Is it possible to find two matrices, $A$ an $B$ such that:
$$A B^0 x = \begin{pmatrix} c_0 \\ ? \\ ? \\ \end{pmatrix}, $$ $$A B^1 x = \begin{pmatrix} c_1 \\ ? \\ ? \\ \end{pmatrix} $$ $$A B^2 x = \begin{pmatrix} c_2 \\ ? \\ ? \\ \end{pmatrix} $$ $$\dots$$ $$A B^{m-1} x = \begin{pmatrix} c_{m-1} \\ ? \\ ? \\ \end{pmatrix} $$ $$(A B^m = A B^1) \text{ or } (A B^m = A B^0) $$
In other words, we would have two $n \times n$ matrices $A$ and $B$, and a vector $x$. We are interested in the topmost coefficient of the results of the above multiplications. Also, $B$ to a high enough power is equal to $B$.
If we can do this, how would we find $A$ and $B$?
I'm primarily interested in the case where we are working modulo integers, but I'm also interested in the case with reals. As for the $c_i$s, they are given ahead of time. The vectors on the right are in $Z^n$.
An idea would be to choose $B$ such that it rotates the coordinates of $x$.
For this consider $B$ as the matrix with $0$ everywhere except a line of $1$ over the diagonal.
Here I supposed $n=m=4$ but the idea may clearly be generalized :
$\quad B:=\quad$$\begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ \end{bmatrix}\quad$ while $\ x:=\quad \begin{bmatrix} c_0\\ c_1\\ c_2\\ c_3\\ \end{bmatrix}$
$A$ is simply the identity.