Let $u$ be an endomorphism of $E$ nilpotent of index $p$ and of rank $r$. Suppose the matrix of $u$ in some basis $\mathcal{B}$ is $N_\sigma = \operatorname{diag}\left(J_{\alpha_1}, \ldots, J_{\alpha_k}\right)$ where $\sigma = (\alpha_1, \ldots, \alpha_k)$ is a partition of $n$.
Assume that there exist a partition $\sigma'$ of the integer $n$ and a basis $\mathcal{B}'$ of $E$ such that the matrix of $u$ in $\mathcal{B}'$ is equal to $N_{\sigma'}$. Show that $\sigma = \sigma'$.