r
5.1 SOLUTIONS
Notes
: Exercises 1–6 reinforce the d
e
eigenvectors and difference equations,
a
example and anticipates discussions of
d
1. The number 2 is an eigenvalue of
A
This equation is equivalent to
(−A
2. The number −3 is an eigenvalue o
f
This equation is equivalent to
(A+
⎡⎤⎡⎤
3. Is
Ax
a multiple of x? Compute
1
⎡
⎢
⎣
eigenvalue
2−
.
3
3
r
r
5
⎡
r
e
finitions of eigenvalues and eigenvectors. The s
u
a
long with Exercises 33 and 34, refers to the chapter
d
ynamical systems in Sections 5.2 and 5.6.
A
if and only if the equation
2A=xx
has a nontrivial
s
2) .=
x
I0
Compute
12
f
A if and only if the equation
3A=−xx
has a nontriv
i
3) .I=
x
0
Compute
24
⎡⎤
1
11 2 1
2.
6
43 6 3
−−
⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤
==−
⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥
−−
⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ So 1
3
⎡⎤
⎢⎥
⎣⎦
is an eigenvector
5
21 3 1
⎡⎤
−− −
⎤⎡ ⎤ ⎡ ⎤
−
⎡⎤
u
bsection on
introductory
s
olution.
i
al solution.
of A with
274 CHAPTER 5 • Eigenvalues and Eigenvectors
4333 15 3
−−
⎡⎤⎡⎤⎡⎤⎡⎤
3
⎡⎤
of A for the eigenvalue
5−
.
⎢
⎢
⎣
367 1 5 1
⎡ ⎤⎡⎤⎡⎤ ⎡⎤
⎢ ⎥⎢⎥⎢⎥ ⎢⎥
1
⎡
⎤
⎢
⎥
7. To determine if 4 is an eigenvalue of A, decide if the matrix
4AI−
is invertible.
30 1 400 1 0 1
−−−
⎡⎤⎡⎤⎡ ⎤
⎣⎦⎣⎦⎣ ⎦
Invertibility can be checked in several ways, but since an eigenvector is needed in the event that one
exists, the best strategy is to row reduce the augmented matrix for
(4)AI−=
x0
:
10 10 10 10 10 10
−− −−
⎡⎤⎡⎤⎡⎤
8. To determine if 1 is an eigenvalue of A, decide if the matrix
AI−
is invertible.
423 100 323
−−
⎡⎤⎡⎤⎡⎤
⎣⎦⎣⎦⎣⎦
Row reducing the augmented matrix
[(A ) ]I−
0
yields:
3230 1230 1000 10 00
−−
⎡⎤⎡⎤⎡⎤⎡ ⎤
5.1 • Solutions 275
⎡
⎢
⎢
2
⎢
⎥
⎣
⎦
9. For 30 10 20
11
21 01 20
AI
λ
⎡
⎤⎡ ⎤⎡ ⎤
=: − = − =
⎢
⎥⎢ ⎥⎢ ⎥
⎣
⎦⎣ ⎦⎣ ⎦
The augmented matrix for
()AI−=
x0
is 200
.
200
⎡
⎤
⎢
⎥
⎣
⎦ Thus
1
0x= and
2
x is free. The general
⎡
⎢
⎣
⎡
⎢
⎣
10. For 42 10 12
55 5 .
31 01 36
AI
λ
−
⎡
⎤⎡ ⎤⎡ ⎤
=−:+=+=
⎢
⎥⎢ ⎥⎢ ⎥
⎣
⎦⎣ ⎦⎣ ⎦
The augmented matrix for
(5)AI+=
x0
is 120 120
.
⎣
⎡
⎢
⎣
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
∼ Thus
12
2xx=− and
2
x is
11. For
1
λ
=−
: 13 10 23
45 01 46
AI −−
⎡
⎤⎡ ⎤⎡ ⎤
+= + =
⎢
⎥⎢ ⎥⎢ ⎥
−−
⎣
⎦⎣ ⎦⎣ ⎦
The augmented matrix for
()AI+=
x0
is 230 13/20
.
460000
−−
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
−
⎣
⎦⎣ ⎦
∼ Thus
12
(3 / 2)xx= and
⎢
⎣
⎡
⎢
x is free. The general solution is
12
(3 / 2) 3/2 .
xx
x
⎡⎤
⎢⎥
⎡⎤
⎡
⎤
==
A basis for the eigenspace
276 CHAPTER 5 • Eigenvalues and Eigenvectors
For
7
λ
=
: 13 70 63
745 07 42
AI −−−
⎡
⎤⎡ ⎤⎡ ⎤
−= − =
⎢
⎥⎢ ⎥⎢ ⎥
−−−
⎣
⎦⎣ ⎦⎣ ⎦
⎡
⎢
⎣
⎡
⎢
⎣
⎡
⎢
⎣
12. For 41 30 11
33
36 03 33
AI
λ
⎡
⎤⎡ ⎤⎡ ⎤
=: − = − =
⎢
⎥⎢ ⎥⎢ ⎥
⎣
⎦⎣ ⎦⎣ ⎦
⎡
⎢
⎣
⎡
⎢
⎣
For 41 70 3 1
77 .
36 07 3 1
AI
λ
−
⎡
⎤⎡ ⎤⎡ ⎤
=:−=−=
⎢
⎥⎢ ⎥⎢ ⎥
−
⎣
⎦⎣ ⎦⎣ ⎦
The augmented matrix for
(7)AI−=
x0
is 310 11/30
.
310 0 00
−−
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
−
⎣
⎦⎣ ⎦
∼ Thus
12
(1 / 3)xx= and
⎡
⎢
⎣
⎡
⎢
⎣
13. For λ = 1:
401 100 301
1210010200
201 001 200
AI
⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
−=− − =−
⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥
−−
⎣⎦⎣⎦⎣⎦
5.1 • Solutions 277
401 200 2 0 1
2210020210
201 002 2 0 1
AI
⎡⎤⎡⎤⎡ ⎤
⎢⎥⎢⎥⎢ ⎥
−=− − =− −
⎢⎥⎢⎥⎢ ⎥
⎢⎥⎢⎥⎢ ⎥
−−−
⎣⎦⎣⎦⎣ ⎦
For λ = 3:
401 300 1 0 1
3 210 030 220
201003 202
⎡⎤⎡⎤⎡ ⎤
⎢⎥⎢⎥⎢ ⎥
−=− − =− −
⎢⎥⎢⎥⎢ ⎥
⎢⎥⎢⎥⎢ ⎥
−−−
⎣⎦⎣⎦⎣ ⎦
AI
14. For
4 0 1 300 1 0 1
3 (3) 3 303 030 333.
2 2 5 003 2 2 2
AIAI
λ
−−
⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
=: − = − = − = −
⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥
−−
⎣⎦⎣⎦⎣⎦
The augmented matrix for
(3)AI−=
x0
is
15. For
1110 1110
5 [( 5) ] 2220 0000.
3330 0000
AI
λ
⎡⎤⎡⎤
⎢⎥⎢⎥
=− : + =⎢⎥⎢⎥
⎢⎥⎢⎥
⎣⎦⎣⎦
0∼
Thus
123
0,xx x++=
with
2
x and
Note:
For simplicity, the text answer omits the set brackets. I permit my students to list a basis without
the set brackets. Some instructors may prefer to include brackets.
16. For
[]
1 0 10 1 0 10 10 10
1100 0110 0110
4(4) .
2110 0110 0000
4 2 20 0 220 0000
AI
λ
−−−
⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
−− −
⎢⎥⎢⎥⎢⎥
=: − = =
⎢⎥⎢⎥⎢⎥
−− −
⎢⎥⎢⎥⎢⎥
−− −
⎣⎦⎣⎦⎣⎦
0∼
Note:
I urge my students always to include the extra column of zeros when solving a homogeneous
system. Exercise 16 provides a situation in which failing to add the column is likely to create problems
for a student, because the matrix
4AI−
itself has a column of zeros.
17. The eigenvalues of
00 0
03 4
00 2
⎡⎤
⎢⎥
⎢⎥
⎢⎥
−
⎣⎦
are 0, 3, and
2−
, the numbers on the main diagonal, by Theorem
18. The eigenvalues of
500
000
⎡⎤
⎢⎥
are 5, 0, and 3, the numbers on the main diagonal, by Theorem 1.
19. The matrix
123
123
⎡⎤
⎢⎥
⎢⎥
is not invertible because its columns are linearly dependent. So the number 0
5.1 • Solutions 279
20. The matrix
222
222
222
A
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
is not invertible because its columns are linearly dependent. So the
21. a. False. The equation
A=λxx
must have a nontrivial solution.
b. True. See the paragraph after Example 5.
22. a. False. The vector x in
A=λxx
must be nonzero.
b. False. See Example 4 for a two-dimensional eigenspace, which contains two linearly independent
23. If a
22×
matrix A were to have three distinct eigenvalues, then by Theorem 2 there would
correspond three linearly independent eigenvectors (one for each eigenvalue). This is impossible
24. A simple example of a
22×
matrix with only one distinct eigenvalue is a triangular matrix with the
same number on the diagonal. By experimentation, one finds that if such a matrix is actually a
diagonal matrix then the eigenspace is two dimensional, and otherwise the eigenspace is only one
25. If is an eigenvalue of A, then there is a nonzero vector
x
such that
.=xxA
Since A is invertible,
11
(),
−−
=xxAA A
and so
1
().
−
=xxA
Since
≠x0
(and since A is invertible), cannot be zero.
280 CHAPTER 5 • Eigenvalues and Eigenvectors
Note:
The Study Guide points out here that the relation between the eigenvalues of A and
1
A
−
is
important in the so-called inverse power method for estimating an eigenvalue of a matrix. See Section 5.8.
26. Suppose that
2
A
is the zero matrix. If
A=xx
for some
,≠x0
then
27. Use the Hint in the text to write, for any
()() .
TT TT
AI A I A I,− =− =−
Since
( )
T
AI−
is
invertible if and only if
AI−
is invertible (by Theorem 6(c) in Section 2.2), it follows that
T
AI−
28. If A is lower triangular, then
T
A
is upper triangular and has the same diagonal entries as A. Hence,
30. Suppose the column sums of an nn× matrix A all equal the same number s. By Exercise 29 applied
31. Suppose T reflects points across (or through) a line that passes through the origin. That line consists
of all multiples of some nonzero vector v. The points on this line do not move under the action of A.
32. Since T rotates points around a given line, the points on the line are not moved at all. Hence 1 is an
33. (The solution is given in the text.)
+
k
5.1 • Solutions 281
34. You could try to write
0
x as linear combination of eigenvectors,
1
.,,vv
p
…
If
1
λ, ,λ
p
…
are
corresponding eigenvalues, and if
011
,
pp
cc=++xv v
then you could define
+
35. Using the figure in the exercise, plot
()Tu
as
2,u
because u is an eigenvector for the eigenvalue 2 of
36. As in Exercise 35,
()
T=−
uu
and
() 3
T=
vv
because u and v are eigenvectors for the eigenvalues
Note
: The matrix programs supported by this text all have an eigenvalue command. In some cases, such
as MATLAB, the command can be structured so it provides eigenvectors as well as a list of the
eigenvalues. At this point in the course, students should not use the extra power that produces
eigenvectors. Students need to be reminded frequently that eigenvectors of A are null vectors of a
translate of A. That is why the instructions for Exercises 35–38 tell students to use the method of Example
4.
37. [M] Let A be the given matrix. Use the MATLAB commands eig and nulbasis (or equivalent
commands). The command
ev = eig(A)
computes the three eigenvalues of A and stores them in a
vector ev. In this exercise,
(10 15 5).=,,
ev
The eigenspace for the eigenvalue 10 is the null space
of
10 .AI−
Use nulbasis to produce a basis for each null space:
⎢
⎢
⎢
⎣
3
−
⎡
⎤
⎣
⎦
Basis for the eigenspace for
2
15 2
⎢
⎣
⎡
⎢
⎢
⎢
⎣
λ
⎧⎫
⎡
⎤
⎪⎪
⎢
⎥
=:
⎨⎬
⎢
⎥
.
38. [M]
(2 2 1 1).=,−,−,
ev = eig(A)
⎪
⎢
⎪
⎢
⎨
⎢
⎪
⎢
⎪
⎣
⎧
22
⎧
⎫
⎡
⎤⎡⎤
For
2
λ
=−
: nulbasis
1.
⎢
⎢
⎣
⎨
⎢
⎪
⎢
⎪
⎣
⎧
⎡
⎤
⎢
⎥
⎢
⎥
(A –ev(2)*eye(4))=
Basis:
1
⎡
⎤
⎪
⎪
⎢
⎥
⎪
⎪
⎢
⎥
For
1
λ
=−
: nulbasis
1.
⎢
⎢
⎣
⎨
⎢
⎪
⎢
⎪
⎣
⎧
⎡
⎤
⎢
⎥
⎢
⎥
(A –ev(3)*eye(4))=
Basis:
1
⎡
⎤
⎪
⎪
⎢
⎥
⎪
⎪
⎢
⎥
For
1
λ
=
: nulbasis
2.
0
⎢
⎣
⎢
⎪
⎣
⎧
⎡
⎪
⎢
⎪
⎢
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
(A –ev(4)*eye(4))=
Basis:
2
0
⎡
⎤
⎪
⎪
⎢
⎥
⎪
⎪
⎢
⎥
⎨
⎬
⎢
⎥
⎪
⎪
39. [M] For
4
λ
=−
, basis:
.
2
⎢
⎪
⎪
⎢
⎪
⎣
⎢⎥
⎨⎬
For
12,
λ=
basis:
12
⎪
⎪
⎢
⎥⎢⎥
,
−
⎨
⎬
. For
8λ=−
, basis:
31
⎢⎥⎢ ⎥
,−
⎨⎬
.
⎡
⎢
⎢
⎢
5.2 • Solutions 283
40. [M] For
14
λ
=−
, basis:
110
06
,.
10
⎢
⎪
⎢
⎪
⎢
⎪
⎣
⎧⎫
−
⎡⎤⎡⎤
⎪⎪
⎢⎥⎢⎥
⎪⎪
⎢⎥⎢⎥
⎪⎪
⎢⎥⎢⎥
⎨⎬
For
42,
λ=
basis:
05
11
23
⎧
⎫
−
⎡
⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢ ⎥
−
⎪
⎪
⎢
⎥⎢ ⎥
⎪
⎪
⎢
⎥⎢ ⎥
,
−−
⎨
⎬
.
5.2 SOLUTIONS
Notes
: Exercises 9–14 can be omitted, unless you want your students to have some facility with
determinants of
33×
matrices. In later sections, the text will provide eigenvalues when they are needed
for matrices larger than
22.×
If you discussed partitioned matrices in Section 2.4, you might wish to
bring in Supplementary Exercises 12–14 in Chapter 5. (Also, see Exercise 14 of Section 2.4.)
Exercises 25 and 27 support the subsection on dynamical systems. The calculations in these exercises
and Example 5 prepare for the discussion in Section 5.6 about eigenvector decompositions.
1. 27 27 0 2 7 .
−
⎡⎤ ⎡⎤⎡ ⎤⎡ ⎤
=,−=− =
AAI
λλ
λλλ
The characteristic polynomial is
5.−
2. 41 4 1
.
AAI
λ
λλ
−− −− −
⎡⎤ ⎡ ⎤
=,−=
The characteristic polynomial is
3. 42 4 2 .
67 6 7
AAI
λ
λλ
−−−
⎡⎤ ⎡ ⎤
=,−=
⎢⎥ ⎢ ⎥
−
⎣⎦ ⎣ ⎦
The characteristic polynomial is
284 CHAPTER 5 • Eigenvalues and Eigenvectors
λ
−
⎡⎤ ⎡ ⎤
5. 84 8 4.
48 4 8
AAI
−λ
⎡⎤ ⎡ ⎤
=,−λ=
⎢⎥ ⎢ ⎥
−λ
⎣⎦ ⎣ ⎦
The characteristic polynomial of A is
6. 92 9 2
.
25 2 5
AAI
λ
λλ
−−−
⎡⎤ ⎡ ⎤
=,−=
⎢⎥ ⎢ ⎥
−
⎣⎦ ⎣ ⎦
The characteristic polynomial is
7. 53 5 3 .
44 4 4
−
⎡⎤ ⎡ ⎤
=,−=
⎢⎥ ⎢ ⎥
−−−
⎣⎦ ⎣ ⎦
AAI
λ
λλ
The characteristic polynomial is
Use the quadratic formula to solve det
()0AI
λ
−=
:
8. 43 4 3.
21 2 1
AAI
λ
λλ
−−−
⎡⎤ ⎡ ⎤
=,−=
⎢⎥ ⎢ ⎥
−
⎣⎦ ⎣ ⎦
The characteristic polynomial is
9.
401
det( ) det 0 4 1 .
102
AI
λ
λλ
λ
−−
⎡⎤
⎢⎥
−= − −
⎢⎥
⎢⎥
−
⎣⎦
Using cofactor expansion down column 1, the
characteristic polynomial is
(4 ) 1 0 1
det( ) (4 ) det 0 1 det
0(2) (4)1
AI
λ
λλ λλ
−− −
⎡⎤⎡⎤
−=−⋅ ++⋅
⎢⎥⎢⎥
−−−
⎣⎦⎣⎦
5.2 • Solutions 285
This matrix has eigenvalue 4 with multiplicity 1, and eigenvalue 3 with multiplicity 2.
10.
311
det( ) det 0 5 0 .
AI
λ
λλ
−
⎡⎤
⎢⎥
−= −
⎢⎥
Using cofactor expansion down the first column, the
characteristic polynomial is
(5 ) 0 1 1
det( ) (3 ) det 0 ( 2)det
0(7) (5)0
AI
λ
λλ λλ
−
⎡⎤⎡⎤
−=−⋅ ++−
⎢⎥⎢⎥
−−
⎣⎦⎣⎦
11. The special arrangements of zeros in A makes a cofactor expansion along the first row highly
effective.
300 14
λλ
−
⎡⎤
12. Make a cofactor expansion along the third row:
102 12 1 0
λλλ
−−
⎡⎤
13. Make a cofactor expansion down the third column:
620 62
λλ
−−
⎡⎤
14. Make a cofactor expansion along the second column:
401 41 41
λλλ
−−
⎡⎤
15. Use the fact that the determinant of a triangular matrix is the product of the diagonal entries:
286 CHAPTER 5 • Eigenvalues and Eigenvectors
5502
λ
−
⎡⎤
16. The determinant of a triangular matrix is the product of its diagonal entries:
3000
λ
−
⎡⎤
17. The determinant of a triangular matrix is the product of its diagonal entries:
3 0000
λ
−
⎡⎤
⎢⎥
18. Row reduce the augmented matrix for the equation (4)AI−=
x0
:
02330 02 3 30 02 3 00
⎡⎤⎡ ⎤⎡ ⎤
19. Since the equation
12
det( ) ( )( ) ( )−λ = λ −λ λ −λ λ −λ
n
AI holds for all
λ
, set
0λ=
and conclude
that
12
det .=λλ λ
n
A
20. det( ) det( )
TTT
AI AI
−λ = −λ
5.2 • Solutions 287
21. a. False. See Example 1.
22. a. False. See the paragraph before Theorem 3.
b. False. See Theorem 3.
23. If ,=AQR with Q invertible, and if
1
,=ARQ
then write
11
1
,
−−
==
AQQRQQAQ
which shows that
1
A is similar to A.
24. First, observe that if P is invertible, then Theorem 3(b) shows that
25. Example 5 of Section 4.9 showed that
11
,=vvA which means that
1
v is an eigenvector of A
corresponding to the eigenvalue 1.
a. Since A is a
22×
matrix, the eigenvalues are easy to find, and factoring the characteristic
polynomial is easy when one of the two factors is known.
The eigenvalues are 1 and .3. For the eigenvalue .3, solve
(3)
AI
−. =x0
:
b. Write
01 2
c=+xv v
: 12 37 1 .
12 47 1
//−
⎡⎤⎡ ⎤ ⎡⎤
=+
⎢⎥⎢ ⎥ ⎢⎥
//
c By inspection, c is
114.−/
(The value of c depends
c. For
12 ,=, ,
k… define
0
.=
xx
k
kA
Then
112121 2
() (3),Ac AcA c=+=+=+.xvvvvv v
because
v and
v are eigenvectors. Again
288 CHAPTER 5 • Eigenvalues and Eigenvectors
26. If
0,≠
a then
1
,
0
−
⎡⎤
⎡⎤
==
⎢⎥
⎢⎥
−
⎣⎦
⎣⎦
∼
ab
ab
AU
cd dcab
and
1
det ( )( ) .
−
=− =−
Aadcabadbc
If
0,=
a
27. a.
11
,A=vv
22
5,A=.vv
33
2.A=.vv
b. The set
123
{},,vv v is linearly independent because the eigenvectors correspond to different
eigenvalues (Theorem 2). Since there are three vectors in the set, the set is a basis for
3
. So there
01 12 23 3 12 23 3 1
kAcAcAcA c c k
28. [M] Answers will vary, but should show that the eigenvectors of A are not the same as the
29. [M] Answers will vary. The product of the eigenvalues of A should equal det A.
30. [M] The characteristic polynomials and the eigenvalues for the various values of a are given in the
following table:
a Characteristic Polynomial Eigenvalues
31.9
23
838 4ttt.−. + − 2.7042, 1, .2958
32.1
23
32 62 4ttt.−. + −
1 5 9747 1
i
.±. ,
r
Notes
: An appendix in Section 5.3 of t
h
with integer coefficients, in case you
w
m
n
p
e
c
5.3 SOLUTIONS
1. 57 20
23 01
−
⎡⎤⎡⎤
=,=,=
⎢⎥⎢⎥
⎣⎦⎣⎦
PDAPDP
2. 12 10
23 03
PDAPDP
−
⎡⎤⎡⎤
=,=,=
⎢⎥⎢⎥
⎣⎦⎣⎦
3.
1
10 0
210
k
kk
k
a
APDP
b
⎡⎤
⎢⎥
−⎢⎥
⎢⎥
⎣⎦
⎡⎤ ⎡
==
⎢⎥ ⎢
−
⎣⎦ ⎣
5. By the Diagonalization Theorem, e
i
correspond respectively to the eige
n
5.3 • Sol
u
r
h
e Study Guide gives an example of factoring a cubi
c
w
ant your students to find integer eigenvalues of si
m
1
,
−
and
441
.
−
=APDP
We compute
1
,
−
and
441
.
−
=APDP
We compute
10 0
.
21 22
k
kk
k
a
abb
⎡⎤
⎢⎥
⎢⎥
⎢⎥
⎣⎦
⎤=
⎥
−
−
⎦
i
genvectors form the columns of the left factor, and t
h
n
values on the diagonal of the middle factor.
u
tions 289
c
polynomial
m
ple
33×
or
h
ey
290 CHAPTER 5 • Eigenvalues and Eigenvectors
6. As in Exercise 5, inspection of the factorization gives:
031
−
⎡⎤ ⎡⎤⎡ ⎤
7. Since A is triangular, its eigenvalues are obviously
1.±
For λ = 1: 00
1.
62
⎡⎤
−=
⎢⎥
−
⎣⎦
AI The equation
(1)AI−=x0
amounts to
12
62 0,xx−=
so
12
(1 3)xx=/
with
2
x free. The general solution is
2
13 ,
1
/
⎡
⎤
⎢
⎥
⎣
⎦
x and a nice basis vector for the eigenspace is
1
1.
3
⎡
⎤
=
⎢
⎥
⎣
⎦
v
⎡
⎢
⎣
⎡
⎢
⎣
⎡
⎢
⎣
8. Since A is triangular, its only eigenvalue is obviously 3.
For λ = 3: 02
3.
00
AI
⎡⎤
−=
⎢⎥
⎣⎦
The equation
(3)AI−=x0
amounts to
2
0,=x so
2
0x= with
1
x free.
9. To find the eigenvalues of A, compute its characteristic polynomial:
22
2 1
det( )det (2 )(4 )(1)(1) 69( 3)
14
AI −−
⎡⎤
− = =− −−− =−+=−
⎢⎥
−
⎣⎦
Thus the only eigenvalue of A is 3.
10. To find the eigenvalues of A, compute its characteristic polynomial:
2
1 3
det( )det (1 )(2 )(3)(4) 310 ( 5)( 2)
42
AI −
⎡⎤
−= =− −− =−−=− +
⎢⎥
−
⎣⎦
⎡
⎢
⎣
⎡
⎢
⎣
2
3.
4
⎡⎤
=⎢⎥
⎣⎦
v
⎢
⎣
⎢
⎣
13
−
⎡
⎤
⎡
⎤
11. The eigenvalues of A are given to be -1, and 5.
For λ = –1:
111
222,
333
AI
⎡⎤
⎢⎥
+=
⎢⎥
⎢⎥
⎣⎦
and row reducing
[]
AI+0
yields
1110
0000.
0000
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
The general
For λ = 5:
511
5242,
333
AI
−
⎡⎤
⎢⎥
−= −
⎢⎥
⎢⎥
−
⎣⎦
and row reducing
[]
5AI−0
yields
10 130
01 2/30.
00 00
−/
⎡
⎤
⎢
⎥
−
⎢
⎥
⎢
⎥
⎣
⎦
The
⎡
⎢
⎢
⎢
⎣
From
12
,vv
and
3
v construct
123
111
102.
⎢
⎣
P
⎡⎤
⎢⎥
⎣⎦
−−
⎡
⎤
⎢
⎥
==
⎢
⎥
vvv
Then set
D
=
100
010,
−
⎡⎤
⎢⎥
−
⎢⎥
12. The eigenvalues of A are given to be 2 and 5.
For λ = 2:
111
2111,
⎢
⎣
AI
⎡⎤
⎢⎥
−=
⎢⎥
and row reducing
[]
2AI−0
yields
1110
0000.
⎡
⎤
⎢
⎥
⎢
⎥
The general
For λ = 5:
21 1
5121,
11 2
AI
−
⎡⎤
⎢⎥
−= −
⎢⎥
⎢⎥
−
⎣⎦
and row reducing
[]
5AI−0
yields
10 10
01 10.
00 00
−
⎡⎤
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
The
⎢
⎢
⎢
⎣
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎣
1
⎡
⎤
1
⎡
⎤
13. The eigenvalues of A are given to be 5 and 1.
For λ = 5:
321
5121,
123
AI
−−
⎡⎤
⎢⎥
−= − −
⎢⎥
⎢⎥
−−−
⎣⎦
and row reducing
[]
5AI−0
yields
1010
0110.
0000
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
The
⎡
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎣
For λ = 1:
121
1121,
121
−
⎡⎤
⎢⎥
−= −
⎢⎥
⎢⎥
−−
⎣⎦
AI
and row reducing
[]
AI−0
yields
12 10
00 00.
00 00
−
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
The
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎣
21
−
⎡
⎤⎡⎤
21
⎧⎫
−
⎡⎤⎡⎤
⎪⎪