Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-31
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A defective Jordan block shows that tiny perturbations can destroy an eigenvector picture even when eigenvalues barely move

Statement refuted

Refuted claim: if eigenvalues move only a little under perturbation, then the eigenvectors stay well conditioned.

Let

J=[1101],Jε=[1101+ε](ε0).

Then the eigenvalues change only by ε, but an eigenbasis matrix for Jε has condition number of order ε1.

Facts & Assumptions

Given: The defective Jordan block J and its perturbation Jε.

Counterexample

technique · direct
1.1

The eigenvalues of Jε are 1 and 1+ε, so they differ from the repeated eigenvalue of J by at most ε. An eigenvector for 1 is e1, and an eigenvector for 1+ε is (1,ε)T.

algebra
2.1

Hence an eigenbasis matrix is Xε=[110ε],Xε1=[1ε10ε1]. So Xε12 is of order ε1, and therefore κ2(Xε) is also of order ε1.

step 1.1algebra
3.1

As ε0, the eigenvalues of Jε coalesce gently, but the eigenvectors become nearly parallel and the eigenbasis becomes badly conditioned. This refutes the claim and illustrates the warning in [L1].

L1step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

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.