# Spectra of a Symmetric Toeplitz Operator

For a physics application, I would like to be able to compute the eigenvalues of the linear operator (acting on the Hilbert space $\ell^2$) given by an infinite matrix of the form

$\begin{bmatrix} a_0 & a_1 & a_2 & a_3 & \dots\cr a_1 & a_0 & a_1 & a_2 & \dots\cr a_2 & a_1 & a_0 & a_1 & \dots\cr a_3 & a_2 & a_1 & a_0 & \dots\cr \vdots & \vdots & \vdots &\vdots & \ddots \end{bmatrix}$

(where the $a_i \in \mathbb{R}$). Some Googling tells me these (or at least, the finite dimensional analogue) are known as symmetric Toeplitz matrices, but I'm having trouble finding answers to the following questions:

1. What conditions must there be on the $a_i$ for the above operator to be compact (so we can apply the spectral theorem). Is it enough for $\sum_{i=0}^{\infty} a_i^2$ to be finite? (This question might be easy, it's just been a long time since I've done any functional analysis).

2. In the case that the above operator is compact (and so the spectral theorem applies), is there any good way to approximate its eigenvalues/eigenvectors? In particular, do the eigenvalues and eigenvectors of the upper-leftmost $N$ by $N$ finite submatrix (which is also symmetric Toeplitz) "approach" the eigenvalues/eigenvectors of the operator in any sense.

3. Is there a known closed form for the eigenvalues/eigenvectors? I really doubt this, but since these operators "kind of" look like circulant matrices (whose eigenvectors do have a fairly nice closed form), perhaps there is some really subtle roots-of-unity trick?

EDIT: I can't seem to get the LaTeX for the above matrix to display properly on my computer. If other people are having the same problem, it is supposed to look like this:

a_0 a_1 a_2 a_3 ...
a_1 a_0 a_1 a_2 ...
a_2 a_1 a_0 a_1 ...
a_3 a_2 a_1 a_0 ...
.   .   .   .  .
.   .   .   .    .

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Use \cr instead of \\ if the latter doesn't work :) –  fedja Jul 3 '11 at 22:36
For a general linear operator in $\ell^2$ given by a matrix $(a_ij)_{i,j=1}^\infty$ the standard sufficient condition for its compactness is $\sum_{i,j} a_{ij}^2<\infty$. You clearly want to relax it for your case; it might work (it can be indeed relaxed for the case of diagonal operators), though I am not sure. –  Nikita Sidorov Jul 3 '11 at 22:56
Böttcher, A.; Silbermann, B. (2006), Analysis of Toeplitz Operators, Springer Monographs in Mathematics (2nd ed.), Springer-Verlag, ISBN 9783540324348 . –  ABC Jul 3 '11 at 23:06
An answer to (2) is given by the Szego(-Tyrtyshnikov-Zamarashkin-Tilli) theorem: they converge (in measure) to the function having $a_i$ as Fourier coefficients. An answer to (3) is "no", up to my knowledge. –  Federico Poloni Jul 4 '11 at 8:45
Apart from the matrix form of the Toeplitz operators it is very convenient to consider their "so called" defining functions (or the symbols). With this function we have a nice form of the spectrum of chosen Toeplitz operator. For the references I can recommend "Basic Classes of Linear Operators" by Gohberg, Goldberg and Kaashoek. Look also at Laurent Operators which are in some sense generalised Toeplitz operators, but they have some nicer properties i.e. they form a commutative Banach algebra. –  Romanov Jul 4 '11 at 17:55
(1) and (2): There are no non-zero compact Toeplitz operators (I'm talking about the infinite case). In fact, for an operator represented by a matrix $(a_{ij})_{i,j=0}^{\infty}$ to be compact (on $l^2$), the necessary condition is that the limit of the entries on each diagonal (parallel to the main diagonal) must be zero: $\lim_{m\to\infty}a_{i+m, j+m}=0$ for each fixed $i$ and $j$.