Jordan decomposition of powers of the Shift Matrix [closed]

Question

Given the upper 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.

Answer 1

$\def\Z{\mathbb Z} $This seems unlikely to be in the form you'd like, but, as you requested, here is a slightly expanded version of my comment. 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), 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, $\sigma^{-1} = (2\ 3\ 5\ 4)$.