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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
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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
.
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This note was uploaded on 05/01/2011 for the course ELECTRIC MAT351 taught by Professor Ahmettantuni during the Spring '10 term at Abant İzzet Baysal University.
 Spring '10
 AHMETTANTUNI

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