This preview shows pages 1–4. Sign up to view the full content.
Chapter
6
SECTION 1
2.
If
A
is triangular then
A

a
ii
I
will be a triangular matrix with a zero entry
in the (
i,i
) position. Since the determinant of a triangular matrix is the
product of its diagonal elements it follows that
det(
A

a
ii
I
)=0
Thus the eigenvalues of
A
are
a
11
,a
22
,...,a
nn
.
3.
A
is singular if and only if det(
A
) = 0. The scalar 0 is an eigenvalue if and
only if
det(
A

0
I
) = det(
A
Thus
A
is singular if and only if one of its eigenvalues is 0.
4.
If
A
is a nonsingular matrix and
λ
is an eigenvalue of
A
, then there exists a
nonzero vector
x
such that
A
x
=
λ
x
A

1
A
x
=
λA

1
x
It follows from Exercise 3 that
λ
±
= 0. Therefore
A

1
x
=
1
λ
x
(
x
±
=
0
)
and hence 1
/λ
is an eigenvalue of
A

1
.
110
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document Section 1
111
5.
The proof is by induction. In the case where
m
=1,
λ
1
=
λ
is an eigenvalue
of
A
with eigenvector
x
. Suppose
λ
k
is an eigenvalue of
A
k
and
x
is an
eigenvector belonging to
λ
k
.
A
k
+1
x
=
A
(
A
k
x
)=
A
(
λ
k
x
λ
k
A
x
=
λ
k
+1
x
Thus
λ
k
+1
is an eigenvalue of
A
k
+1
and
x
is an eigenvector belonging to
λ
k
+1
. It follows by induction that if
λ
an eigenvalue of
A
then
λ
m
is an
eigenvalue of
A
m
, for
m
=1
,
2
,...
.
6.
If
A
is idempotent and
λ
is an eigenvalue of
A
with eigenvector
x
, then
A
x
=
λ
x
A
2
x
=
λA
x
=
λ
2
x
and
A
2
x
=
A
x
=
λ
x
Therefore
(
λ
2

λ
)
x
=
0
Since
x
±
=
0
it follows that
λ
2

λ
=0
λ
=0 or
λ
7.
If
λ
is an eigenvalue of
A
, then
λ
k
is an eigenvalue of
A
k
(Exercise 5). If
A
k
=
O
, then all of its eigenvalues are 0. Thus
λ
k
= 0 and hence
λ
.
9.
det(
A

λI
) = det((
A

λI
)
T
) = det(
A
T

λI
). Thus
A
and
A
T
have the same
characteristic polynomials and consequently must have the same eigenvalues.
The eigenspaces however will not be the same. For example
A
=
11
01
and
A
T
=
10
both have eigenvalues
λ
1
=
λ
2
The eigenspace of
A
corresponding to
λ
= 1 is spanned by (1
,
0)
T
while
the eigenspace of
A
T
is spanned by (0
,
1)
T
. Exercise 27 shows how the
eigenvectors of
A
and
A
T
are related.
10.
det(
A

λI
λ
2

(2 cos
θ
)
λ
+ 1. The discriminant will be negative unless
θ
is a multiple of
π
. The matrix
A
has the eﬀect of rotating a real vector
x
about the origin by an angle of
θ
. Thus
A
x
will be a scalar multiple of
x
if
and only if
θ
is a multiple of
π
.
12.
Since tr(
A
) equals the sum of the eigenvalues the result follows by solving
n
±
i
=1
λ
i
=
n
±
i
=1
a
ii
for
λ
j
.
112
Chapter 6
13.
±
±
±
±
a
11

λa
12
a
21
a
22

λ
±
±
±
±
=
λ
2

(
a
11
+
a
22
)
λ
+(
a
11
a
22

a
21
a
12
)
=
λ
2

(tr
A
)
λ
+ det(
A
)
14.
If
x
is an eigenvector of
A
belonging to
λ
, then any nonzero multiple of
x
is also an eigenvector of
A
belonging to
λ
. By Exercise 5 we know that
A
m
x
=
λ
m
x
,so
A
m
x
must be an eigenvector of
A
belonging to
λ
.
Alternatively we could have proved the result by noting that
A
m
x
=
λ
m
x
±
=
0
and
A
(
A
m
x
)=
A
m
+1
x
=
A
m
(
A
x
A
m
(
λ
x
λ
(
A
m
x
)
15.
If
A

λ
0
I
has rank
k
then
N
(
A

λ
0
I
) will have dimension
n

k
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 04/18/2010 for the course FINANCE 1231854365 taught by Professor Wuyiling during the Spring '10 term at Nashville State Community College.
 Spring '10
 wuyiling

Click to edit the document details