Prove that $V = \ker(\phi^n) \oplus \text{image}(\phi^n)$ 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?
 A: We consider the chains 
$$V\subset\phi(V)\subset\cdots\subset\phi^n(V)\subset\cdots$$ 
and 
$$\dim(V)\geq\dim(\phi(V))\geq\cdots\geq\dim(\phi^n(V))\geq\cdots$$


*

*If $\dim(\phi^n(V))=1$, it is easy to prove.

*If $\dim(\phi^n(V))\geq2$, there exist $k\leq n$ such that $\phi^k(V)=\phi^{k+1}(V)=\cdots$. Actually, $\phi:\phi^n(V)\rightarrow\phi^n(V)$ is an isomorphism.


*

*For any $a\in V$, there exists $b$ such that $\phi^n(a)=\phi^{2n}(b)$. Then we can prove $a=c+\phi^n(b)$ and $\phi^n(c)=0$.

*We assume $a\in\ker(\phi^n)\cap\text{image}(\phi^n)$. That is to say, $\phi^n(a)=0$ and $\phi^n(b)=a$. If $a\ne0$, we get $\phi^{2n}(b)=0$ and $\phi^n(b)\ne0$ which will lead $\dim(\phi^n(V))>\dim(\phi^{2n}(V))$. That is a contraction.




Look at an another question Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$.



*

*In your question, $\phi:\phi^n(V)\rightarrow\phi^n(V)$ is an isomorphism. Also $\phi^n:\phi^n(V)\rightarrow\phi^n(V)$ is an isomorphism.

*In another question, $f=id: f(V)\rightarrow f(V)$ is an isomorphism.

*Lemma Let $f:V\rightarrow V$ be a linear map with $f:f(V)\rightarrow f(V)$ being an isomorphism. Then we have
$$V=\ker(f)\oplus\text{image}(f)$$
A: 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.
