TITLE: Prove that $V = \ker(\phi^n) \oplus \text{image}(\phi^n)$
QUESTION [11 upvotes]: Let $V$ be a $n$-dimensional complex vector space and $\phi:V\to V$ a linear mapping. Prove that $$V = \ker(\phi^n) \oplus \text{image}(\phi^n)$$
Here is my attempt:
Since $\phi^n$ is also a linear mapping of $V$ into $V$, we have that $$\dim V = \dim \ker(\phi^n) + \dim \text{image}(\phi^n).$$
We only need only to show that this sum is direct, in other words, that $$\ker(\phi^n) \cap \text{image}(\phi^n) = \{0\}.$$ since this would imply $$V = \ker(\phi^n) + \text{image}(\phi^n)$$
We let $v \in \ker(\phi^n) \cap \text{image}(\phi^n)$ be arbitrary and aim to show that $v=0$. $\ker(\phi^n)$ is the generalized eigenspace of $\phi$ for the eigenvalue $0$, so there is a $k \leq n$ such that $\phi^k(v) = 0$.
This is where I'm stuck. How do I proceed from here? Is there a different way to do this?

REPLY [2 votes]: My answer is just a cleaned up version of the answer by gaoxinge.
$\newcommand{\inters}{\cap}
$$\newcommand{\im}{\mathrm{im}}
$The following chain of subspaces of the $n$-dimensional space $V\!$,
$$
V\supseteq\phi V\supseteq\phi^2V\supseteq\cdots~,
$$
cannot strictly decrease for ever,
thus there is the least $k$ such that $\phi^kV=\phi^{k+1}V\!$,
and then $\phi^kV=\phi^lV$ for all $l\geq k$.
Set $U:=\phi^kV\!$.
In the strictly descending chain
$$
V\supset\phi V\supset\cdots\supset\phi^k V
$$
the dimensions of subspaces decrease by at least one at each step,
so we must have $k\leq n$.
It follows that $\phi^{n+1}V=\phi^nV=\phi^kV=U$, so that $\phi\, U=U$,
which means that the restriction of $\phi$ to $\phi_U\colon U\to U$ is an isomorphism.
As you observed it suffices to prove that $\im(\phi^n)\inters\ker(\phi^n)=0$,
so let us consider any $y$ in this intersection.
Since $y\in\im(\phi^n)=\phi^nV=U$, and $\phi_U^ny=\phi^ny=0$,
and $\phi_U^n\colon U\to U$ is an isomorphism, it follows that $y=0$,
and we are done.