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

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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.

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Singular values contract under left and right multiplication

Statement

Let $X \in F^{m \times n}$ , $Y \in F^{n \times q}$ , and $Z \in F^{r \times m}$ over $F = R$ or $C$ . For every $1 \leq i \leq \min (m, n, q)$ and $1 \leq j \leq \min (r, m, n)$ , $σ_{i} (X Y) \leq σ_{i} (X) ∥ Y ∥_{2}, σ_{j} (Z X) \leq ∥ Z ∥_{2} σ_{j} (X) .$

Facts & Assumptions

Given: Compatible finite matrices $X, Y, Z$ and indices in the displayed ranges.

[F1]

The best rank-at-most-k approximation in operator norm is the rank-k truncation of a singular value decomposition: the spectral-norm distance from a matrix $M$ to matrices of rank at most $i - 1$ is $σ_{i} (M)$ , attained by its SVD truncation.

[F2]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: $∥ Y ∥_{2} = \sup_{∥ z ∥_{2} = 1} ∥ Y z ∥_{2}$ , so $∥ Z Y ∥_{2} \leq ∥ Z ∥_{2} ∥ Y ∥_{2}$ for every compatible $Z$ .

Proof

1.1F1construct

Let $X_{i - 1}$ be the rank-at-most- $(i - 1)$ SVD truncation in [F1]. Then $X_{i - 1} Y$ has rank at most $i - 1$ , and [F1] applied to $X Y$ gives $σ_{i} (X Y) \leq ∥ (X - X_{i - 1}) Y ∥_{2}$ .

2.1step 1.1F1F2algebra

For each unit vector $z$ , [F2] gives $∥ (X - X_{i - 1}) Y z ∥_{2} \leq ∥ X - X_{i - 1} ∥_{2} ∥ Y z ∥_{2} \leq ∥ X - X_{i - 1} ∥_{2} ∥ Y ∥_{2}$ . Taking the supremum over such $z$ and using [F1] yields $∥ (X - X_{i - 1}) Y ∥_{2} \leq σ_{i} (X) ∥ Y ∥_{2}$ . Combining with step 1.1 proves the right-multiplication inequality.

3.1F1F2construct∎

Let $X_{j - 1}$ be the rank-at-most- $(j - 1)$ SVD truncation of $X$ . Then $Z X_{j - 1}$ has rank at most $j - 1$ , so [F1] gives $σ_{j} (Z X) \leq ∥ Z (X - X_{j - 1}) ∥_{2}$ . For every unit vector $v$ , [F2] gives $∥ Z (X - X_{j - 1}) v ∥_{2} \leq ∥ Z ∥_{2} ∥ (X - X_{j - 1}) v ∥_{2}$ . Taking the supremum and using [F1] yields $∥ Z (X - X_{j - 1}) ∥_{2} \leq ∥ Z ∥_{2} σ_{j} (X)$ , proving the left-multiplication inequality.

Depends on

Used by

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Wen-Wei Lin, Lecture Notes of Matrix Computations, §3.1 (standard reference, not scraped)