Math 221, Solutions for assignment due Nov. 11
Section 7.2
Problem 10
det
3

λ
0
4
0

1

λ
0

2
7
3

λ
= (

3

λ
)(

1

λ
)(3

λ
)

2(

4(

1

λ
)) =

(
λ
+1)
2
(
λ

1)
So the matrix has eigenvalues
λ
= 1, with algebraic multiplicity 1 and
λ
=

1, with
algebraic multiplicity 2.
Problem 12
det
2

λ

2
0
0
1

1

λ
0
0
0
0
3

λ

4
0
0
2

3

λ
= det
bracketleftbigg
2

λ

2
1

1

λ
bracketrightbigg
det
bracketleftbigg
3

λ

4
2

3

λ
bracketrightbigg
= (
λ
2

λ
)(
λ
2

1) =
λ
(
λ
+ 1)(
λ

1)
2
.
So The eigenvalues are 0
,

1 (with algebraic multiplicity 1) and

1 (with algebraic multi
plicity 2).
Problem 19
This is true. The characteristip polynomial is
λ
2

(
trace
)
λ

det
with
det <
0, so the
discriminant of this polynomial is
D
=
√
trace
2

4
det
. Since the term under the square
root sign is strictly positive, the characteristic polynomial has two distinct roots:
trace
+
D
2
and
trace

D
2
.
Problem 22
det(
A

λI
) = det(
A

λI
)
T
= det(
A
T

λI
). Thus the characteristic polynomials of
A
and
A
T
are the same, which implies that the eigenvalues are the same.
Problem 22
If
B
=
S

1
AS
, then
det(
A

λI
) = det
S

1
(
A

λI
)
S
= det(
S

1
AS

S

1
λIS
)
= det(
S

1
AS

λI
) = det(
B

λI
)
.
Thus
A
and
B
have the same characteristic polynomials, and the same eigenvalues with
the same multiplicities.
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 Fall '05
 PANTANO
 Linear Algebra, Algebra, Equations, Characteristic polynomial, Eigenvalue, eigenvector and eigenspace, Det

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