Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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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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If two commuting endomorphisms are diagonalisable, then every finite linear combination of products of their powers is diagonalisable; in particular, their sum and product are diagonalisable

Statement

Let S,T be commuting diagonalisable endomorphisms. Every endomorphism of the form

j<mcjSajTbj

with m,aj,bjN and cjF is diagonalisable. In particular, S+T and ST are diagonalisable.

Facts & Assumptions

Given: Commuting diagonalisable endomorphisms S,T and the displayed finite polynomial expression.

Proof

technique · direct
1.1

By [L1], choose a basis in which both S and T are diagonal. Every power, product of powers, and finite linear combination in the Statement remains diagonal in that same basis.

L1algebra
2.1

Thus every displayed expression is diagonalisable. Taking the expressions S+T and ST gives the stated special cases; the empty sum and zero space cause no exception.

step 1.1

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: 32 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