CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-31

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A defective Jordan block can split under perturbation at square-root scale

Statement refuted

Every eigenvalue varies differentiably to first order through a defective point.

Consider

$A_{ε} = (\begin{array}{cc} λ & 1 \\ ε & λ \end{array}) .$

Its eigenvalues are $λ \pm \sqrt{ε}$ , so the splitting occurs at square-root scale rather than linearly.

Facts & Assumptions

Given: The perturbed Jordan block $A_{ε} = (\begin{array}{cc} λ & 1 \\ ε & λ \end{array})$ .

[F1]

Eigenvalues are roots of the characteristic polynomial (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism).

Counterexample

technique · direct

1.1

By [F1], $\det (z I - A_{ε}) = (z - λ)^{2} - ε .$ Therefore the eigenvalues are exactly $z_{\pm} (ε) = λ \pm \sqrt{ε}$ .

F1algebra

2.1

The functions $λ \pm \sqrt{ε}$ do not admit ordinary linear first-order expansions at $ε = 0$ . Hence a defective eigenvalue need not possess differentiable ordered branches through the perturbation, refuting the claim.

step 1.1algebra∎

Depends on

Eigenvalues, eigenvectors, eigenspaces $E_\lambda(T)=\ker(T-\lambda I)$, and the spectrum $\sigma_F(T)$ of an endomorphism

Used by

A 2×2 Jordan block splits into two eigenvalues separated by a square root Example

Dependency tree · two levels

4 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

David Bindel, CS 6210: Matrix Computations - Perturbation theory (standard reference, not scraped)
Benjamin Texier, Basic matrix perturbation theory (standard reference, not scraped)