# Questions on Toeplitz Matrices

These questions are probably very basic but I'll dare to ask them anyway since I didn't have much luck in MathStackExchange.

Let $A$ be an $n\times n$ Hermitian Toeplitz matrix:

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

My questions are:

• Is there a relatively "simple" criteria to determine if $A$ is invertible by analyzing the sequence $\{a_0, \ldots, a_{n-1} \}$?

• Idem as before with positive definite?

• In the invertible case, what is known about the structure of the inverse matrix? I seem to recall that this is well known.

Thanks!

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A few hours is not enough of a wait to cross post. Be more patient; I would say not everybody knows the Gohberg-Semencul formula for Toeplitz inversion, and it takes me a long time to type. – J. M. May 11 '11 at 18:21
I am interested in whether the positive definiteness of $A$ can be easily told with only the information of $\{a_0,...,a_{n-1}\}$. – Sunni May 13 '11 at 17:35
@JM: How this helps from the theoretical point of view? What you are saying is essentially check that all the eigenvalues are positive. The interesting question is to check positive definite by analyzing only the first row. – ght May 14 '11 at 12:18
@JM: I know but I still believe you are missing the point of the question, we are not looking for a "fast" or "good" algorithm in this case. – ght May 14 '11 at 13:25
... the criterion you are looking for can be checked with a finite algorithm. What exactly do you have against an algorithm? – J. M. May 14 '11 at 13:38

This is a bit related:

P. Schmidt, F. Spitzer, The Toeplitz matrices of an arbitrary Laurent polynomial, Math. Scand., 8 (1960), 15–38.

For banded Toeplitz matrices, you can get nice asymptotic results for the roots of its characteristic polynomial. This is very much related to computing the determinant. For example, for large enough $n$, there should be a simple criterion for invertibility, in terms of the $a_i$:s.

I actually gave combinatorial proof of their result, by interpreting the determinant as evaluation of Schur polynomials in certain points. This technique might help you. The trick is to use something called the Jacobi-Trudi identities, and consider Schur polynomials for rectangular Young diagrams.

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A boring answer to some of your questions occurred to me. Here it is.

Just look at the coefficients and check whether $|a_0| \ge \sum_{i \neq 0} |a_i|$. If this holds, then the matrix is diagonally dominant, so that if further, $a_0 \ge 0$, then the matrix will be positive (semidefinite). Also note that if the first inequality stated above is strict, then the matrix is guaranteed to be non-singular.

@Rojo: I think it is correct as written (albeit too strong)---here each diagonal entry is $a_0$, and DD stipulates that the diagonal entry be larger than the sum of the remaining entries in that row (btw. the OPs matrix is Hermitian Toeplitz) – Suvrit Jan 4 '14 at 2:27