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A Toeplitz matrix or diagonal-constant matrix is a matrix in which each descending diagonal from left to right is constant. For instance, the following matrix is an $n\times n$ Toeplitz matrix:

$ A = \begin{bmatrix} a_{0} & a_{-1} & a_{-2} & \ldots & \ldots &a_{-n+1} \\\ a_{1} & a_0 & a_{-1} & \ddots & & \vdots \\ a_{2} & a_{1} & \ddots & \ddots & \ddots& \vdots \\\ \vdots & \ddots & \ddots & \ddots & a_{-1} & a_{-2}\\\ \vdots & & \ddots & a_{1} & a_{0}& a_{-1} \\ a_{n-1} & \ldots & \ldots & a_{2} & a_{1} & a_{0} \end{bmatrix} $

I'm interested in the self-adjoint case ($a_{-k}=a_{k}\in\mathbb{R}$).

My question are:

  • Is there a relatively simple criterion to know when these matrices are invertible by just analyzing the sequence $\{a_{0},\ldots,a_{n-1}\}$?
  • In the invertible case, what is known about its inverse?
  • About its determinant?

Thanks!

  • 0
    If you're interesting in inferring something from the sequence of elements, consider the sequential [Levinson-Durbin algorithm](https://en.wikipedia.org/wiki/Levinson_recursion) for inverting Toeplitz matrices ($\Theta(n^2)$); it also works on block Toeplitz matrices, according to Wikipedia. There appear to be even faster variants cited therein.2013-12-28

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About the last point of you question I think it's not really simple to state a closed simple formula for the determinants. I tried to see if there are symmetries. There are but not really useful (at least as far as I can see). Just to have an idea the first 3 steps of the induction you have the following determinants.

If n=2 then\begin{equation} detA^{(2)}=(a_0^2 - a_1^2) \end{equation} If n=3 then \begin{equation} detA^{(3)}=(a_0 - a_2) (a_0^2 - 2 a_1^2 + a_0 a_2) \end{equation} Already at n=4 the formula is not so simple. Indeed if we have n=4 we the determinant become: \begin{equation} detA^{(4)}=a_0^4 + a_1^4 + a_2^4 - 2 a_1 a_2^2 a_3 +4 a_0 a_1^2 a_2 - 3 a_1^2a_0^2 -2 a_2^2a_1^2 - 2 a_2^2a_0^2 + a_3^2a_1^2 - a_3^2a_0^2 - 2 a_1^3 a_3 + 4 a_0 a_1 a_2 a_3 \end{equation} What happens is that when you have to calculate $detA^{(n)}$ the minor determinants of order $n-1$ are matrices where the condition of symmetry that alouded huge semplifications disappears. With n=2,3 the problem is not so big since the minors are trivial, but when n gets bigger the problems arise. In fact the minors are not really Toeplitz matrix, but "block Toeplitz Matriz" (sort of saying). So maybe there could be a way of enclosing the writing in a simple notation formula, but it wouldn't be a real computational gain...