False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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FALSE: complexification creates a real eigenvector whenever it creates a complex one

Statement

If the complexification of a real operator acquires a complex eigenvector, then the original real operator acquires a real eigenvector.

Facts & Assumptions

Given: The quarter-turn $T : R^{2} \to R^{2}$ with matrix $(\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ and its complexification $T_{C}$ .

[L1]

The complexification $T_{C}$ has the nonreal eigenvalues $\pm i$ with eigenvectors $(1, - i)$ and $(1, i)$ , while $T$ itself has no real eigenvector (The real quarter-turn diagonalises after complexification but has no real eigenvector).

Refutation

technique · direct

1.1

By [L1], complexification creates a complex eigenvector: $(1, - i)$ is an eigenvector of $T_{C}$ with eigenvalue $i$ .

1.2

By [L1], $T$ has no real eigenvector: any real eigenvector $v \neq 0$ would carry a real eigenvalue $λ$ with $λ^{2} + 1 = 0$ , which is impossible in $R$ .

2.1

Steps 1.1 and 1.2 provide a case where a complex eigenvector is created with no accompanying real eigenvector, contradicting the claimed implication.

step 1.1step 1.2∎

Depends on

The real quarter-turn diagonalises after complexification but has no real eigenvector

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

Keith Conrad, Complexification (notes) (standard reference, not scraped)