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Let $A$ and $B$ be two Toeplitz matrices with $A\in \mathbb R^{n\times n}$ symmetric and positive definite, and $B \in \mathbb R^{n\times k}$.

I am searching for an elegant proof (or a counterexample) for the following two statements

  1. $A^{-1}$ is symmetric and Toeplitz as well
  2. $AB$ is Toeplitz

I played around with a few random numerical examples and it seems to hold. However, I was not able to get a simple proof and I was a bit reluctant to start digging in indices. Any hint is appreciated.

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1 Answer 1

up vote 2 down vote accepted

For any invertible matrix $X$, we have $(X^{-1})^T=(X^T)^{-1}$. So, if $X$ is symmetric, $(X^{-1})^T=(X^T)^{-1}=X^{-1}$ and hence $X^{-1}$ is symmetric. Now your $A$ is symmetric and invertible. Hence its inverse is symmetric. This has nothing to do with the property that $A$ is Toeplitz.

The inverses or products of positive definite Toeplitz matrices are not Toeplitz in general. Examples: \begin{align*} &\begin{pmatrix}3&2&1\\2&3&2\\1&2&3\end{pmatrix}^{-1} =\frac18\begin{pmatrix}5&-4&1\\-4&8&-4\\1&-4&5\end{pmatrix},\\ &\begin{pmatrix}3&2&1\\2&3&2\\1&2&3\end{pmatrix}^2 =\begin{pmatrix}14&14&10\\14&17&14\\10&14&14\end{pmatrix}. \end{align*} I am surprised to read that $A^{-1}$ and $AB$ are Toeplitz in all your random numerical examples.

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Apparently, my choice of examples was not very good. Thanks a lot for your answer. –  fabee Mar 11 '13 at 13:31

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