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!