PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

Clearing a common denominator preserves rational eigenpairs only away from poles

Statement

Let $R = q^{- 1} N$ and $Ω$ be as in Rational matrix eigenproblem on its pole-free domain. For every $λ \in Ω$ and every nonzero $x$ , $R (λ) x = 0$ if and only if $N (λ) x = 0$ . No point of $P = C ∖ Ω$ is a rational eigenvalue, regardless of whether $N (λ)$ is singular there.

Facts & Assumptions

Given: A rational matrix $R = q^{- 1} N$ with reduced common denominator $q$ , its pole-free domain $Ω$ , and $x \neq 0$ .

[F1]

Rational matrix eigenproblem on its pole-free domain: $λ \in Ω$ exactly when $q (λ) \neq 0$ , and on that domain $R (λ) = q (λ)^{- 1} N (λ)$ ; rational eigenpairs require $x \neq 0$ and $λ \in Ω$ .

Proof

technique · direct scalar multiplication

1.1givenF1algebra

If $λ \in Ω$ , then $q (λ) \neq 0$ by [F1]. Multiplying $R (λ) x = 0$ by the nonzero scalar $q (λ)$ gives $N (λ) x = 0$ . Conversely, if $N (λ) x = 0$ , division by the same nonzero scalar gives $R (λ) x = 0$ . This proves the stated equivalence exactly on $Ω$ .

2.1givenF1step 1.1algebra∎

If $λ \in P$ , then at least one entry of $R$ has a pole by the reduced common-denominator definition in [F1], so $R (λ)$ is undefined and $(λ, x)$ is outside the rational eigenproblem's domain. A singular value of $N (λ)$ cannot change that domain condition; hence every cleared root at such a point must be rejected.

Depends on

Rational matrix eigenproblem on its pole-free domain

Used by

A transfer-function pole becomes a spurious root after denominator clearing Example

Dependency tree · two levels

2 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

Timo Betcke, Nicholas J. Higham, Volker Mehrmann, Christian Schröder, and Françoise Tisseur, NLEVP: A Collection of Nonlinear Eigenvalue Problems, §2.1 (standard reference, not scraped)