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

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Singular values are Lipschitz in the spectral norm

Statement

Let $A, E \in F^{m \times n}$ , where $F = R$ or $C$ , and put $p = \min (m, n)$ . Then for every $1 \leq i \leq p$ , $∣ σ_{i} (A + E) - σ_{i} (A) ∣ \leq ∥ E ∥_{2} .$

Facts & Assumptions

Given: Matrices $A, E \in F^{m \times n}$ and $1 \leq i \leq p$ .

[F1]

The best rank-at-most-k approximation in operator norm is the rank-k truncation of a singular value decomposition: $σ_{i} (A)$ is the minimum of $∥ A - Z ∥_{2}$ over matrices $Z$ of rank at most $i - 1$ ; the same holds for $A + E$ .

[F2]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: the spectral norm is a norm and satisfies the triangle inequality.

Proof

1.1F1F2choose

Choose a rank-at-most- $(i - 1)$ SVD truncation $A_{i - 1}$ of $A$ . By [F1] and [F2], $σ_{i} (A + E) \leq ∥ A + E - A_{i - 1} ∥_{2} \leq ∥ A - A_{i - 1} ∥_{2} + ∥ E ∥_{2} = σ_{i} (A) + ∥ E ∥_{2}$ .

2.1step 1.1algebra∎

Apply step 1.1 with $A + E$ in place of $A$ and $- E$ in place of $E$ . This gives $σ_{i} (A) \leq σ_{i} (A + E) + ∥ E ∥_{2}$ . The two one-sided inequalities give the stated absolute-value bound.

Depends on

Used by

Dependency tree · two levels

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Sources

Lecture Notes on Matrix Computations, Theorem 6.1.12 (standard reference, not scraped)
James Demmel and W. Kahan, Accurate Singular Values of Bidiagonal Matrices, Theorem 3 (standard reference, not scraped)