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

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Building a Jordan-string basis for a nilpotent operator on $F^{6}$

Example

Let $(e_{1}, \dots, e_{6})$ be the standard basis of $F^{6}$ and define $N e_{1} = 0, N e_{2} = e_{1}, N e_{3} = e_{2}, N e_{4} = 0, N e_{5} = e_{4}, N e_{6} = 0.$ Then $(e_{1}, e_{2}, e_{3})$ , $(e_{4}, e_{5})$ , and $(e_{6})$ are Jordan strings whose concatenation is a basis. Thus $[N] = J_{3} (0) \oplus J_{2} (0) \oplus J_{1} (0)$ and $N^{3} = 0 \neq N^{2}$ .

Facts & Assumptions

Given: The displayed endomorphism on the standard basis of $F^{6}$ .

[L1]

A nilpotent Jordan string satisfies $N v_{1} = 0$ and $N v_{j} = v_{j - 1}$ for $j \geq 2$ (Jordan blocks, Jordan strings, and their endpoints).

[L2]

Every finite-dimensional nilpotent endomorphism admits a basis of Jordan strings (Every finite-dimensional nilpotent endomorphism has a basis of Jordan strings).

Verification

technique · computation

1.1

The six displayed images verify [L1] separately for the three listed strings, and their concatenation is the standard basis.

L1algebra

2.1

In that order the matrix has blocks $J_{3} (0)$ , $J_{2} (0)$ , and $J_{1} (0)$ ; direct iteration gives $N^{3} = 0$ and $N^{2} e_{3} = e_{1} \neq 0$ , consistently with [L2].

step 1.1L2algebra∎

Depends on

Used by

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Building a Jordan-string basis for a nilpotent operator on F6F6