The index of nilpotency of a nilpotent matrix

Question

Let $A$ a matrix in $\mathcal{M}_5(\mathbb C)$ such that $A^5=0$ and $\mathrm{rank}(A^2)=2$, how prove that $A$ is nilpotent with index of nilpotency $4$? Thanks in advance.

Answer 1

In the followings I'll use only words, but you should visualize the matrices I mention.

Jordan normal form tells you that $A=P^{-1}JP$, for some Jordan matrix $J$ and some invertible matrix $P$.

As you know, $\text{rank}(A)=\text{rank}(J)$, therefore $\text{rank}(J^2)=2$.

This tells you that the index of the only eigenvalue (which is $0$) is either $3,4$ or $5$.

Recall that JNF also tells you that the largest Jordan block on $J$ has size equal to the index of $0$.

Now consider the possibilities for $J$ if the index is $5$. Square it and find a contradiction. Similarly if the index is $3$.

Conclude.