$n=\dim V$. Then $V=\ker(T^n)\oplus\mathrm{range}(T^n)$ I trying to solve the following problem. The question is from a past exam.

Suppose that $V$ is a finite dimensional vector space over a field $K$. Let $T: V\rightarrow V$ be a linear operator. If $n=\dim V$, then $V=\ker(T^n)\oplus\mathrm{range}(T^n)$

Attempt/Thoughts.
If I can show that $\ker(T^n)\cap \mathrm{range}T^n={0}$ then we are done. To show this I need the following claim. (At least this is what I think). I claim that $\ker(T^{2n}) \subseteq \ker(T^n)$. I am stuck on trying to prove this. I know that

$\ker(T) \subseteq \ker(T^2) \subseteq \ker(T^3) \subseteq \cdots$

So this set of inclusions has to stop at some point since we are in the finite dimensional case(?). If there is some $r\geq0$ such that $\ker(T^{r})=\ker(T^{r+1})$ then I can easily prove my claim. But I don't know how to make this precise. 
Can somebody help?
Thanks 
 A: Instead of considering the chain $\ker(T) \subseteq \ker(T^2) \subseteq \ker(T^3) \subseteq \cdots$, you may consider the ascending chain of proper subset inclusion $\ker(T) \subset \ker(T^2) \subset \ker(T^3) \subset \cdots$. Since $V$ is finite dimensional, the dimension of $\ker(T^r)$ cannot keep strictly increasing. Therefore there must exist some $r$ that is the first index such that $\ker(T^r)=\ker(T^{r+1})$.
We also have the descending chain $\mathrm{range}(T) \color{green}{\supseteq} \mathrm{range}(T^2) \color{green}{\supseteq} \mathrm{range}(T^3) \color{green}{\supseteq} \cdots$. Therefore, by the rank-nullity theorem,
$$
\mathrm{range}(T) \color{red}{\supset} \mathrm{range}(T^2) \color{red}{\supset} \mathrm{range}(T^3) \color{red}{\supset} \cdots \color{red}{\supset}\mathrm{range}(T^r)=\mathrm{range}(T^{r+1}).
$$
Note that the rank-nullity theorem implies that the inclusions preceding $\mathrm{range}(T^r)$ are strict. It remains to show that


*

*$\ker(T^r)\cap\mathrm{range}(T^r)=0$ and hence $V=\ker(T^r)\oplus\mathrm{range}(T^r)$.

*$r\le n$.

*$V=\ker(T^n)\oplus\mathrm{range}(T^n)$.

A: This was answered by Robert Israel here. Here is his proof with some added detail: From the dimension formulas it suffices to show that $$\ker T^{n}\cap\mbox{Im}T^{n}=\{0\}.$$ Assume for contradiction that $0\neq y\in\ker T^{n}\cap\mbox{Im}T^{n}$.
Then $\exists x,T^{n}x=y$, and $0=T^{n}y=T^{2n}x$. Consider the
set of vectors
$$\{\underset{\neq0}{\underbrace{x,Tx,T^{2}x,...,T^{n}x}},T^{n+1}x,...,\underset{=0}{\underbrace{T^{2n}x}\}}$$
Thus, there is an $1\leq\ell\leq n-1$ such that $T^{n+\ell}x\neq0$
but $T^{n+\ell+1}x=0$. Set $z=T^{\ell}x$. Then 
$$T^{n}z\neq0,\, T^{n+1}z=0.$$
This last property implies that $A=\{z,Tz,...,T^{n}z\}$ is linearly
independent: if $${\displaystyle 0=\sum_{i=0}^{n}\alpha_{i}T^{i}z},$$
then we apply $T^{n}$ to both sides to obtain $0=\alpha_{0}T^{n}z$
and therefore $\alpha_{0}=0$. This argument can be repeated to show
that all the $\alpha_{i}$ are zero, as required for linear independence.
Now since $A$ is linearly independent it has at most $n$ elements.
But $|A|=n+1$, which is a contradiction. As Robert Israel notes in his reply, this argument works over any field. 
Note also jspecter's reply on the same page,  which is the kind of proof your instructor probably expected if the exercise was given "over $\mathbb{C}$".
