PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.

Spectral numerical rank counts singular values above threshold over R or C

Statement

Let $m, n \geq 1$ , $F \in {R, C}$ , $A \in F^{m \times n}$ , $p = \min (m, n)$ , and $τ \geq 0$ . For the spectral norm, the numerical rank defined by Numerical rank relative to a declared norm, scale, and tolerance satisfies ${rank}_{τ} (A) = # {i \in {1, \dots, p} : σ_{i} (A) > τ} .$

Facts & Assumptions

Given: A real or complex matrix $A$ , its ordered singular values, and a threshold $τ \geq 0$ .

[F1]

Numerical rank relative to a declared norm, scale, and tolerance: ${rank}_{τ} (A)$ is the least rank of a matrix within spectral-norm distance $τ$ of $A$ .

[F2]

The best rank-at-most-k approximation in operator norm is the rank-k truncation of a singular value decomposition: for $0 \leq k < p$ , the least spectral-norm distance from $A$ to a matrix of rank at most $k$ is $σ_{k + 1} (A)$ ; the rank- $k$ SVD truncation attains it.

[F3]

The rank of a linear map is the number of its nonzero singular values: the rank of $A$ is the number of its positive singular values, so it is at most $p$ .

Proof

1.1F2F3algebra

Put $q = # {i \leq p : σ_{i} (A) > τ}$ . For every $k < q$ , one has $k < p$ and $σ_{k + 1} (A) \geq σ_{q} (A) > τ$ (when $q > 0$ ), so [F2] shows that no rank-at-most- $k$ matrix lies within distance $τ$ . If $q < p$ , then $σ_{q + 1} (A) \leq τ$ and the rank- $q$ truncation from [F2] is within distance $τ$ . If $q = p$ , [F3] gives $rank A \leq p = q$ , so $A$ itself is an admissible matrix at distance zero. Thus the least admissible rank is $q$ .

2.1step 1.1F1∎

By [F1], the least admissible rank is ${rank}_{τ} (A)$ ; step 1.1 identifies it with the number of singular values strictly greater than $τ$ . This proves the formula, including $q = 0$ and thresholds equal to a singular value.

Depends on

Used by

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Sources

Wen-Wei Lin, Lecture Notes of Matrix Computations, §3.2.2 (standard reference, not scraped)