Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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=(1101) and the family {I2,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
1.1algebra

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

2.1L1L2algebra∎

Since (J−I)2=0 but J−I≠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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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