Jordan form — determining how large $n \times n$ matrix is

Question

Can you determine how large $n$ can be if you have an $n \times n$ matrix $A$ such that $A^3 = 0$ and $A$ has a Jordan form of exactly $4$ blocks? If so, how?

Answer 1

Yes (assuming the matrix has entries in $\mathbb C$).

Since $A^3=0$ the only eigenvalue is $0$. For every Jordan block $J$ you have to have $J^3=0$, since generalised eigenspaces are invariant under $A$. What size can any block thus have?