CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A symmetric indefinite matrix can make the CG denominator vanish or change sign before convergence

Statement refuted

Symmetry alone is enough for the CG denominator test: for every symmetric invertible matrix, every nonconverged CG step has $p_{k}^{T} A p_{k} > 0$ .

Facts & Assumptions

Given: The CG recurrence and the positive-denominator theorem for Hermitian positive-definite matrices.

[F1]

To test the proposed extension outside the positive-definite domain, define the attempted first search direction by $p_{0} : = r_{0} : = b - A x_{0}$ and inspect the attempted denominator $p_{0}^{T} A p_{0}$ . These are the same algebraic formulas used by CG on its legitimate domain (The conjugate-gradient recurrence); this fact does not assert that the cited definition applies to an indefinite matrix.

[L1]

For Hermitian positive-definite matrices, every nonconverged CG denominator is positive (Before convergence, every CG denominator $p_{k}^{*} A p_{k}$ is positive).

Counterexample

technique · direct calculation

1.1

Take $A = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}), b = (\binom{1}{1}), x_{0} = (\binom{0}{0}) .$ Then $A$ is symmetric and invertible but indefinite. By [F1], $r_{0} = p_{0} = b = (\binom{1}{1}),$ so $p_{0}^{T} A p_{0} = (1, 1) (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}) (\binom{1}{1}) = (1, 1) (\binom{1}{- 1}) = 0.$ Yet $r_{0} \neq 0$ , so the method has not converged.

F1algebra

2.1

Step 1.1 shows that the first CG denominator can fail to be positive even before convergence. Therefore the refuted statement is false. The contrast with [L1] isolates the missing hypothesis: positive definiteness, not mere symmetry, is load-bearing.

L1step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Jonathan Richard Shewchuk, An Introduction to the Conjugate Gradient Method Without the Agonizing Pain (standard reference, not scraped)