CounterexampleConstruction: 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 prescribed column can miss the dominant direction

Statement refuted

For every real matrix $A \in R^{m \times n}$ with $m \geq n \geq 2$ , every integer $1 \leq k < n$ , and every prescribed ordered list $J$ of $k$ distinct valid column indices, there is a matrix $T \in R^{k \times n}$ with $T (:, J) = I_{k}$ , $\max_{i, j} ∣ T_{i j} ∣ \leq 1$ , and $∥ A - A (:, J) T ∥_{2} \leq \sqrt{1 + k (n - k)} σ_{k + 1} (A)$ , where $∥ \cdot ∥_{2}$ is the Euclidean operator norm and $σ_{j} (A)$ is the $j$ th singular value.

Facts & Assumptions

Given: The claimed universal bound above.

[A1]

Consider $A = [\begin{array}{cc} 10 & 0 \\ 0 & 1 \end{array}]$ , $k = 1$ , and the prescribed index set $J = (2)$ .

Given: The matrix, rank parameter, and prescribed selected column in [A1].

Counterexample

1.1givenA1algebra

Any $1 \times 2$ matrix satisfying $T (:, J) = 1$ has the form $T = (t, 1)$ . The coefficient condition gives $∣ t ∣ \leq 1$ , and direct multiplication gives $A - A (:, J) T = [\begin{array}{cc} 10 & 0 \\ - t & 0 \end{array}]$ . Its spectral norm is the Euclidean norm of its first column, $\sqrt{100 + ∣ t ∣^{2}} \geq 10$ .

2.1A1step 1.1algebra∎

The singular values of the diagonal matrix in [A1] are $10$ and $1$ , so the proposed upper bound is $\sqrt{1 + 1 \cdot (2 - 1)} σ_{2} (A) = \sqrt{2}$ . Since $10 > \sqrt{2}$ , no such $T$ satisfies the claim.

Used by

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Dependency tree · 0 levels

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