ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

How statement and proof provenance work

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Computing commuting diagonal and nilpotent parts of a split Jordan matrix

Example

Over any field, put $T = J_{2} (0) \oplus J_{2} (1)$ . Then $D = 0 I_{2} \oplus I_{2} = 3 T^{2} - 2 T^{3}, N = T - D = J_{2} (0) \oplus J_{2} (0) .$ Thus $T = D + N$ , the operator $D$ is diagonal, $N^{2} = 0$ , and $D N = N D$ .

Facts & Assumptions

Given: The displayed block diagonal operator $T$ .

[L1]

For a split minimal polynomial, the primary projections are polynomials in the endomorphism (Each projection in the primary decomposition is a polynomial in the endomorphism).

[L2]

On each generalised eigenspace, the operator is the eigenvalue scalar plus a nilpotent operator (Split characteristic polynomials decompose into generalised eigenspaces of the algebraic multiplicities).

Verification

technique · computation

1.1L1algebra

For $h (x) = 3 x^{2} - 2 x^{3}$ , one has $h (0) = h^{'} (0) = 0$ and $h (1) = 1$ , $h^{'} (1) = 0$ ; evaluating on the two size-two blocks therefore gives $h (J_{2} (0)) = 0$ and $h (J_{2} (1)) = I_{2}$ . These identities remain valid in characteristics $2$ and $3$ by direct reduction.

2.1step 1.1L2

Hence $D = h (T)$ is the scalar part on the two generalised eigenspaces, while $N = T - D$ is the direct sum of their nilpotent parts as in [L2].

3.1step 2.1algebra∎

The displayed blocks give $N^{2} = 0$ and $D N = N D$ ; commutation also follows because both $D$ and $N = T - h (T)$ are polynomials in $T$ .

Depends on

Used by

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Dependency tree · two levels

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