# Endomorphism of finite dimensional vector space composes to give equal kernels and images

Let $\varphi: V\rightarrow V$ be a homomorphism with $V$ a finite dimensional vector space. To state the question in the title more precisely, I want to show that there's some $n\in\mathbb{Z}$ such that $\ker \varphi^n=\ker\varphi^{n+1}$ and $\text{im} \varphi^{n}=\text{im}\varphi^{n+1}$. I think I have to look at what $\varphi$ does to basis elements but I'm pretty lost.

Edit: Does rank-nullity help me at all?

• This is called Fitting's lemma. Nov 16, 2016 at 6:30

It suffices to make the following observation. Let $$\phi^k$$ denote $$\phi\circ\ldots\circ\phi$$ k-times and suppose $$\dim V=n$$, then $$\ker\phi\subset\ker\phi^2\subset\ldots\subset\ker\phi^n,$$ and $$\text{Im}\phi\supset\text{Im}\phi^2\supset\ldots\supset\text{Im}\phi^n.$$
Since the dimension is finite, notice that if at any point $$\dim\ker\phi^i=\dim\ker\phi^{i+1}$$, then we must have $$\ker\phi^i=\ker\phi^{i+1}$$, and similarly if $$\dim\text{Im}\phi^i=\dim\text{Im}\phi^{i+1}$$, we must have $$\text{Im}\phi^i=\text{Im}\phi^{i+1}$$, in either of these cases, the chain will remain constant.
Furthermore, we know the chains will have terminated by the $$n$$'th application since if $$\ker\phi^i$$ or $$\text{Im}\phi^i$$ is not constant, then the dimension must change by at least one at every application of $$\phi$$, but this cannot happen more than $$n$$ times.