TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

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A certified computed smallest singular value encloses pseudospectrum membership

Statement

Let $A \in C^{n \times n}$ with $n \geq 1$ , $z \in C$ , and $ϵ > 0$ . Suppose a computed singular-value path returns $\hat{s} \geq 0$ and a certified $δ \geq 0$ with

$∣ \hat{s} - σ_{\min} (z I - A) ∣ \leq δ .$

Then the rule “inside” when $\hat{s} + δ \leq ϵ$ is sound, and the rule “outside” when $\hat{s} - δ > ϵ$ is sound for the closed set Closed epsilon-pseudospectrum in the spectral norm. In every other case the correct status is undecided.

Facts & Assumptions

Given: The matrix, point, tolerance, and a certified pair $(\hat{s}, δ)$ satisfying the displayed error bound.

[F1]

Closed epsilon-pseudospectrum in the spectral norm: $z$ is inside exactly when $σ_{\min} (z I - A) \leq ϵ$ .

[A1]

Computed singular-value error is bounded by backward perturbation plus readout error supplies the absolute singular-value bound conditionally on its successful nearby-factorization hypothesis; its upstream premise is Local stable reflectors and rotations give an aggregate SVD backward error. For its specified real/complex scaled arithmetic, that theorem now instantiates the local Householder/Givens inequalities and the corollary adds any rounded final modulus-readout allowance. Thus a successful certified path supplies the required interval; the exact premise here still requires all arithmetic, deflation, readout and input-representation allowances, including any error in forming the stored matrix $z I - A$ . Budget exhaustion or an unavailable bound remains undecided.

Proof

1.1givenF1A1algebra

By [A1] and the stated certificate, $σ_{\min} (z I - A) \leq \hat{s} + δ$ . If $\hat{s} + δ \leq ϵ$ , then [F1] gives $z \in Λ_{ϵ} (A)$ .

2.1givenF1A1step 1.1algebra∎

The same absolute-error inequality gives $σ_{\min} (z I - A) \geq \hat{s} - δ$ . If $\hat{s} - δ > ϵ$ , then [F1] gives $z \notin Λ_{ϵ} (A)$ . If neither test passes, the interval $[\max (0, \hat{s} - δ), \hat{s} + δ]$ meets both sides of the threshold or includes the boundary, so these data alone justify only undecided.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Youcef Saad, Numerical Methods for Large Eigenvalue Problems, 2nd ed., §3.5 (standard reference, not scraped)
Wen-Wei Lin, Lecture Notes of Matrix Computations, §6.3 (standard reference, not scraped)