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An invertible indefinite Toeplitz matrix triggers the pivot-failure branch

Statement

For the Hermitian Toeplitz matrix $T = (\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}),$ the general-RHS solver reports a nonpositive second pivot, even though $T$ is invertible. The status records that the solver's HPD domain condition failed; it does not claim that the indefinite system has no solution.

Facts & Assumptions

Given: The matrix $T$ above and the exact-arithmetic failure rules of General-RHS Hermitian Toeplitz solve by reflection recursion and Gohberg–Semencul.

[F1]

A square Toeplitz matrix has entries constant on each diagonal, and Hermitian positive definiteness requires $x^{*} T x > 0$ for every nonzero $x$ . Toeplitz, circulant, and finite-index conventions for structured solvers

[F2]

The algorithm initializes $ν_{0} = t_{0}$ and returns nonpositive-pivot $(k)$ when a computed pivot is nonpositive. General-RHS Hermitian Toeplitz solve by reflection recursion and Gohberg–Semencul

Proof

1.1givenF1algebra

The matrix is Hermitian Toeplitz with $t_{0} = 1$ and $t_{1} = t_{- 1} = 2$ . Direct multiplication gives $T (- \frac{1}{3} (\begin{array}{cc} 1 & - 2 \\ - 2 & 1 \end{array})) = I_{2},$ so $T$ is invertible. For $z = (1, - 1)^{T}$ , $z^{*} T z = - 2 < 0$ , hence it is not positive definite by [F1].

2.1givenF2step 1.1algebra∎

The solver starts with $ν_{0} = 1$ , computes $α_{1} = t_{1} / ν_{0} = 2$ and then $ν_{1} = ν_{0} (1 - ∣ α_{1} ∣^{2}) = - 3$ . Since this pivot is nonpositive, [F2] returns nonpositive-pivot $(1)$ before constructing an inverse-column denominator.

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Sources

Ralf Hiptmair, Numerical Methods for Computational Science and Engineering, Lecture Document, §4.5.3 (standard reference, not scraped)
MIT OpenCourseWare 6.341, Discrete-Time Signal Processing, Lecture 13: The Levinson-Durbin Recursion (standard reference, not scraped)