Chapter 7
ISM:
Linear Algebra
Chapter 7
7.1
1. If
v
is an eigenvector of
A
, then
Av
=
λ
v
.
Hence
A
3
v
=
A
2
(
Av
) =
A
2
(
λ
v
) =
A
(
A
λ
v
) =
A
(
λ
Av
) =
A
(
λ
2
v
) =
λ
2
Av
=
λ
3
v
, so
v
is an
eigenvector of
A
3
with eigenvalue
λ
3
.
2. We know
Av
=
λ
v
so
v
=
A

1
Av
=
A

1
λ
v
=
λ
A

1
v
, so
v
=
λ
A

1
v
or
A

1
v
=
1
λ
v
.
Hence
v
is an eigenvector of
A

1
with eigenvalue
1
λ
.
3. We know
Av
=
λ
v
, so (
A
+ 2
I
n
)
v
=
Av
+ 2
I
n
v
=
λ
v
+ 2
v
= (
λ
+ 2)
v
, hence
v
is an
eigenvector of (
A
+ 2
I
n
) with eigenvalue
λ
+ 2.
4. We know
Av
=
λ
v
, so 7
Av
= 7
λ
v
, hence
v
is an eigenvector of 7
A
with eigenvalue 7
λ
.
5. Assume
Av
=
λ
v
and
Bv
=
β
v
for some eigenvalues
λ
,
β
. Then (
A
+
B
)
v
=
Av
+
Bv
=
λ
v
+
β
v
= (
λ
+
β
)
v
so
v
is an eigenvector of
A
+
B
with eigenvalue
λ
+
β
.
6. Yes. If
Av
=
λ
v
and
Bv
=
μv
, then
ABv
=
A
(
μv
) =
μ
(
Av
) =
μ
λ
v
7. We know
Av
=
λ
v
so (
A

λ
I
n
)
v
=
Av

λ
I
n
v
=
λ
v

λ
v
= 0 so a nonzero vector
v
is in
the kernel of (
A

λ
I
n
) so ker(
A

λ
I
n
) =
{
0
}
and
A

λ
I
n
is not invertible.
8. We want all
a
b
c
d
such that
a
b
c
d
1
0
= 5
1
0
hence
a
c
=
5
0
, i.e. the desired
matrices must have the form
5
b
0
d
.
9. We want
a
b
c
d
1
0
=
λ
1
0
for any
λ
. Hence
a
c
=
λ
0
, i.e., the desired matrices
must have the form
λ
b
0
d
, they must be upper triangular.
10. We want
a
b
c
d
1
2
= 5
1
2
, i.e. the desired matrices must have the form
5

2
b
b
10

2
d
d
.
11. We want
a
b
c
d
2
3
=

2

3
. So, 2
a
+ 3
b
=

2 and 2
c
+ 3
d
=

3. Thus,
b
=

2

2
a
3
,
and
d
=

3

2
c
3
. So all matrices of the form
a

2

2
a
3
c

3

2
c
3
will fit.
12. Solving
2
0
3
4
v
1
v
2
= 2
v
1
v
2
we get
v
1
v
2
=
t

3
2
t
(with
t
= 0) and
328
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ISM:
Linear Algebra
Section 7.1
solving
2
0
3
4
v
1
v
2
= 4
v
1
v
2
we get
v
1
v
2
=
0
t
(with
t
= 0).
13. Solving

6
6

15
13
v
1
v
2
= 4
v
1
v
2
, we get
v
1
v
2
=
3
5
t
t
(with
t
= 0).
14. We want to find all 4
×
4 matrices
A
such that
Ae
2
=
λ
e
2
, i.e. the second column of
A
must be of the form
0
λ
0
0
, so
A
=
a
0
c
d
e
λ
f
g
h
0
i
j
k
0
l
m
.
15. Any vector on
L
is una
ff
ected by the reflection, so that a nonzero vector on
L
is an
eigenvector with eigenvalue 1. Any vector on
L
⊥
is flipped about
L
, so that a nonzero
vector on
L
⊥
is an eigenvector with eigenvalue

1. Picking a nonzero vector from
L
and
one from
L
⊥
, we obtain a basis consisting of eigenvectors.
16. Rotation by 180
◦
is a flip about the origin so every nonzero vector is an eigenvector with
the eigenvalue

1. Any basis for
R
2
consists of eigenvectors.
17. No (real) eigenvalues
18. Any nonzero vector in the plane is unchanged, hence is an eigenvector with the eigenvalue
1. Since any nonzero vector in
V
⊥
is flipped about the origin, it is an eigenvector with
eigenvalue

1. Pick any two noncollinear vectors from
V
and one from
V
⊥
to form a
basis consisting of eigenvectors.
19. Any nonzero vector in
L
is an eigenvector with eigenvalue 1, and any nonzero vector in
the plane
L
⊥
is an eigenvector with eigenvalue 0. Form a basis consisting of eigenvectors
by picking any nonzero vector in
L
and any two nonparallel vectors in
L
⊥
.