$\def\b{\begin{bmatrix}}\def\e{\end{bmatrix}}$
I suppose I can say that $\b 0&1&0\\0&0&1\\0&0&0\e$ has nilpotency of index $3$, and all of these have characteristic $\lambda^k$ and minimal polynomial is the same, and hence we have all $k\times k$ Jordan blocks of eigenvalue $0$ of index $k$.
I just typed that all up at once so I will rigorise that, but its the idea
