Math 152 Chapter 4 solutions
1
Problem 4.1:
Show that
1
1
and
1

1
are eigenvectors for the matrix
1
1
1
1
. What are the corresponding
eigenvalues?
Answer:
We compute
1
1
1
1
1
1
=
2
2
= 2
1
1
, so
1
1
is an eigenvector with eigenvalue 2.
Also
1
1
1
1
1

1
=
0
0
= 0
1

1
, so
1

1
is an eigenvector with eigenvalue 0.
Problem 4.2:
Suppose
P
is a projection matrix. What are the eigenvalues and eigenvectors of
P
?
Answer:
If
P
projects onto some line, then a vector
x
lying on that line doesn’t get changed by
P
so
P
x
=
x
and
x
is an eigenvector with eigenvalue 1. On the other hand, if
x
is perpendicular to the line, then
P
x
=
0
= 0
x
so
x
is an eigenvector with eigenvalue 0.
Problem 4.3:
Find the eigenvalues and eigenvectors for
a
)
0
3
3
0
b
)

2

8
4
10
c
)
29

10
105

36
d
)

9

14
7
12
Answer:
a) det(
A

λI
) =
λ
2

9 = (
λ

3)(
λ
+ 3) so the eigenvalues are
λ
1
= 3 and
λ
2
=

3.
To find the eigenvector for
λ
1
= 3 we must solve the homegeneous equation with matrix

3
3
3

3
. The
matrix reduces to

3
3
0
0
and the eigenvector is
x
1
=
1
1
.
To find the eigenvector for
λ
2
=

3 we must solve the homegeneous equation with matrix
3
3
3
3
.
The
matrix reduces to
3
3
0
0
and the eigenvector is
x
2
=
1

1
.
b) det(
A

λI
) =
λ
2

8
λ
+ 12 = (
λ

2)(
λ

6) so the eigenvalues are
λ
1
= 3 and
λ
2
=

3.
To find the eigenvector for
λ
1
= 2 we must solve the homegeneous equation with matrix

4

8
4
8
. The
matrix reduces to

4

8
0
0
and the eigenvector is
x
1
=
2

1
.
To find the eigenvector for
λ
2
= 6 we must solve the homegeneous equation with matrix

8

8
4
4
. The
matrix reduces to

8

8
0
0
and the eigenvector is
x
2
=
1

1
.
c) det(
A

λI
) =
λ
2
+ 7
λ
+ 6 = (
λ
+ 6)(
λ
+ 1) so the eigenvalues are
λ
1
=

6 and
λ
2
=

1.
To find the eigenvector for
λ
1
=

6 we must solve the homegeneous equation with matrix
35

10
105

30
.
The matrix reduces to
35

10
0
0
and the eigenvector is
x
1
=
2
7
.
To find the eigenvector for
λ
2
=

1 we must solve the homegeneous equation with matrix
30

10
105

35
.
The matrix reduces to
30

10
0
0
and the eigenvector is
x
2
=
1
3
.
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2
Math 152 Chapter 4 solutions
d) det(
A

λI
) =
λ
2

3
λ

10 = (
λ
+ 2)(
λ

5) so the eigenvalues are
λ
1
=

2 and
λ
2
= 5.
To find the eigenvector for
λ
1
=

2 we must solve the homegeneous equation with matrix

7

14
7
14
. The
matrix reduces to

7

14
0
0
and the eigenvector is
x
1
=
2

1
.
To find the eigenvector for
λ
2
= 5 we must solve the homegeneous equation with matrix

14

14
7
7
. The
matrix reduces to

14

14
0
0
and the eigenvector is
x
2
=
1

1
.
Problem 4.4:
Find the eigenvalues and eigenvectors for
a
)
0

1
1
1
0
2
2
0
2
b
)
1
1
1
1
0

2
1

1
1
c
)
7

9

15
0
4
0
3

9

11
d
)
31

100
70
18

59
42
12

40
29
Answer:
a) det(
A

λI
) =

λ
3
+ 2
λ
2
+
λ

2 =

(
λ

1)(
λ

2)(
λ
+ 1) so the eigenvalues are
λ
1
= 1 and
λ
2
= 2 and
λ
3
=

1.
To find the eigenvector for
λ
1
= 1 we must solve the homegeneous equation with matrix

1

1
1
1

1
2
2
0
1
.
The matrix reduces to

1

1
1
0

2
3
0
0
0
and the eigenvector is
x
1
=
1

3
2
.
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 Spring '08
 Caddmen
 Linear Algebra, Eigenvectors, Vectors, Boundary value problem, Eigenvalue, eigenvector and eigenspace, Complex number, 1 l

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