Finding the rank of an endomorphism Recently I tried to prove a statement I know should be easy, but for some reason I just can't prove it. The statement is: given a $9 \times 9$ matrix $N$ sucht that $N^3 = 0$ and that $rk(N^2)$ = 3, proof that $rk(N)=6$.
I tried to proof this by using the dimension kernel formula, but I got stuck. Help would be greatly appreciated!

That the rank of $N^2$ is $3$ means that the image  of $N^2$ is three dimensional. That $N^3$ is $0$ means that $N$ restricted to the image of $N^2$ is $0$. Thus the kernel of $N$ restricted to the image of $N^2$ has dimension $3$. 
This implies that the kernel of $N$ has dimension at least $3$, it also implies that the kernel of $N$ restricted to the image of $N$ has dimension at least $3$. 
From the fact that the rank of $N^2$ is $3$ we know that the kernel of $N^2$ has dimension $6$. 
Yet the dimension of the kernel of $N^2$ is the dimension of the  kernel of $N$  plus the dimension of the kernel of $N$ restricted to the image of $N$. 
Thus they are both $3$ and the claim follows.
