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.

Exact range recovery on a rank-k matrix

Example

Assume Countable Choice for the finite-dimensional Gaussian probability law. Let $A = diag (4, 1, 0, 0) \in R^{4 \times 4}$ , take target rank $k = 2$ , oversampling $p = 2$ , and draw $Ω \in R^{4 \times 4}$ with iid standard real Gaussian entries. The randomized range finder returns $Q Q^{*} A = A$ almost surely, so its spectral error is zero.

Facts & Assumptions

Given: Countable Choice for the finite Gaussian law, and the fixed diagonal matrix and the Gaussian range-finder draw.

[F1]

The algorithm returns an orthonormal basis of $Ran (A Ω)$ and then $Q Q^{*} A$ (Exact Gaussian randomized range finder).

[F2]

Orthogonal SVD coordinates preserve the iid Gaussian law (Orthogonal and unitary invariance of Gaussian matrix blocks).

[F3]

A wide Gaussian matrix has full row rank almost surely (A wide Gaussian matrix has full row rank almost surely).

Verification

1.1givenF2F3algebra

The displayed diagonal matrix has rank $2$ and SVD $A = U Σ V^{*}$ with $U = V = I$ and $Σ_{2} = 0$ . Write $Ω_{1}$ for its first two rows. By [F2]–[F3], $Ω_{1}$ has row rank $2$ almost surely.

2.1step 1.1F1givenalgebra∎

For every such draw, $A Ω = [\begin{array}{cc} 4 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{array}] Ω_{1}$ has range exactly $span (e_{1}, e_{2})$ , since the first two rows of $Ω$ have rank $2$ and multiplication by $diag (4, 1)$ preserves rank. Thus $Q Q^{*}$ is the projector onto this span and $Q Q^{*} A = A$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Halko, Martinsson, and Tropp, Finding Structure with Randomness, §10.2 (standard reference, not scraped)
Martinsson and Tropp, Randomized Numerical Linear Algebra: Foundations & Algorithms, §11.4.2 (standard reference, not scraped)