$\dim V<\infty$. Show there exists $m$ so that $\ker T^m \cap T^m(V)=0$

Question

Here's a question.

M7. Let $V$ be a finite dimensional vector space and $T$ be a linear transformation on $V$. Prove there exists an $m\ge1$ such that
(a) $\ker T^n=\ker T^m$ for all $n\ge m$.
(b) $\ker T^m\cap T^m(V)=0$.
Note: $T^n$ denotes the composite map $T\circ T\circ T\circ\cdots\circ T$ ($n$ times); and $\ker$ denotes kernel.

Now I have done the first part. Here's what I have done Now I'm stuck with the last part. Any help is appreciated.

Answer 1

Your proof for (a) is correct.

(b): Let $x \in ker(T^m) \cap T^m(V)$. Then $T^mx=0$ and $x= T^mv$ for some $v \in V.$

Hence $0=T^mx=T^{2m}v$. This gives $v\in ker(T^{2m}).$ By (a) we have $ker(T^m)=ker(T^{2m})$, thus $v \in ker(T^m)$. We get $x=T^mv=0.$