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.

Absolute error can erase a tiny singular value while relative error preserves it

Example

Let $B = diag (1, ϵ)$ with $0 < ϵ < 1$ . A perturbation of spectral norm $ϵ$ can change the small singular value to zero, giving relative error one. A componentwise relative perturbation of size $η$ instead changes it by relative error exactly $η$ , independently of how small $ϵ$ is. Thus the absolute guarantee and the stronger bidiagonal relative guarantee answer different questions.

Facts & Assumptions

Given: $0 < ϵ < 1$ , $0 < η < 1$ , and the diagonal matrix $B$ .

[F1]

Singular values are Lipschitz in the spectral norm: each singular-value change under a matrix perturbation is bounded by the spectral norm of that perturbation.

[F2]

Componentwise relative perturbations of bidiagonal matrices control every singular value relatively: a componentwise relative perturbation of an upper bidiagonal matrix bounds every positive singular value relatively, with a factor independent of its absolute magnitude.

Verification

1.1F1algebra

The singular values of $B$ are $1$ and $ϵ$ . Set $E_{0} = diag (0, - ϵ)$ . Then $B + E_{0} = diag (1, 0)$ , $∥ E_{0} ∥_{2} = ϵ$ , and the small singular value changes by absolute error $ϵ$ and relative error $ϵ / ϵ = 1$ . This attains the absolute bound in [F1].

1.2F2algebra

Set $E_{1} = η B$ . Each nonzero diagonal entry is multiplied by $1 + η$ and the zero superdiagonal remains zero, so [F2] applies. Directly, $B + E_{1} = (1 + η) B$ , whose small singular value is $(1 + η) ϵ$ and whose relative error is exactly $η$ .

2.1step 1.1step 1.2∎

Since $η$ can be fixed while $ϵ$ tends to zero, step 1.2 gives a relative guarantee independent of the small singular value; step 1.1 shows that the absolute error scale alone permits complete loss of that value.

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