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Exercise 2.3.16

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1.

Since we know

R (T)

is a

T

-invariant space, we can view

T

as a mapping from

R (T)

R (T)

and call this restricted mapping

T |_{R (T)}

. So now we have that

\dim R (T) = rank (T) = rank (T^{2}) = \dim (T (T (V))

= \dim (T (R (T)) = rank (T |_{R (T)}) .

And so the mapping $T |_{R (T)}$ is surjective and hence injective with the help of the fact $R (T)$ is finite dimensional. This also means $N (T |_{R (T)} = R (T) \cap N (T) = 0$ . This complete the proof of the first statement. For the other, it’s sufficient to say that $R (T) + N (T) = V$ . But this is instant conclusion of the fact that $R (T) + N (T) \subset V$ and that

\dim (R (T) + N (T)) = \dim (R (T)) + \dim (N (T)) - \dim (R (T) \cap N (T))

= \dim (R (T)) + \dim (N (T)) = \dim (V) .

2.

In general we have rank

(T^{s + 1}) \leq

rank

(T^{s})

since the fact

T^{s + 1} (V) = T^{s} (R (T)) \subset T^{s} (V)

. But the integer rank

(T^{s})

can only range from

0

to dim

(V)

. So there must be some integer

k

such that

rank (T^{k}) = rank (T^{k + 1}) .

And this means $T^{k + 1} (V) = T^{k} (V)$ and hence $T^{s} (V) = T^{k} (V)$ for all $s \geq k$ . Since $2 k \geq k$ , we can conclude that rank $(T^{k}) =$ rank $(T^{2 k})$ and hence we have $V = R (T^{k}) \oplus N (T^{k})$ by the previous exercise.

jephianlin

2011-06-27 00:00

Exercise 2.3.16

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