ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

A nearly rank-deficient two-column design changes numerical rank with tolerance

Example

For $ϵ = 10^{- 8}$ , let the design matrix be $A_{ϵ} = (\begin{array}{cc} 1 & 1 \\ 0 & ϵ \end{array}) .$ Its exact rank is two, but its spectral numerical rank is one at tolerance $10^{- 7}$ and two at tolerance $10^{- 9}$ .

Facts & Assumptions

Given: The displayed real matrix $A_{ϵ}$ and the two absolute spectral-norm tolerances.

[F1]

Spectral numerical rank counts singular values above threshold over R or C: spectral numerical rank counts singular values strictly greater than the declared tolerance.

[F2]

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition: $A$ has an SVD $A = U Σ V^{*}$ with diagonal entries of $Σ$ equal to its nonnegative singular values.

[F3]

Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison: $∥ A ∥_{2} = σ_{1} (A)$ and $∥ A ∥_{F}^{2} = σ_{1} (A)^{2} + σ_{2} (A)^{2}$ .

[F4]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: $∥ A ∥_{2} = \sup_{∥ x ∥_{2} = 1} ∥ A x ∥_{2}$ .

Verification

1.1F2F3F4algebra

The determinant of $A_{ϵ}$ is $ϵ \neq 0$ , so its exact rank is two. By [F2], write $A_{ϵ} = U Σ V^{*}$ ; taking determinants and using $∣ \det U ∣ = ∣ \det V ∣ = 1$ (from $U^{*} U = V^{*} V = I$ ) gives $σ_{1} σ_{2} = ϵ$ . Its first column has norm one, hence [F4] gives $σ_{1} = ∥ A_{ϵ} ∥_{2} \geq 1$ ; by [F3], $σ_{1} \leq ∥ A_{ϵ} ∥_{F} = \sqrt{2 + ϵ^{2}} < 2$ . Therefore $ϵ / 2 < σ_{2} \leq ϵ$ .

2.1step 1.1algebra

At $τ = 10^{- 7}$ , $σ_{2} \leq 10^{- 8} < τ < 1 \leq σ_{1}$ , so exactly one singular value exceeds $τ$ . At $τ = 10^{- 9}$ , $σ_{2} > 5 \cdot 10^{- 9} > τ$ and $σ_{1} \geq 1 > τ$ , so both exceed $τ$ .

3.1step 2.1F1∎

Applying [F1] to the two counts in step 2.1 gives numerical ranks one and two, respectively, although the exact rank remains two.

Depends on

Used by

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