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.

A split $3 \times 3$ operator that is triangularisable but not diagonalisable

Example

Over any field, let $A = J_{2} (0) \oplus J_{1} (0) = (\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}) .$ Then $χ_{A} = x^{3}$ and $μ_{A} = x^{2}$ . Hence $A$ is triangularisable over its base field but is not diagonalisable.

Facts & Assumptions

Given: The displayed matrix $A$ .

[L1]

An endomorphism is triangularisable exactly when its characteristic polynomial splits over the base field ( $T$ is triangularisable iff its minimal polynomial splits iff its characteristic polynomial splits).

[L2]

An endomorphism is diagonalisable exactly when its minimal polynomial is a product of distinct linear factors (An endomorphism is diagonalisable if and only if its minimal polynomial is a product of distinct linear factors).

Verification

technique · computation

1.1algebra

The matrix is strictly upper triangular, so direct determinant computation gives $χ_{A} = x^{3}$ ; also $A \neq 0$ but $A^{2} = 0$ , so $μ_{A} = x^{2}$ .

2.1step 1.1L1L2∎

The polynomial $x^{3}$ splits, so [L1] confirms triangularisability, while the repeated factor in $x^{2}$ makes [L2] rule out diagonalisability.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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.

A split 3×33×3 operator that is triangularisable but not diagonalisable