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Is there a reliable method for testing how invertible a Hermitian Toeplitz matrix is without going through the work of actually inverting it?

A determinant is obviously easy to compute, but I'm not sure what threshold to compare it against.

The algorithm I am using is Levinson recursion. The function can detect the divide-by-zero condition, but I am interested in also detecting/scoring the matrices that are almost singular.

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up vote 2 down vote accepted

It's the condition number $\kappa$ you should be looking at, not the determinant. Since Levinson is an O(n²) method, you can hook it up to the Hager-Higham condition estimator for determining how near to singularity your matrix is. The classical heuristic is that one loses $\log_{b}\kappa$ base-$b$ significant figures in solving a linear system.

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See also Mathworld and Wikipedia's descriptions of the condition number. If you are using some linear algebra library, note that many calculate the reciprocal of the condition number by default. – Larry Wang Sep 15 '10 at 21:12

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