Since
W
is a subspace, we know that
g
(
T
)(
v
)
is always an element is
W
.
5. Let
{
W
i
}
i
∈
I
be the collection of
T
invatirant subspaces and
W
be the
intersection of them. For every
v
∈
W
, we have
T
(
v
)
∈
W
i
for every
i
∈
I
,
since
v
is an element is each
W
i
. This means
T
(
v
)
is also an element in
W
.
6. Follow the prove of Theorem 5.22.
And we know that the dimension is
the maximum number
k
such that
{
z,T
(
z
)
,T
2
(
z
)
,...,T
k

1
(
z
)}
is independent and the set is a basis of the subspace.
(a) Calculate that
z
=
(
1
,
0
,
0
,
0
)
,T
(
z
)
=
(
1
,
0
,
1
,
1
)
,
T
2
(
z
)
=
(
1
,

1
,
2
,
2
)
,T
3
(
z
)
=
(
0
,

3
,
3
,
3
)
.
So we know the dimension is 3 and the set
{
z,T
(
z
)
,T
2
(
z
)}
is a basis.
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(b) Calculate that
z
=
x
3
,T
(
z
)
=
6
x,T
2
(
z
)
=
0
.
So we know that the dimension is 2 and the set
{
z,T
(
z
)}
is a basis.
(c) Calculate that
T
(
z
)
=
z
. So we know that the dimension is 1 and
{
z
}
is a basis.
(d) Calculate that
z
=
0
1
1
0
,T
(
z
)
=
1
1
2
2
,
T
2
(
z
)
=
3
3
6
6
.
So we know that the dimension is 2 and the set
{
z,T
(
z
)}
is a basis.
7. Let
W
be a
T
invariant subspace and
T
W
be the restricted operator on
W
. We have that
R
(
T
W
)
=
T
W
(
W
)
=
T
(
W
)
⊂
W.
So at least it’s a welldefined mapping. And we also have
T
W
(
x
)
+
T
W
(
y
)
=
T
(
x
)
+
T
(
y
)
=
T
(
x
+
y
)
=
T
W
(
x
+
y
)
and
T
W
(
cx
)
=
T
(
cx
)
=
cT
(
x
)
=
cT
W
(
x
)
.
So the restriction of
T
on
W
is also a lineaer operator.
8. If
v
is an eigenvector of
T
W
corresponding eigenvalue
λ
, this means that
T
(
v
)
=
T
W
(
v
)
=
λv
. So the same is true for
T
.
9. See Example 5.4.6.
(a) For the first method, we may calculate
T
3
(
z
)
=
(
3
,

3
,
3
,
3
)
and rep
resent it as a linear combination of the basis
T
3
(
z
)
=
0
z

3
T
(
z
)
+
3
T
2
(
z
)
.
So the characteristic polynomial is

t
3
+
3
t
2

3
t
.
For the second
method, denote
β
to be the ordered basis
{
z,T
(
z
)
,T
2
(
z
)
,T
3
(
z
)}
.
134
And we may calculate the matrix representation
[
T
W
]
β
=
0
0
0
1
0

3
0
1
3
and directly find the characteristic polynomial of it to get the same
result.
(b) For the first method, we may calculate
T
3
(
z
)
=
0 and represent it as
a linear combination of the basis
T
2
(
z
)
=
0
z
+
0
T
(
z
)
.
So the characteristic polynomial is
t
2
. For the second method, denote
β
to be the ordered basis
{
z,T
(
z
)}
.
And we may calculate the matrix representation
[
T
W
]
β
=
0
0
1
0
and directly find the characteristic polynomial of it to get the same
result.
(c) For the first method, we may calculate
T
(
z
)
=
z
. So the characteristic
polynomial is

t
+
1.
For the second method, denote
β
to be the
ordered basis
{
z
}
. And we may calculate the matrix representation
[
T
W
]
β
=
1
and directly find the characteristic polynomial of it to get the same
result.
(d) For the first method, we may calculate
T
2
(
z
)
=
3
T
(
z
)
. So the char
acteristic polynomial is
t
2

3
t
. For the second method, denote
β
to
be the ordered basis
{
z,T
(
z
)}
.
And we may calculate the matrix representation
[
T
W
]
β
=
0
0
1
3
and directly find the characteristic polynomial of it to get the same
result.