TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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An interval determinant excluding zero certifies nonsingularity and sign

Statement

Let $n \geq 1$ and let $A$ be a nonempty family of real $n \times n$ matrices. Suppose a real interval $D$ contains $\det A$ for every $A \in A$ . If $0 \notin D$ , then every $A \in A$ is nonsingular; if $D$ is positive or negative, its sign is the common determinant sign. If $D = {0}$ , every member is singular. When neither condition holds, the certificate makes no singularity or sign assertion.

Facts & Assumptions

Given: $A$ and a sound real interval determinant enclosure $D$ as stated.

[F1]

The determinant is the finite Leibniz sum for a real square matrix (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F2]

A finite square real matrix is invertible exactly when its determinant is nonzero (A finite square real matrix is invertible if and only if its determinant is nonzero).

Proof

1.1givenF1F2

For any $A \in A$ , the assumed enclosure gives $\det A \in D$ . If $0 \notin D$ , then $\det A \neq 0$ , and [F2] makes $A$ nonsingular. If $D$ is a positive or negative interval, every $\det A$ has its corresponding sign.

2.1givenF1F2∎

If $D = {0}$ , then $\det A = 0$ for every $A \in A$ , so [F2] makes every member singular. In the remaining case the stated output is inconclusive and no singularity or sign claim is made.

Depends on

Used by

Dependency tree · two levels

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Sources

Siegfried M. Rump, Verification methods: Rigorous results using floating-point arithmetic, Acta Numerica 19 (2010), §10.10 (standard reference, not scraped)