Given the upper [Shift Matrix](https://en.wikipedia.org/wiki/Shift_matrix), which for e.g. dimension $5$ is
$$
{\bf E}\_{\,{\bf 5}} = \left( {\matrix{
 0 & 1 & 0 & 0 & 0 \cr 
 0 & 0 & 1 & 0 & 0 \cr 
 0 & 0 & 0 & 1 & 0 \cr 
 0 & 0 & 0 & 0 & 1 \cr 
 0 & 0 & 0 & 0 & 0 \cr 
 } } \right)
$$
then its non-negative integral powers are just given by a shift 
of the non-null diagonal, till ${\bf E}\_{\,{\bf 5}} ^{\,{\bf 5}} ={\bf 0}$.


I know that the Jordan decomposition of ${\bf E}\_{\,h} ^{\,{\bf n}}$ is given by 
$$
{\bf E}\_{\,h} ^{\,{\bf n}} = {\bf P}\_{\,h} (n)\;{\bf C}\_{\,h} (n)\;{\bf P}\_{\,h} (n)^{\, - \;{\bf 1}} 
$$
where  

 - ${\bf P}\_{\,h} (n)$ is a permutation matrix;  

 - ${\bf C}\_{\,h} (n)$ is actually a ${\bf E}\_{\,h}$ with $n-1$ ones missing in certain positions, i.e.
it is also expressible as a permutation.


For instance
$$
\eqalign{
 & {\bf E}\_{\,{\bf 5}} ^{\,{\bf 2}} = \left( {\matrix{
 0 & 0 & 1 & 0 & 0 \cr 
 0 & 0 & 0 & 1 & 0 \cr 
 0 & 0 & 0 & 0 & 1 \cr 
 0 & 0 & 0 & 0 & 0 \cr 
 0 & 0 & 0 & 0 & 0 \cr 
 } } \right) = {\bf P}\_{\,{\bf 5}} (2)\;{\bf C}\_{\,{\bf 5}} (2)\;{\bf P}\_{\,{\bf 5}} (2)^{\, - \;{\bf 1}} = \cr 
 & = \left( {\matrix{
 1 & 0 & 0 & 0 & 0 \cr 
 0 & 0 & 0 & 1 & 0 \cr 
 0 & 1 & 0 & 0 & 0 \cr 
 0 & 0 & 0 & 0 & 1 \cr 
 0 & 0 & 1 & 0 & 0 \cr 
 } } \right)\left( {\matrix{
 0 & 1 & 0 & 0 & 0 \cr 
 0 & 0 & 1 & 0 & 0 \cr 
 0 & 0 & 0 & 0 & 0 \cr 
 0 & 0 & 0 & 0 & 1 \cr 
 0 & 0 & 0 & 0 & 0 \cr 
 } } \right)\left( {\matrix{
 1 & 0 & 0 & 0 & 0 \cr 
 0 & 0 & 1 & 0 & 0 \cr 
 0 & 0 & 0 & 0 & 1 \cr 
 0 & 1 & 0 & 0 & 0 \cr 
 0 & 0 & 0 & 1 & 0 \cr 
 } } \right) \cr} 
$$


But, after various trials, I could not yet find an effective way to express the type of P and C permutations wrt $h$ and $n$.  

Most probably the subject has been already studied or it is easily assessable with the appropriate approach, and I am asking
for hints in this respect.



Jordan decomposition of powers of the Shift Matrix
$\def\Z{\mathbb Z}
$This seems unlikely to be in the form you'd like, but, as you [requested](https://mathoverflow.net/questions/290029/jordan-decomposition-of-powers-of-the-shift-matrix#comment718905_290029), here is a slightly expanded version of my [comment](https://mathoverflow.net/questions/290029/jordan-decomposition-of-powers-of-the-shift-matrix#comment718903_290029). There's no idea here, just computation, and you should check it yourself to make sure I haven't made (another) silly error.


For $k \in \Z$, put $\overline k = \{1, \dotsc, k\}$. For $i \in \overline h$, let $e\_i$ be the $i$th standard basis vector, so that $E\_h^n e\_i = e\_{i + n}$ for all $i \in \overline{h - n}$ and $E\_h^n e\_i = 0$ for all $i \in \overline h \setminus \overline{h - n}$. Let $\sigma \in \mathrm S\_h$ be the permutation so that
$$
\sigma^{-1}(1, \dotsc, h) = (\underbrace{1, 1 + n, 1 + 2n, \dotsc}\_{\lfloor(h - 1)/n\rfloor + 1}, \underbrace{2, 2 + n, 2 + 2n, \dotsc}\_{\lfloor(h - 2)/n\rfloor + 1}, \dotsc, \underbrace{n, 2n, 3n, \dotsc}\_{\lfloor(h - n)/n\rfloor + 1})
$$
(I made a fencepost error counting the size of each block in my [comment](https://mathoverflow.net/questions/290029/jordan-decomposition-of-powers-of-the-shift-matrix#comment718903_290029)), and $P$ the $h$-square permutation matrix with $P\_{i j} = [\sigma(i) = j]$ for all $i, j \in \overline h$. Then $P^{-1}E\_h^n P e\_j = e\_{j + 1}$ for all $j \in \sigma(\overline{h - n})$, and $P^{-1}E\_h^n P^{-1}e\_j = 0$ for all $j \in \overline h \setminus \sigma(\overline{h - n})$. That is, if, for $k \in \mathbb Q\_{\ge 0}$, we write $J\_k$ for the $(\lfloor k\rfloor + 1)$-square Jordan block
$$
J\_k = \begin{pmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & \ddots & \ddots & \vdots \\
\vdots & \ddots & \ddots & 1 & 0 \\
0 & \cdots & 0 & 0 & 1 \\
0 & \cdots & 0 & 0 & 0
\end{pmatrix}
$$
(with $\lfloor k\rfloor$ $1$'s), then $C \mathrel{:=} P^{-1}E\_h^n P$ equals $J\_{(h - 1)/n} \oplus J\_{(h - 2)/n} \oplus \dotsb \oplus J\_{(h - n)/n}$.


Notice this gives the obvious right answer when $n = 1$, in which case $\sigma$ is the identity permutation.


For your example with $h = 5$ and $n = 2$, we have that
$$
\sigma^{-1}(1, 2, 3, 4, 5) = (\underbrace{1, 3, 5}\_{\lfloor(h - 1)/n\rfloor + 1 = 3}, \underbrace{2, 4}\_{\lfloor(h - 2)/n\rfloor + 1 = 2});
$$
that is, in cycle notation, $\sigma^{-1} = (2\ 3\ 5\ 4)$.


