# Powers of bidiagonal Toeplitz matrix

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 $$0$$.

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 $$k \ge n$$

You can use the spectral symbol that generates $$A_n$$, to get the symbol that generates $$A_n^m$$

Symbol that generates $$A_n$$: $$f(\theta)=1-p+pe^{\mathbf{i}\theta}$$

Symbol that generates $$A_n^m$$: $$(f(\theta))^m$$

For example $$m=3$$ we have

$$(1-p+pe^{\mathbf{i}\theta})^3=-(p - 1)^3+3p(p - 1)^2e^{\mathbf{i}\theta}-3p^2(p - 1)e^{2\mathbf{i}\theta}+p^3e^{3\mathbf{i}\theta}$$.

Thus you have $$-(p - 1)^3$$ on the main diagonal, $$3p(p - 1)^2$$ on first sub diagonal, $$-3p^2(p - 1)$$ on second sub diagonal, and finally $$p^3$$ on the third sub diagonal.