TITLE: The kernel and image of $T^n$
QUESTION [5 upvotes]: I need help with this question:
Let $V$ be a finite vector space where $ \dim V = n $, over the complex numbers
and let $ T: V\to V $ be a linear transformation.
Prove that $ V = \ker(T^n) \oplus Im(T^n) $

REPLY [3 votes]: This is just an attempt to make explicit what the answer by Daniel Fischer seems to suggest. Consider
$$
  \{0\}=\ker(T^0)\subseteq\ker(T^1)\subseteq\cdots\subseteq\ker(T^n)\subseteq\ker(T^{n+1})
$$
There are $n+1$ symbols '$\subseteq$', so for dimension reasons they cannot all be strict inclusions. Now $\ker(T^i)=\ker(T^{i+1})$ means that there does not exist $v$ with $T^i(v)\neq 0$ and $T^{i+1}(v)=0$, and this is equivalent to 
$\def\im{\operatorname{im}}\im(T^i)\cap\ker(T)=\{0\}$. Since the subspace $\im(T^i)$ decreases as $i$ increases, we see that once one instance of  '$\subseteq$' is nonstrict (one has equality there), the ones further to the right will be as well. It follows that the last '$\subseteq$' is certainly an equality, in other words $\im(T^n)\cap\ker(T)=\{0\}$. This means that $T$ restricted to $\im(T^n)$ is injective, and so it implies $\im(T^n)\cap\ker(T^n)=\{0\}$.
The desired result now follows using rank-nullity.