎦⎣⎦ ⎣⎦
and
1
1
⎡⎤
⎢⎥
⎢⎥
is an eigenvector of A with corresponding eigenvalue λ = 5. The eigenvector may be
7.1 • Solutions 413
projection to an orthogonal basis
11
1, 1
02
⎧⎫−−
⎡⎤⎡⎤
⎪⎪
⎢⎥⎢⎥
−
⎨⎬
⎢⎥⎢⎥
⎪⎪
⎢⎥⎢⎥
⎣⎦⎣⎦
⎩⎭
for the eigenspace, and these vectors can be
24. Let
542
452.
222
A
−−
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
−
⎣⎦
One may compute that
220 2
220102
110 1
A
−− −
⎡
⎤⎡ ⎤ ⎡⎤
⎢
⎥⎢ ⎥ ⎢⎥
==
⎢
⎥⎢ ⎥ ⎢⎥
⎢
⎥⎢ ⎥ ⎢⎥
⎣
⎦⎣ ⎦ ⎣⎦
, so
1
2
2
1
−
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
v is an
computes that a basis for the eigenspace is
11
1,0 .
02
⎧
⎫
⎡
⎤⎡⎤
⎪
⎪
⎢
⎥⎢⎥
⎨
⎬
⎢
⎥⎢⎥
⎪
⎪
⎢
⎥⎢⎥
⎣
⎦⎣⎦
This basis may be converted via orthogonal
2
v
, and
3
v
may be normalized to get the vectors
1
2/3
2/3 ,
1/3
−
⎡
⎤
⎢
⎥
=
⎢
⎥
⎢
⎥
⎣
⎦
u
2
1/ 2
1/ 2 ,
0
⎡
⎤
⎢
⎥
=
⎢
⎥
⎢
⎥
⎢
⎥
u
and
3
1/ 18
1/ 18 .
4/ 18
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
⎢⎥
u
25. a. True. See Theorem 2 and the paragraph preceding the theorem.
414 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
26. a. True. See Theorem 2.
b . True. See the displayed equation in the paragraph before Theorem 2.
27. Since A is symmetric,
()
TTTTTTT
BAB BAB BAB==
, and
T
BAB
is symmetric. Applying this result
28. Let A be an n × n symmetric matrix. Then
29. Since A is orthogonally diagonalizable,
1
APDP
−
=
, where P is orthogonal and D is diagonal. Since
30. If A and B are orthogonally diagonalizable, then A and B are symmetric by Theorem 2. If AB = BA,
31. The Diagonalization Theorem of Section 5.3 says that the columns of P are linearly independent
32. If
1
,A PRP
−
=
then
1
.PAP R
−
=
Since P is orthogonal,
T
R
PAP=
. Hence
33. It is previously been found that A is orthogonally diagonalized by P, where
1/ 2 1/ 6 1/ 3 800
⎡⎤
−−
⎡
⎤
Thus the spectral decomposition of A is
7.1 • Solutions 415
1/ 2 1/ 2 0 1/6 1/ 6 2/6 1/3 1/3 1/3
−−
⎡⎤⎡ ⎤⎡⎤
34. It is previously been found that A is orthogonally diagonalized by P, where
1/ 2 1/ 18 2/3 70 0
⎡⎤
−−
⎡
⎤
Thus the spectral decomposition of A is
111 2 2 2 3 3 3 11 2 2 3 3
77 2
TTTTTT
A=+ + =+−uu u u u u uu u u uu
1/ 2 0 1/2 1/18 4/18 1/18 4/9 2/9 4/9
−− −
⎡⎤⎡ ⎤⎡ ⎤
35. a. Given x in
n
,
() ()(),
TTT
b===xuuxuux uxu
because
T
ux
is a scalar. So Bx = (x ⋅ u)u. Since
u is a unit vector, Bx is the orthogonal projection of x onto u.
36. Given any y in
n
, let
ˆ
y
= By and z = y –
ˆ
y
. Suppose that
T
BB=
and
2
BB=
. Then
.
T
BB BB B==
b. Any vector in W = Col B has the form Bu for some u. Noting that B is symmetric, Exercise 28
gives
( y –
ˆ
y
) ⋅ (Bu) = [B(y –
ˆ
y
)] ⋅ u = [By – BBy] ⋅ u = 0
416 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
37. [M] Let
5296
2569
.
9652
6925
A
−
⎡⎤
⎢⎥
−
⎢⎥
=⎢⎥
−
⎢⎥
−
⎢⎥
⎣⎦
The eigenvalues of A are 18, 10, 4, and –12. For λ = 18, one
λ = 10, one computes that a basis for the eigenspace is
1
1
1
1
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
, which can be normalized to get
1/ 2
⎡
⎤
⎢
⎥
1
⎡⎤
⎢⎥
which can be normalized to get
4
1/2
1/2 .
1/2
1/2
⎡⎤
⎢⎥
−
⎢⎥
=⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
u
Let
⎣
⎦
orthogonally diagonalizes A, and
1
APDP
−
=
.
38. [M] Let
.38 .18 .06 .04
.18 .59 .04 .12 .
.06 .04 .47 .12
A
−−−
⎡⎤
⎢⎥
−−
⎢⎥
=⎢⎥
−− −
The eigenvalues of A are .25, .30, .55, and .75. For λ =
⎣
⎦
⎣
⎦
For
7.1 • Solutions 417
λ = .30, one computes that a basis for the eigenspace is
1
2,
2
−
⎡
⎤
⎢
⎥
−
⎢
⎥
⎢
⎥
which can be normalized to get
⎣⎦
⎣
⎦
normalized to get
3
.4
.2 .
.8
.4
⎡
⎤
⎢
⎥
−
⎢
⎥
=
⎢
⎥
−
⎢
⎥
⎢
⎥
⎣
⎦
u
For λ = .75, one computes that a basis for the eigenspace is
2
4,
1
2
−
⎡⎤
⎢⎥
⎢⎥
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
.4
−
⎡⎤
⎢⎥
.8 .2 .4 .4
−−
⎡
⎤
⎢
⎥
39. [M] Let
.31 .58 .08 .44
.58 .56 .44 .58 .
.08 .44 .19 .08
A
⎡⎤
⎢⎥
−−
⎢⎥
=⎢⎥
−
The eigenvalues of A are .75, 0, and –1.25. For λ = .75, one
⎣
⎦⎣⎦
⎩⎭
13
13
⎧⎫
⎡⎤⎡ ⎤
⎪⎪
⎢⎥⎢ ⎥
⎢⎥⎢ ⎥
⎪⎪
−
⎢⎥⎢ ⎥
⎣⎦⎣ ⎦
⎩⎭
1/ 2
⎡⎤
3/ 50
⎡⎤
2
−
⎡⎤
418 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
.4
−
⎡⎤
⎣⎦
eigenspace is
2
4,
1
2
−
⎡⎤
⎢⎥
⎢⎥
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
which can be normalized to get
4
.4
.8 .
.2
.4
−
⎡
⎤
⎢
⎥
⎢
⎥
=
⎢
⎥
−
⎢
⎥
⎢
⎥
⎣
⎦
u
1/ 2 3/ 50 .4 .4
⎡⎤
−−
.75 0 0 0
⎡
⎤
Then P orthogonally diagonalizes A, and
1
APDP
−
=
.
10226 9
210 2 6 9
−
⎡⎤
⎢⎥
−
one computes that a basis for the eigenspace is
11
10
,.
01
00
00
⎧
⎫
−
⎡
⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢ ⎥
−
⎪
⎪
⎢
⎥⎢ ⎥
⎪
⎪
⎢
⎥⎢ ⎥
⎨
⎬
⎢
⎥⎢ ⎥
⎪
⎪
⎢
⎥⎢ ⎥
⎪
⎪
⎢
⎥⎢ ⎥
⎪
⎪
⎣
⎦⎣ ⎦
⎩⎭
This basis may be converted via
1/ 2
⎡⎤
1/ 6
⎡
⎤
1
⎡⎤
7.1 • Solutions 419
1/ 12
⎡⎤
⎢⎥
1
1
⎡⎤
⎢⎥
⎢⎥
1/ 20
1/ 20
⎡
⎤
⎢
⎥
⎢
⎥
1
1
1
⎡⎤
⎢⎥
⎢⎥
⎣⎦
1/ 5
1/ 5
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
1/ 2 1/ 6 1/ 12 1/ 20 1/ 5
1/ 2 1/ 6 1/ 12 1/ 20 1/ 5
⎡⎤
⎢⎥
−
⎢⎥
80 0 0 0
08 0 0 0
.
0032 0 0
D
⎡⎤
⎢⎥
⎢⎥
⎢⎥
=⎢⎥
Then P orthogonally diagonalizes A, and
1
APDP
−
=
.
7.2 SOLUTIONS
Notes:
This section can provide a good conclusion to the course, because the mathematics here is widely
used in applications. For instance, Exercises 23 and 24 can be used to develop the second derivative test
for functions of two variables. However, if time permits, some interesting applications still lie ahead.
Theorem 4 is used to prove Theorem 6 in Section 7.3, which in turn is used to develop the singular value
decomposition.
1. a.
[]
2
12
12 1 122
2
51/3 5(2/3)
1/3 1
Tx
Axx x xxx
x
⎡⎤
⎡⎤
==++
⎢⎥
⎢⎥
⎣⎦
⎣⎦
xx
b . When
6,
1
⎡⎤
=⎢⎥
⎣⎦
x
22
5(6) (2 / 3)(6)(1) (1) 185.
T
A=+ +=xx
2. a.
[]
1
222
123 2 1 231223
3
430
321 4 2 6 2
011
T
x
Axxx x x xx xxxx
x
⎡⎤⎡⎤
⎢⎥⎢⎥
==++++
⎢⎥⎢⎥
⎢⎥⎢⎥
⎣⎦⎣⎦
xx
b . When
2
1,
5
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
⎣⎦
x
222
4(2) 2( 1) (5) 6(2)( 1) 2( 1)(5) 21.
T
A=+−++−+−=xx
3. a. The matrix of the quadratic form is
10 3 .
33
−
⎡
⎤
⎢
⎥
−−
⎣
⎦
4. a. The matrix of the quadratic form is
15/ 2 10
⎡
⎤
⎢
⎥
−
⎣
⎦
832
−
⎡
⎤
⎣
⎦
7.2 • Solutions 421
023
⎡
⎤
6. a. The matrix of the quadratic form is
55/2 3/2
5/2 1 0 .
3/2 0 7
−
⎡
⎤
⎢
⎥
−
⎢
⎥
⎢
⎥
−
⎣
⎦
7. The matrix of the quadratic form is
15
.
51
A
⎡
⎤
=
⎢
⎥
⎣
⎦
The eigenvalues of A are 6 and –4. An eigenvector
for λ = 6 is
1,
1
⎡⎤
⎢⎥
⎣⎦
which may be normalized to
1
1/ 2 .
1/ 2
⎡
⎤
=
⎢
⎥
⎢
⎥
⎣
⎦
u
An eigenvector for λ = –4 is
1,
1
−
⎡⎤
⎢⎥
⎣⎦
the new quadratic form is
8. The matrix of the quadratic form is
944
470.
4011
A
−
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
The eigenvalues of A are 3, 9, and 15. An
is
1
2,
2
−
⎡⎤
⎢⎥
⎢⎥
⎢⎥
⎣⎦
which may be normalized to
2
1/3
2/3 .
2/3
−
⎡
⎤
⎢
⎥
=
⎢
⎥
⎢
⎥
⎣
⎦
u An eigenvector for λ = 15 is
2
1,
2
⎡⎤
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
which may
be normalized to
3
2/3
1/3 .
2/3
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
⎣
⎦
u Then
1
APDP
−
=
, where
422 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
9. The matrix of the quadratic form is
32
.
26
A
−
⎡
⎤
=
⎢
⎥
−
⎣
⎦
The eigenvalues of A are 7 and 2, so the
quadratic form is positive definite. An eigenvector for λ = 7 is
1,
2
−
⎡
⎤
⎢
⎥
⎣
⎦
which may be normalized to
10. The matrix of the quadratic form is
94
.
43
A
−
⎡
⎤
=
⎢
⎥
−
⎣
⎦
The eigenvalues of A are 11 and 1, so the
quadratic form is positive definite. An eigenvector for λ = 11 is
2,
1
⎡
⎤
⎢
⎥
−
⎣
⎦
which may be normalized to
1
2/ 5 .
1/ 5
⎡⎤
=⎢⎥
−
⎢⎥
⎣⎦
u
An eigenvector for λ = 1 is
1
2
⎡
⎤
⎢
⎥
⎣
⎦
, which may be normalized to
2
1/ 5 .
2/ 5
⎡⎤
=⎢⎥
⎢⎥
⎣⎦
u
Then
11. The matrix of the quadratic form is
25
.
52
A
⎡
⎤
=
⎢
⎥
⎣
⎦
The eigenvalues of A are 7 and –3, so the quadratic
form is indefinite. An eigenvector for λ = 7 is
1,
1
⎡
⎤
⎢
⎥
which may be normalized to
1
1/ 2 .
⎡⎤
=⎢⎥
u
An
7.2 • Solutions 423
12. The matrix of the quadratic form is
52
.
22
A
−
⎡
⎤
=
⎢
⎥
−
⎣
⎦
The eigenvalues of A are –1 and –6, so the
quadratic form is negative definite. An eigenvector for λ = –1 is
1,
2
⎡
⎤
⎢
⎥
⎣
⎦
which may be normalized to
13. The matrix of the quadratic form is
13
.
39
A
−
⎡
⎤
=
⎢
⎥
−
⎣
⎦
The eigenvalues of A are 10 and 0, so the
quadratic form is positive semidefinite. An eigenvector for λ = 10 is
1,
3
⎡
⎤
⎢
⎥
−
⎣
⎦
which may be
14. The matrix of the quadratic form is
83
.
30
A
⎡
⎤
=
⎢
⎥
⎣
⎦
The eigenvalues of A are 9 and –1, so the quadratic
form is indefinite. An eigenvector for λ = 9 is
3,
1
⎡
⎤
⎢
⎥
⎣
⎦
which may be normalized to
1
3/ 10 .
1/ 10
⎡⎤
=⎢⎥
⎢⎥
⎣⎦
u
An
424 CHAPTER 7 • Symmetric Matrices and Quadratic Forms
15. [M] The matrix of the quadratic form is
2222
2600
.
2093
2039
A
−
⎡
⎤
⎢
⎥
−
⎢
⎥
=
⎢
⎥
−
⎢
⎥
−
⎢
⎥
⎣
⎦
The eigenvalues of A are 0, –6, –
8, and –12, so the quadratic form is negative semidefinite. The corresponding eigenvectors may be
computed:
These eigenvectors may be normalized to form the columns of P, and
1
APDP
−
=
, where
3/ 12 0 1/2 0 000 0
⎡⎤
−
⎡
⎤
⎢⎥
The desired change of variable is x = Py, and the new quadratic form is
16. [M] The matrix of the quadratic form is
43/2 0 2
3/2420
.
0243/2
203/24
A
−
⎡
⎤
⎢
⎥
⎢
⎥
=
⎢
⎥
⎢
⎥
−
⎢
⎥
⎣
⎦
The eigenvalues of A are
13/2 and 3/2, so the quadratic form is positive definite. The corresponding eigenvectors may be
computed:
43 4 3
⎧⎫⎧ ⎫−
⎡⎤⎡⎤ ⎡⎤⎡⎤
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 13/2 000
⎡⎤
−
⎡
⎤
The desired change of variable is x = Py, and the new quadratic form is
