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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 $0$.

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2 Answers 2

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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 $k \ge n$

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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.

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