Consider the following bidiagonal $n \times n$ Toeplitz matrix $A$
$$A = \begin{bmatrix}
1-p & 0 & 0 & \cdots & 0\\
p & 1-p & 0 && \vdots \\
0 & \ddots & \ddots & \ddots & 0 \\
\vdots && p & 1-p & 0\\
0 & \cdots & 0 & p & 1-p
\end{bmatrix}$$where $0 < p < 1$. What is $A^m$ for any $m \ge 2$?
It's easy to show what the matrix is when $n = 2$ for all $m$, but not for general $n$. I have seen several papers on powers of tridiagonal Toeplitz matrices but they assume that the off-by-$1$ diagonals are all nonzero, but the "upper" diagonal here is all $0$.
Best Answer
You can rewrite $A$ as $$A = (1-p)I + p D $$ and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.
Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices: $$A^m = \sum_k {m \choose k} (1-p)^k p^{m-k} D^{m-k} = \sum_k {m \choose k} (1-p)^{m-k} p^{k} D^{k}$$
Note: $D^k = 0$ for $k \ge n$