ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

Shift-invert isolates an interior vibration mode and records solve error

Example

Use the exact shift-invert branch with start $v = e_{2}$ , Arnoldi dimension $m = 1$ and residual tolerance $τ = 0$ . Use the mass-normalized three-mode vibration matrix $A = diag (1, 4, 9)$ and $B = I$ , and seek its interior eigenvalue $4$ with shift $σ = 19 / 5$ . The shifted matrix is invertible because $σ \notin {1, 4, 9}$ . In a separate harmonic-branch invocation, use the same $A$ and $σ$ , trial basis $U = e_{2}$ , QZ sweep budget $N = 1$ , and QZ tolerance $τ_{Q Z} = 0$ .

Verification

Given: $A$ , $B$ , $σ$ , the exact target vector $u = e_{2}$ , and the separate harmonic invocation inputs $U = e_{2}$ , $N = 1$ , and $τ_{Q Z} = 0$ .

[F1] Shift-invert and harmonic Ritz extraction with solve and pencil status: in the exact shift-invert branch, a Ritz value $θ \neq 0$ maps to $λ = σ + 1 / θ$ .

[F2] Shift-invert and harmonic Ritz extraction with solve and pencil status: the harmonic branch uses the projected pencil $G - η C$ .

[F3] Shift-invert and harmonic Ritz extraction with solve and pencil status: a finite QZ diagonal pair $(α, β)$ maps to $σ + α / β$ .

[F4] Shift-invert maps eigenvalues exactly and transfers an inexact solve residual to the original eigenpair: $(A - λ I) u = θ^{- 1} (s_{0} - M r_{T})$ , with $s_{0} = M y - u$ and $r_{T} = y - θ u$ .

1.1givenF1F2F3algebra

Here $M = diag (- 14 / 5, 1 / 5, 26 / 5)$ and $M^{- 1} = diag (- 5 / 14, 5, 5 / 26)$ . Thus the interior mode 4 maps to $θ = 5$ , whose modulus exceeds the other transformed eigenvalue moduli $5 / 14$ and $5 / 26$ ; $19 / 5 + 1 / 5 = 4$ . In the separate harmonic invocation, $W = M U = e_{2} / 5$ , so $G = W^{*} W = 1 / 25$ and $C = W^{*} U = 1 / 5$ . The QZ input is this regular one-by-one pencil, already upper triangular with a nonzero diagonal pair, so the QZ procedure's computed condition holds and its status is qz-computed; $(α, β) = (1 / 25, 1 / 5)$ . The scalar equation $G - η C = 0$ gives $η = 1 / 5$ and the same value $σ + η = 4$ .

2.1givenF4step 1.1algebra∎

For an inexact solve take $y = (49 / 10) e_{2}$ and $θ = 49 / 10$ . Then $s_{0} = M y - u = - e_{2} / 50$ and $r_{T} = 0$ , while the mapped value is $λ = 19 / 5 + 10 / 49 = 981 / 245$ . Its original residual is $(A - λ I) u = - e_{2} / 245$ , which equals $θ^{- 1} (s_{0} - M r_{T}) = - e_{2} / 245$ by [F4].

Depends on

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Sources

Yousef Saad, Numerical Methods for Large Eigenvalue Problems, second edition, §8.1 (standard reference, not scraped)
Peter Arbenz, Lecture Notes on Solving Large Scale Eigenvalue Problems, §§11.4 and 12.6 (standard reference, not scraped)