ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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.

One low-rank ADI step gives an exact residual certificate

Example

Let $A = - I_{2}$ , $B = e_{1} = (1, 0)^{T}$ , and take the single shift $p_{1} = - 1$ . One low-rank ADI step returns $V_{1} = - e_{1} / 2$ , $X_{1} = e_{1} e_{1}^{T} / 2$ , and $R_{1} = 0$ . The returned factor residual certificate is exact.

Facts & Assumptions

Given: The matrices and shift in the Example.

[F1]

Low-rank ADI for a continuous Lyapunov equation: $V_{1} = (A + p_{1} I)^{- 1} B$ , $c_{1} = - 2 Re (p_{1})$ , $X_{1} = c_{1} V_{1} V_{1}^{*}$ and $R_{1} = B + c_{1} V_{1}$ .

[F2]

Low-rank ADI has an exact residual factor identity: $A X_{1} + X_{1} A^{*} + B B^{*} = R_{1} R_{1}^{*}$ .

Verification

1.1givenF1algebra

$A + p_{1} I = - 2 I_{2}$ is invertible and $c_{1} = 2$ , so $V_{1} = (- 2 I_{2})^{- 1} e_{1} = - e_{1} / 2$ .

2.1F1step 1.1algebra

The updates in [F1] give $X_{1} = 2 (e_{1} e_{1}^{T} / 4) = e_{1} e_{1}^{T} / 2$ and $R_{1} = e_{1} + 2 (- e_{1} / 2) = 0$ .

3.1F2step 2.1algebra∎

Consequently $R_{1} R_{1}^{*} = 0$ , and directly $A X_{1} + X_{1} A^{*} + B B^{*} = - X_{1} - X_{1} + e_{1} e_{1}^{T} = 0$ , agreeing with [F2].

Depends on

Used by

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Dependency tree · two levels

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