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

$J = [\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}], J_{ε} = [\begin{array}{cc} 1 & 1 \\ 0 & 1 + ε \end{array}] (ε \neq 0) .$

Then the eigenvalues change only by $ε$ , but an eigenbasis matrix for $J_{ε}$ has condition number of order $∣ ε ∣^{- 1}$ .

Facts & Assumptions

Given: The defective Jordan block $J$ and its perturbation $J_{ε}$ .

[L1]

Small residuals certify nearby exact eigenpairs only in backward-error terms (For a unit vector $x$ , the smallest perturbation making $(μ, x)$ an exact eigenpair has spectral norm $∥ A x - μ x ∥_{2}$ ).

Counterexample

technique · direct

1.1algebra

The eigenvalues of $J_{ε}$ are $1$ and $1 + ε$ , so they differ from the repeated eigenvalue of $J$ by at most $∣ ε ∣$ . An eigenvector for $1$ is $e_{1}$ , and an eigenvector for $1 + ε$ is $(1, ε)^{T}$ .

2.1step 1.1algebra

Hence an eigenbasis matrix is $X_{ε} = [\begin{array}{cc} 1 & 1 \\ 0 & ε \end{array}], X_{ε}^{- 1} = [\begin{array}{cc} 1 & - ε^{- 1} \\ 0 & ε^{- 1} \end{array}] .$ So $∥ X_{ε}^{- 1} ∥_{2}$ is of order $∣ ε ∣^{- 1}$ , and therefore $κ_{2} (X_{ε})$ is also of order $∣ ε ∣^{- 1}$ .

3.1L1step 2.1∎

As $ε \to 0$ , the eigenvalues of $J_{ε}$ coalesce gently, but the eigenvectors become nearly parallel and the eigenbasis becomes badly conditioned. This refutes the claim and illustrates the warning in [L1].

Depends on

For a unit vector $x$, the smallest perturbation making $(\mu,x)$ an exact eigenpair has spectral norm $\|Ax-\mu x\|_2$

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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