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Sorry if this question has been asked already but I didn't find it. Given a symmetric Toeplitz matrix of the form

$$\left[\begin{array}{llll} a_0 & a_1 & \dots & a_n\\ a_1 & a_0 & \dots & a_{n-1}\\ \vdots& & & \\ a_n & a_{n-1} & \dots & a_0\\ \end{array} \right]$$

Suppose we have the relation $a_0\geq a_1\geq\dots\geq a_n$. A simple $3 \times 3$ example will show you that the inverse need not have this property (even in terms of absolute value). But does it at least have the property that away from the main diagonal, the difference decreases? That is, do we have $|a_1-a_0|\geq|a_2-a_1|\geq\dots\geq|a_n-a_{n-1}|$?

After some experimenting, it seems like we might be able to expect this, but I am wondering if this is known already as I know very little about Toeplitz forms? A followup question I have is if this is true, then is it "easy" to see that it should remain true for bi-infinite Toeplitz matrices of the same form?

The actual matrix I am interested in, for what it's worth, is $A_\lambda = (e^{-\lambda|j-k|^2})_{j,k\in\mathbb{Z}}$.

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I have a trouble understanding the question. The inverse of a Topelitz matrix is not (necessarily) a Topelitz matrix. As a matter of fact, $A_\lambda^{-1}$ is not Topelitz (at least for the finite dimensions and for some $\lambda$). So, what are $a_0,\dots,a_n$ in $|a_1-a_0| \ge \cdots \ge |a_n-a_{n-1}|$? –  Vedran Šego May 27 '13 at 1:43
    
In the bi-infinite case, my $A_\lambda$ is called a Laurent operator in much of the literature, and one can show that its inverse is again a Laurent operator (meaning it has actually a stronger form that $a_{ij}=a_{ji}$. You are correct of course in the finite case which I didn't realize could be significantly different at the time of posting this. There is a paper of Jaffard in the Annales de l'Institut Henri Poincare that partly answers this question in the following way: if an operator from $\ell_2\to\ell_2$ is such that $|a_{ij}|\leq C e^{-\gamma|i-j|}$, then the inverse satisfies this.. –  Keaton May 27 '13 at 20:47
    
type of upper bound, and similarly if $|a_{ij}|\leq c|1+|i-j||^{-\alpha}$ for some $\alpha>1$, then the inverse operator again satisfies this polynomial bound on the diagonal. –  Keaton May 27 '13 at 20:48
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It seems like you've answered your own question. Maybe you could make an answer (and accept it) from our three comments and accept it, in case someone else gets a similar problem. –  Vedran Šego May 27 '13 at 22:44

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