The desired basis is
12
14
,

7

8
.
57. If we can find a basis
B
= (
v
1
, v
2
, v
3
) such that the
B
matrix of
A
is
B
=
1
0
0
0
1
0
0
0

1
, then
A
must be similar to
1
0
0
0
1
0
0
0

1
. Because of the entries in
the matrix
B
, it is required that
Av
1
=
v
1
, Av
2
=
v
2
and
Av
3
=

v
3
. So, all we need for
our basis is to pick independent
v
1
, v
2
in the plane, and
v
3
perpendicular to the plane.
58. a. Consider a linear relation
c
1
A
2
v
+
c
2
Av
+
c
3
v
= 0.
Multiplying
A
2
with the vectors on both sides and using that
A
3
v
= 0 and
A
4
v
= 0,
we find that
c
3
A
2
v
= 0 and therefore
c
3
= 0, since
A
2
v
= 0.
170
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ISM:
Linear Algebra
Section 3.4
Therefore, our relation simplifies to
c
1
A
2
v
+
c
2
Av
= 0.
Multiplying
A
with the vectors on both sides we find that
c
2
A
2
v
= 0 and therefore
c
2
= 0. Then
c
1
= 0 as well.
We have shown that there is only the trivial relation
among the vectors
A
2
v, Av
, and
v
, so that these three vectors from a basis of
R
3
, as
claimed.
b.
T
(
A
2
v
) =
A
3
v
= 0 so [
T
(
A
2
v
)]
B
=
0
0
0
.
T
(
Av
) =
A
2
v
so [
T
(
Av
)]
B
=
1
0
0
.
T
(
v
) =
Av
so [
T
(
v
)]
B
=
0
1
0
.
Hence, by Fact 3.4.3, the desired matrix is
0
1
0
0
0
1
0
0
0
.
59. First we find the matrices
S
=
x
y
z
t
such that
2
0
0
3
x
y
z
t
=
x
y
z
t
2
1
0
3
, or
2
x
2
y
3
z
3
t
=
2
x
x
+ 3
y
2
z
z
+ 3
t
. The solutions are of the form
S
=

y
y
0
t
, where
y
and
t
are arbitrary constants. Since there are
invertible
solutions
S
(for example, let
y
=
t
= 1),
the matrices
2
0
0
3
and
2
1
0
3
are indeed similar.
60. First we find the matrices
S
=
x
y
z
t
such that
1
0
0

1
x
y
z
t
=
x
y
z
t
0
1
1
0
,
or,
x
y

z

t
=
y
x
t
z
. The solutions are of the form
S
=
y
y

t
t
, where
y
and
t
are arbitrary constants. Since there are
invertible
solutions
S
(for example, let
y
=
t
= 1),
the matrices
1
0
0

1
and
0
1
1
0
are indeed similar.
61. We seek a basis
v
1
=
x
z
, v
2
=
y
t
such that the matrix
S
= [
v
1
v
2
] =
x
y
z
t
satisfies
the equation

5

9
4
7
x
y
z
t
=
x
y
z
t
1
1
0
1
. Solving the ensuing linear system
171
Chapter 3
ISM:
Linear Algebra
gives
S
=

3
z
2
z
4

3
t
2
z
t
.
We need to choose
z
and
t
so that
S
will be invertible. For example, if we let
z
= 6 and
t
= 1, then
S
=

9
0
6
1
, so that
v
1
=

9
6
, v
2
=
0
1
.
62. We seek a basis
v
1
=
x
z
, v
2
=
y
t
such that the matrix
S
= [
v
1
v
2
] =
x
y
z
t
satisfies
the equation
1
2
4
3
x
y
z
t
=
x
y
z
t
5
0
0

1
.
Solving the ensuing linear system
gives
S
=
z
2

t
z
t
. We need to choose both
z
and
t
nonzero to make
S
invertible. For
example, if we let
z
= 2 and
t
= 1, then
S
=
1

1
2
1
, so that
v
1
=
1
2
, v
2
=

1
1
.
63. First we find the matrices
S
=
x
y
z
t
such that
p

q
q
p
x
y
z
t
=
x
y
z
t
p
q

q
p
,
or,
px

qz
py

qt
qx
+
pz
qy
+
pt
=
px

qy
qx
+
py
pz

qt
qz
+
pt
. If
q
= 0, then the solutions are of the
form
S
=

t
z
z
t
, where
z
and
t
are arbitrary constants.
Since there are
invertible
solutions
S
(for example, let
z
=
t
= 1), the matrices
p

q
q
p
and
p
q

q
p
are indeed
similar. (If
q
= 0, then the two matrices are equal.)
64. If
b
and
c
are both zero, then the given matrices are equal, so that they are similar, by
Fact 3.4.6.a. Let’s now assume that at least one of the scalars
b
and
c
is nonzero; reversing
the roles of
b
and
c
if necessary, we can assume that
c
= 0.