For any linear operator $\phi$ on $V$, prove such an integer $m$ exists. 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!
 A: 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
$$
V=\operatorname{Im} (\phi^m)\oplus \operatorname{Ker} (\phi^m)
$$
