Proving $V=\text{KerT}^n \oplus \text{Im}T^n$ for $V$ in $\mathbb C$ 
Let $V$ be a vector space over $\mathbb C$ with dimension $n$.
Let $T:V\to V$ be a linear transformation. I tried to show that
  $V=\text{KerT}^n \oplus \text{Im}T^n$ .

This is my solution, would like you to verify it:
Since $T$ is over $\mathbb C$, we know that there exists a jordan form $J$ for which $T=M^{-1}JM$.
Therefore $T^n = M^{-1}J^nM$.  So I want to show that  $V=\text{KerJ}^n \oplus \text{Im}J^n$.
$J^n$ is just each jordan block $n$'th power. Therefore, if the eigenvalue of the block is $\lambda \ne 0$ then the block stays inversible, having a rank of the size of the block.
If the eigenvalue of the block is $\lambda = 0$ then since it is the $n$'th power, and because the block is nilpotent, the block becomes $0$.
Now, since each block operates on a different subspace in $V$, we get that for all blocks with eigenvalues $\ne 0$ the relevant subspace is in $\text{Im}T$, and for all blocks with eigenvalues $=0$ the relevant subpace is in $\text{Ker}T$. This is why they sum $\oplus$, and by using the dimensions equations it is easy to show that $\text{dim}V=\text{dimKer}T + \text{dimIm}T$.
Is the direction right? I know it is not very formal. Would also love to see other ideas on this, but mainly comments on my solution, Thanks!
 A: Your solution has the right idea, but the last two paragraphs need a little work.
Here is one idea:
The Jordan form decomposes $V$ into a product of $J$ invariant subspaces $V= V_1 \oplus \cdots \oplus V_m $ and
each subspace $V_k$ is invariant under the corresponding Jordan block $J_k$.
Pick some $x \in V$ and write $x = x_1+\cdots x_m$ with $x_k \in V_k$.
If $J_k$ is invertible, let $v_k = (J_k^n)^{-1} x_k$ and $n_k = 0$.
If $J_k$ is not invertible (and hence $J_k^n = 0$) let $v_k = 0$
and $n_k = x_k$.
Let $v = v_1+\cdots+v_k, n = n_1+\cdots+n_k$. Note that $n \in \ker J^n$.
Then $x = J^n v + n$ is a suitable decomposition.
A: I propose the following solution.
a) Let $P(x)$ the characteristic polynomial of $\phi$, and suppose that $P(x)=x^lQ(x)$ with $Q(0)\not =0$ (we have $l\leq n$). Let $k\geq n$. There exists polynomials $U,V$ such that $U(x)P(x)+V(x)x^k =x^l$. Hence $ U(\phi)P(\phi)+V(\phi)\phi^k =\phi^l=V(\phi)\phi^k$, using Cayley-Hamilton's theorem. Hence, if $x$ is such that $\phi^k(x)=0$, we get $\phi^l(x)=0$, hence $\phi^n(x)=0$.
b) Let $x\in E={Ker}(\phi^n)\cap {Im}(\phi^n)$. Then $\phi^n{x}=0$, and there exist $y$ such that $\phi^n(y)=x$. We have $\phi^{2n}(y)=0$, hence by a) $\phi^n(y)=x=0$.
c) as ${Dim}({Ker}(\phi^n))+{Dim} ({Im}(\phi^n))=n$, we are done.
