False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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FALSE: Every normal operator is diagonalisable over its base field

Statement

Every normal operator is diagonalisable over its base field.

Facts & Assumptions

Given: The quarter-turn matrix $A = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$ on $R^{2}$ .

[L1]

The real quarter-turn is normal and has complex eigenvalues $\pm i$ (The real quarter-turn is normal and appears as a single 2x2 block in the real normal classification).

Refutation

technique · direct

1.1

By [L1], the quarter-turn $A$ is a normal real operator.

2.1

The same example [L1] shows that $A$ has no real eigenvalue, so it cannot have a real eigenbasis and is not diagonalisable over $R$ . This refutes the field-free claim.

L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Nicholas Hu, The Schur decomposition (standard reference, not scraped)