Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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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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A companion block for (xλ)e is similar to the Jordan block Je(λ)

Statement

For e1 and λF, C((xλ)e) is similar to Je(λ).

Facts & Assumptions

[F1]

For m1 and λF, the Jordan block Jm(λ) has λ on the diagonal, 1 on the superdiagonal, and 0 elsewhere (Jordan blocks, Jordan strings, and their endpoints).

Proof

technique · direct
1.1

In F[x]/((xλ)e), the residue classes 1,(xλ),,(xλ)e1 form a basis. Reversing their order gives the basis (xλ)e1,,xλ,1.

algebra
2.1

Multiplication by x=λ+(xλ) in the reversed basis has λ on the diagonal and sends each basis vector except the first to itself times λ plus the preceding basis vector. Its matrix therefore has ones on the superdiagonal and is Je(λ) by [F1]. For e=1 it is the matrix (λ).

step 1.1F1
3.1

In the ordinary power basis 1,x,,xe1, the same multiplication operator has companion matrix C((xλ)e). The two matrices represent one operator in two bases, so they are similar.

step 2.1given

Depends on

Used by

Dependency tree · two levels

14 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