CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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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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Commuting alone does not imply simultaneous diagonalisability

Statement refuted

Pairwise commutation alone forces a family of endomorphisms to be simultaneously diagonalisable.

Facts & Assumptions

Given: Over any field, $J = (\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array})$ and the family ${I_{2}, J}$ .

[L1]

A family of diagonalisable endomorphisms is simultaneously diagonalisable exactly when it is pairwise commuting (A family of diagonalisable endomorphisms of a finite-dimensional space is simultaneously diagonalisable if and only if its members commute pairwise).

[L2]

A repeated linear factor in the minimal polynomial prevents diagonalisability (An endomorphism is diagonalisable if and only if its minimal polynomial is a product of distinct linear factors).

Counterexample

technique · direct

1.1

The identity commutes with $J$ , so the family is pairwise commuting.

algebra

2.1

Since $(J - I)^{2} = 0$ but $J - I \neq 0$ , the minimal polynomial of $J$ is $(x - 1)^{2}$ . Thus [L2] says $J$ is not diagonalisable, so the family cannot be simultaneously diagonalisable. The missing hypothesis in [L1] is diagonalisability of every member.

L1L2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 49 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Keith Conrad, The Minimal Polynomial and Some Applications, Remark 5.3 (standard reference, not scraped)