# 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!

-
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
... 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
Actually, I am feeling slightly confused. JM's answer suggests using LDL' as the "efficient / simple check"---in what respect is this not acceptable? thanks for clarifications. –  Suvrit May 21 '11 at 7:36

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.