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.

A PSD Nyström residual remains positive semidefinite

Example

For $A = diag (4, 1)$ and $S = e_{1}$ , the Nyström approximation is $diag (4, 0)$ and its residual is the positive-semidefinite matrix $diag (0, 1)$ .

Facts & Assumptions

Given: $A = diag (4, 1)$ and the one-column sketch $S = e_{1}$ .

[A1]

The matrix $A$ and sketch $S$ are those stated in the Example.

[F1]

The positive-semidefinite Nyström approximation: ${\hat{A}}_{n y s} = A S (S^{*} A S)^{†} S^{*} A$ .

[F2]

Nyström is positive semidefinite, dominated by A, and exact when it captures the range: $0 ⪯ {\hat{A}}_{n y s} ⪯ A$ for every PSD $A$ and conformal sketch.

[F3]

The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations: the Moore-Penrose inverse is characterized by the four Penrose equations.

[F4]

Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse: every finite matrix has a unique Moore-Penrose inverse.

Given: $A, S$ from [A1].

Verification

1.1givenA1F1F3F4algebra

We have $A S = (4, 0)^{T}$ and $S^{*} A S = [4]$ . The scalar $[1 / 4]$ satisfies all four equations [F3], so it is the Moore-Penrose inverse by uniqueness [F4]. Substitution in [F1] gives ${\hat{A}}_{n y s} = diag (4, 0)$ .

2.1A1F2step 1.1algebra∎

Subtracting from $A$ gives $A - {\hat{A}}_{n y s} = diag (0, 1)$ , which is positive semidefinite by its nonnegative diagonal entries. The general order statement [F2] also gives $0 ⪯ {\hat{A}}_{n y s} ⪯ A$ in this instance.

Depends on

Used by

Nothing in the library uses this result yet.

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