CounterexampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Indefinite and semidefinite matrices can both fail positive-diagonal Cholesky

Statement refuted

Indefinite matrices, and even positive-semidefinite singular matrices, admit Cholesky factorisations with positive diagonal.

Facts & Assumptions

Given: The matrices $A = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}), B = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}) .$

[L1]

A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite (A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique).

Counterexample

technique · direct

1.1

The matrix $A$ is Hermitian, but with $x = (0, 1)^{T}$ one has $x^{*} A x = - 1 < 0$ , so $A$ is not positive definite. Therefore [L1] forbids a positive-diagonal Cholesky factorisation of $A$ .

givenL1algebra

1.2

The matrix $B$ is positive semidefinite but singular. If $B = L L^{*}$ with positive diagonal, then every diagonal entry of $L$ is nonzero, so $L$ would be invertible and $B$ would be invertible as well, a contradiction. Hence $B$ also has no such Cholesky factorisation.

givenL1algebra

2.1

Steps 1.1-1.2 refute the statement in both the indefinite and the merely semidefinite cases.

step 1.1step 1.2∎

Depends on

A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, Section 2.9 (standard reference, not scraped)