Suppose $V$ is an $n$-dimensional vector space over some infinite number field $K$, $\phi\in\mathcal L(V)$, prove there exists such a (positive) integer $m$ that $$\text{Im} \phi^m=\text{Im} \phi^{m+1},\,\text{Ker}\phi^m=\text{Ker}\phi^{m+1},\,V=\text{Im} \phi^m\oplus \text{Ker} \phi^m$$
Honestly I have no idea where to start. Using matrix representation doesn't help much for me.
Can you help me? Any kind of help or hint will be appreciated. Thanks!
Since $\phi$ maps each $x$ to $0$ or nonzero, $\dim(\phi^{k+1})\leqslant\dim(\phi^{k})$. If inequalities hold for all $n$, i.e, $$ \dim(\phi^{})>\dim(\phi^{2})>\cdots>\dim(\phi^{n}) $$ Then $\dim(\phi^{n})=0$, or $\operatorname{Im}(\phi^{n})=0$ and $\dim(\operatorname{Ker}(\phi^{n}))=n$. So $$ V=\operatorname{Ker}(\phi^{n})=\operatorname{Im}(\phi^{n})\oplus \operatorname{Ker}(\phi^{n}) $$ If for some $m$ such that $\dim(\phi^{m+1})=\dim(\phi^{m})$, then $\phi^{m}$ is invariant under $\phi$. So there is $$ \operatorname{Im}(\phi^{m})\cap \operatorname{Ker}(\phi^{m})=\{0\}\tag{1} $$ for if not, let $x\in \operatorname{Im}(\phi^{m})\cap \operatorname{Ker}(\phi^{m})$, then $\phi(x)=0$ and $\dim(\phi^{m+1})<\dim(\phi^{m})$.
Since $\dim(\operatorname{Im}(\phi^{m}))+\dim(\operatorname{Ker}(\phi^{m}))=n$, by $(1)$, $\phi$ can be decomposed into the direct sum of $\operatorname{Im}(\phi^{m})$ and $\operatorname{Ker}(\phi^{m})$. So $$ V=\operatorname{Im} (\phi^m)\oplus \operatorname{Ker} (\phi^m) $$