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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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 (x−λ)e is similar to the Jordan block Je(λ)

Statement

For e≥1 and λ∈F, C((x−λ)e) is similar to Je(λ).

Facts & Assumptions

[F1]

For m≥1 and λ∈F, 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
1.1algebra

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

2.1step 1.1F1

Multiplication by x=λ+(x−λ) 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).

3.1step 2.1given∎

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

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