ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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.

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The closed epsilon-pseudospectrum of the two-by-two Jordan block is a disk

Example

Let $J = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ and $ϵ > 0$ . Then

$Λ_{ϵ} (J) = {z \in C : ∣ z ∣^{2} \leq ϵ + ϵ^{2}},$

the closed disk centered at $0$ with radius $\sqrt{ϵ + ϵ^{2}}$ . In particular, its radius is of order $\sqrt{ϵ}$ as $ϵ ↓ 0$ , while $σ (J) = {0}$ .

Facts & Assumptions

Given: The matrix $J$ and $ϵ > 0$ .

[F1]

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

[F2]

The singular values of a linear map as the eigenvalues of the positive square root of T^*T: the singular values are the nonnegative square roots of the eigenvalues of $M^{*} M$ .

Verification

1.1givenF2algebra

Put $x = ∣ z ∣^{2}$ and $M = z I - J = (\begin{array}{cc} z & - 1 \\ 0 & z \end{array})$ . Then $M^{*} M = (\begin{array}{cc} x & - \overline{z} \\ - z & x + 1 \end{array})$ has trace $1 + 2 x$ and determinant $x^{2}$ , so its smaller eigenvalue is $f (x) = σ_{\min} (M)^{2} = (1 + 2 x - \sqrt{1 + 4 x}) / 2$ by [F2].

2.1givenF1step 1.1algebra∎

For $y \geq x \geq 0$ , $f (y) - f (x) = (y - x) (1 - \frac{2}{\sqrt{1 + 4 y} + \sqrt{1 + 4 x}}) \geq 0$ . The parenthesized factor is strictly positive when $y > x$ , since then $\sqrt{1 + 4 y} + \sqrt{1 + 4 x} > 2$ ; thus $f$ is strictly increasing on $[0, \infty)$ . Also $\sqrt{1 + 4 (ϵ + ϵ^{2})} = 1 + 2 ϵ$ , hence $f (ϵ + ϵ^{2}) = ϵ^{2}$ . By strict monotonicity and [F1], $f (∣ z ∣^{2}) \leq ϵ^{2}$ exactly when $∣ z ∣^{2} \leq ϵ + ϵ^{2}$ .

Depends on

Used by

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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)
Mark Embree, Matrix Theory course notes, §5.3 (standard reference, not scraped)