CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-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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The identity on $F^{2}$ has no cyclic vector

Statement refuted

Every endomorphism of a finite-dimensional vector space has a cyclic vector.

Facts & Assumptions

Given: The identity endomorphism $I$ of $F^{2}$ .

[L1]

An endomorphism has a cyclic vector exactly when its minimal and characteristic polynomials are equal (A cyclic vector exists exactly when the minimal and characteristic polynomials agree).

Counterexample

technique · counterexample

1.1

For any $v \in F^{2}$ , every power $I^{k} v$ equals $v$ , so $Z (v; I) = F v$ has dimension at most one and cannot equal $F^{2}$ .

algebra

2.1

Equivalently, $μ_{I} = x - 1$ while $χ_{I} = (x - 1)^{2}$ , so [L1] also rules out a cyclic vector. Thus the identity on $F^{2}$ refutes the universal claim.

L1algebra∎

Depends on

A cyclic vector exists exactly when the minimal and characteristic polynomials agree

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 46 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

K. Hoffman and R. Kunze, Linear Algebra, 2nd ed., Section 7.1, Example 1 (standard reference, not scraped)

The identity on F2F2 has no cyclic vector