CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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For a nonnormal matrix, a start vector orthogonal to the dominant left eigendirection can defeat power iteration

Statement refuted

Refuted claim: once a matrix has a unique dominant eigenvalue, every start vector makes power iteration converge to its eigendirection.

Let

$A = [\begin{array}{cc} 2 & 1 \\ 0 & 1 \end{array}], x_{0} = \frac{1}{\sqrt{2}} [\begin{array}{c} 1 \\ - 1 \end{array}] .$

Then $2$ is the unique dominant eigenvalue of $A$ , but the power iteration started from $x_{0}$ stays fixed at the eigendirection of the smaller eigenvalue $1$ .

Facts & Assumptions

Given: The matrix $A$ and start vector $x_{0}$ in the statement.

[L1]

The convergence theorem requires a nonzero component in the dominant eigendirection (If a diagonalisable matrix has a simple eigenvalue of strictly largest modulus and the start vector has a nonzero component in that eigendirection, power iteration converges projectively at the eigenvalue-ratio rate).

Counterexample

technique · direct

1.1

The right eigenvectors are $e_{1}$ for eigenvalue $2$ and $(1, - 1)^{T}$ for eigenvalue $1$ . A left eigenvector for eigenvalue $2$ is $(1, 1)^{T}$ , and $[\begin{array}{cc} 1 & 1 \end{array}] x_{0} = 0.$ So the start vector is orthogonal to the dominant left eigendirection.

algebra

2.1

Direct multiplication gives $A x_{0} = \frac{1}{\sqrt{2}} [\begin{array}{c} 1 \\ - 1 \end{array}] = x_{0} .$ Hence every normalised power iterate equals $x_{0}$ .

step 1.1algebra

3.1

The power iteration never approaches the dominant eigendirection $e_{1}$ , so the start condition in [L1] is genuinely necessary.

L1step 2.1∎

Depends on

If a diagonalisable matrix has a simple eigenvalue of strictly largest modulus and the start vector has a nonzero component in that eigendirection, power iteration converges projectively at the eigenvalue-ratio rate

Used by

Nothing in the library uses this result yet.

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