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?

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

1 Answer 1


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.


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