This is the form of matrices I am concerned about recently,

Denote by $A=(a_{ij})_{e \times e}$, where $a_{ij}$ is block matrix of dimension $n_i \times n_j$. Now given $n, 0 \leq n \leq e-1 $. Let $a_{ij}=0$, except for $j-i=n$ mod $e$. For example, $e=3,n=2$, then

$A= \left( \begin{matrix} 0 & 0 & a_{13} \\ a_{21} & 0 & 0 \\ 0 & a_{32} & 0 \\ \end{matrix} \right) $

My question is, what kind of Jordan decomposition can we get here, especially for its nilpotent part? For another example, $e=2, n=1$ case, we cannot directly use the construction way, because the characteristic polynomial for $ A= \left( \begin{matrix} 0 & a_{12} \\ a_{21} & 0 \\ \end{matrix} \right) $ is $x^{n_1-n_2} \times |xI_{n_2}-a_{21}a_{12}|$.

Since there is no further restriction on $A$, I think the answer should be general.