## A fact about linear transformations on a finite dimensional vector space

Let $F$ be a field and let $V$ be a finite dimensional vector space over $F$. Let $\varphi : V \rightarrow V$ be a linear transformation. Prove that there exists a positive natural number $m$ such that $\mathsf{im}\ \varphi^m \cap \mathsf{ker}\ \varphi^m = 0$, where $\varphi^m$ denotes the $m$th iterate of $\varphi$.

Note that if $\varphi^a(x) = 0$, then $\varphi^{a+1}(x) = 0$. Similarly, if $x = \varphi^{a+1}(y)$, then $x = \varphi^a(\varphi(y))$. This gives us two chains of subspaces in $V$, one ascending and one descending: $\mathsf{ker}\ \varphi^1 \subseteq \mathsf{ker}\ \varphi^2 \subseteq \cdots \subseteq \mathsf{ker}\ \varphi^s \subseteq \mathsf{ker}\ \varphi^{s+1} \subseteq \cdots$ and $\mathsf{im}\ \varphi^1 \supseteq \mathsf{im}\ \varphi^2 \supseteq \cdots \supseteq \mathsf{im}\ \varphi^t \supseteq \mathsf{im}\ \varphi^{t+1} \supseteq \cdots$.

Note that because $V$ is finite dimensional, these chains must stabilize. Let $s$ and $t$ be minimal such that $\mathsf{ker}\ \varphi^s =\mathsf{ker}\ \varphi^{s+1}$ and $\mathsf{im}\ \varphi^t = \mathsf{im}\ \varphi^{t+1}$, respectively, and let $m = \mathsf{max}(s,t)$. We claim that $\mathsf{ker}\ \varphi^m \cap \mathsf{im}\ \varphi^m = 0$.

Let $x \in \mathsf{ker}\ \varphi^m \cap \mathsf{im}\ \varphi^m$. Now $x = \varphi^m(y)$ for some $y \in V$, and so $0 = \varphi^{m}(x) = \varphi^{2m}(y)$. In particular, $y \in \mathsf{ker}\ \varphi^{2m} = \mathsf{ker}\ \varphi^m$. So $x = \varphi^m(y) = 0$. Thus $\mathsf{ker}\ \varphi^m \cap \mathsf{im}\ \varphi^m = 0$, as desired.