Does there exist a species of matrix, say $M$, such that
$$M^{\alpha} = M, \quad M \neq I$$
for some $\alpha > 2$. I'm not looking for idempotent matrices (square to themselves), but in essence a kind of generalisation of them.
A construction I considered was by isomorphism – the involutive matrices are isomorphic to the idempotent so I wondered if there was a similar isomorphism between matrices of general finite order and the ones I seek above. So far, I haven't found such a thing.
Best Answer
What you are describing is often called a periodic matrix.
A common example is any $(3\times 3)$ normalized $(\|N\|^2_F = 2)$ skew-symmetric $(N^T = -N)$ matrix.
Such a matrix is rank deficient $({\rm rank}(N)=2)$ and its nullspace consists of a single vector $(n\implies Nn=0)$, which is usually normalized $(\|n\|^2={\tt1})$.
Here is its multiplication table $$\eqalign{ N^2 &= nn^T-I \\ N^3 &= -N \\ N^4 &= I-nn^T \\ N^5 &= N \\ }$$ This table can be represented by a variety of formulas $(k\ge 0)$ $$\eqalign{ \def\o{{\tt1}} N^{k+3} &= -N^{k+1} \\ N^{2k+1} &= (-\o)^{k}N \\ N^{2k+2} &= (-\o)^{k}N^2 \\ N^{4k+1} &= N \\ }$$
Another recipe for constructing a periodic matrix is to use a permutation matrix $P$ as the leading diagonal block and a zero matrix as the lower diagonal block, e.g. $$B = \pmatrix{ P & 0 \\ 0 & 0 \\ }$$ This matrix will have the same period as $P$ but $B^k\ne I$ for all $k\ge\o$.