TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

How statement and proof provenance work

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.

Closed pseudospectrum equals the resolvent, singular-value, and perturbation tests

Statement

Let $A \in C^{n \times n}$ with $n \geq 1$ and $ϵ > 0$ . With the closed convention of Closed epsilon-pseudospectrum in the spectral norm, the following sets are equal:

$Λ_{ϵ} (A) = {z \in C : ∥ (z I - A)^{- 1} ∥_{2} \geq ϵ^{- 1}} = ⋃_{∥ E ∥_{2} \leq ϵ} σ (A + E) .$

where the resolvent norm is defined to be $+ \infty$ when $z I - A$ is singular. Equivalently, $z \in Λ_{ϵ} (A)$ exactly when $σ_{\min} (z I - A) \leq ϵ$ . The non-strict inequalities include the boundary.

Facts & Assumptions

Given: A complex square matrix $A$ , a point $z \in C$ , and $ϵ > 0$ .

[F1]

Closed epsilon-pseudospectrum in the spectral norm: membership means $σ_{\min} (z I - A) \leq ϵ$ .

[F2]

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition: $z I - A$ has unitary left and right singular-vector bases, and its smallest singular value is attained on a unit right singular vector.

[F3]

The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector: the spectral operator norm is the largest singular value.

[F4]

Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism: $z$ is an eigenvalue of a matrix exactly when that matrix has a nonzero eigenvector for $z$ .

Proof

1.1givenF1F2F4algebra

Suppose $z \in Λ_{ϵ} (A)$ . By [F1] and [F2], choose a unit right singular vector $v$ of $z I - A$ with $r = (z I - A) v$ and $∥ r ∥_{2} = σ_{\min} (z I - A) \leq ϵ$ . Define $E = r v^{*}$ . Then $E v = r$ , so $(A + E) v = z v$ . Moreover $∥ E ∥_{2} = ∥ r ∥_{2}$ : for every unit $x$ , $∥ E x ∥_{2} = ∥ r ∥_{2} ∣ v^{*} x ∣ \leq ∥ r ∥_{2}$ , with equality at $x = v$ . Thus $∥ E ∥_{2} \leq ϵ$ and [F4] gives $z \in σ (A + E)$ .

1.2givenF1F4algebra

Conversely, suppose $z \in σ (A + E)$ for some $∥ E ∥_{2} \leq ϵ$ . By [F4], choose a unit eigenvector $v$ with $(A + E) v = z v$ . Then $(z I - A) v = E v$ , hence $σ_{\min} (z I - A) \leq ∥ (z I - A) v ∥_{2} = ∥ E v ∥_{2} \leq ∥ E ∥_{2} \leq ϵ$ . By [F1], $z \in Λ_{ϵ} (A)$ .

2.1givenF1F2F3algebra∎

If $z \notin σ (A)$ , write $z I - A = U Σ V^{*}$ by [F2], with every diagonal singular value positive. Its inverse is $V Σ^{- 1} U^{*}$ , so [F3] and unitary invariance give $∥ (z I - A)^{- 1} ∥_{2} = 1 / σ_{\min} (z I - A)$ . If $z \in σ (A)$ , then $σ_{\min} (z I - A) = 0$ and the adopted resolvent convention is $+ \infty$ . Therefore the resolvent test agrees with [F1], including equality at $ϵ^{- 1}$ .

Depends on

Used by

A Jordan eigenpair residual can be much smaller than its eigenvalue error Example

Dependency tree · two levels

16 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

Youcef Saad, Numerical Methods for Large Eigenvalue Problems, 2nd ed., Proposition 3.7 (standard reference, not scraped)
Mark Embree, Matrix Theory course notes, Theorem 5.2 (standard reference, not scraped)