Section 2.3, #16:
Let
V
be a finitedimensional vector space, and let
T
:
V
→
V
be linear.
(a) If rank(
T
) = rank(
T
2
), prove that
R
(
T
)
∩
N
(
T
) =
{
0
}
. Deduce that
V
=
R
(
T
)
⊕
N
(
T
).
Proof.
Assume that rank(
T
) = rank(
T
2
), i.e., that dim(
R
(
T
)) = dim(
R
(
T
2
)).
We want to show first that
R
(
T
)
∩
N
(
T
) =
{
0
}
. The proof takes a cou
ple of steps. First, we note the following two facts:
Claim
1
.
R
(
T
2
)
⊆
R
(
T
).
Proof of claim 1.
Let
v
∈
R
(
T
2
). Then
v
=
T
2
(
x
) for some
x
∈
V
, so
v
=
T
(
T
(
x
)), so
v
∈
R
(
T
). Thus
R
(
T
2
)
⊆
R
(
T
).
Claim
2
.
N
(
T
)
⊆
N
(
T
2
).
Proof of claim 2.
Let
v
∈
N
(
T
). Then
T
(
v
) = 0, so
T
2
(
v
) =
T
(
T
(
v
)) =
T
(0) = 0, so
v
∈
N
(
T
2
). Thus
N
(
T
)
⊆
N
(
T
2
).
Now recall (from Theorem 1.11) that if
W
is a subspace of
V
, then
dim(
W
)
≤
dim(
V
), and if dim(
W
) = dim(
V
) then
W
=
V
. Applying
this to the subspace
R
(
T
2
) of
R
(
T
), since we are assuming dim(
R
(
T
)) =
dim(
R
(
T
2
)), we have that
R
(
T
) =
R
(
T
2
).
But also, by the Rank
Nullity Theorem, we have rank(
T
)+nullity(
T
) = dim(
V
). We can also
apply this to the linear map
T
2
, so we have rank(
T
2
) + nullity(
T
2
) =
dim(
V
). Thus
rank(
T
) + nullity(
T
) = rank(
T
2
) + nullity(
T
2
)
.
Since rank(
T
) = rank(
T
2
), we must also have nullity(
T
) = nullity(
T
2
).
Combining this fact with claim 2 and Theorem 1.11, since
N
(
T
)
⊆
N
(
T
2
) and dim(
N
(
T
)) = dim(
N
(
T
2
)), we must have
N
(
T
) =
N
(
T
2
)
.
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 Spring '10
 FUCKHEAD
 Rank, Natural number

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