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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Computed sensitivity errors perturb every singular value by at most their operator-norm error

Statement

Let $J, J_{h} \in R^{m \times n}$ , let $r = \min {m, n}$ , and let $σ_{1} (A) \geq \dots \geq σ_{r} (A) \geq 0$ denote the singular values of a real matrix $A$ . Then for every $1 \leq i \leq r$ , $∣ σ_{i} (J_{h}) - σ_{i} (J) ∣ \leq ∥ J_{h} - J ∥_{2},$ where $∥ \cdot ∥_{2}$ is the Euclidean operator norm.

Facts & Assumptions

Given: Real matrices $J, J_{h} \in R^{m \times n}$ and an index $1 \leq i \leq r = \min {m, n}$ .

[F1]

The best rank-at-most-k approximation in operator norm is the rank-k truncation of a singular value decomposition: the best operator-norm error among matrices of rank at most $i - 1$ is $σ_{i} (A)$ , attained by an SVD truncation, for both $A = J$ and $A = J_{h}$ .

[F2]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: $∥ A ∥_{2} = \max_{∥ v ∥_{2} = 1} ∥ A v ∥_{2}$ for a nonzero domain, and the Euclidean vector norm satisfies the triangle inequality (The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product).

Proof

technique · direct

1.1F1F2givenchoosealgebra

Let $J_{i - 1}$ be the rank-at-most- $(i - 1)$ SVD truncation of $J$ . By [F1], $∥ J - J_{i - 1} ∥_{2} = σ_{i} (J)$ , while the minimal-error characterization in [F1] for $J_{h}$ gives $σ_{i} (J_{h}) \leq ∥ J_{h} - J_{i - 1} ∥_{2}$ . For every unit vector $v$ , the vector triangle inequality gives $∥ (J_{h} - J_{i - 1}) v ∥_{2} \leq ∥ (J - J_{i - 1}) v ∥_{2} + ∥ (J_{h} - J) v ∥_{2} \leq ∥ J - J_{i - 1} ∥_{2} + ∥ J_{h} - J ∥_{2}$ ; taking the maximum over unit $v$ proves $σ_{i} (J_{h}) \leq σ_{i} (J) + ∥ J_{h} - J ∥_{2}$ .

2.1step 1.1algebra∎

Apply step 1.1 with $J_{h}$ in place of $J$ and $J - J_{h}$ in place of $J_{h} - J$ . This gives $σ_{i} (J) \leq σ_{i} (J_{h}) + ∥ J - J_{h} ∥_{2}$ . The two one-sided inequalities imply the asserted absolute-value bound.

Depends on

Used by

A small sensitivity solve error can reverse a two-design ranking Example

Dependency tree · two levels

9 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

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)