CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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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The commutant of a cyclic endomorphism consists of its polynomials

Statement

Let $T$ be a cyclic endomorphism of a finite-dimensional vector space. An endomorphism $S$ commutes with $T$ if and only if $S = p (T)$ for some $p \in F [x]$ . Thus the commutant of $T$ is $F [T]$ .

Facts & Assumptions

Given: A cyclic endomorphism $T$ with cyclic vector $v$ and an endomorphism $S$ .

[L1]

Cyclicity means $Z (v; T) = V$ (Cyclic subspaces, cyclic vectors, and vector annihilators).

[L2]

The powers $v, T v, \dots, T^{d - 1} v$ form a basis of $Z (v; T)$ , where $d = \deg m_{T, v}$ (A vector annihilator gives a power basis and its companion matrix).

[L3]

Polynomial evaluation is $p (T) = \sum_{k} a_{k} T^{k}$ (Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ ).

Proof

technique · direct

1.1L1L2choose

Suppose $S T = T S$ . By [L1], $S v \in V = Z (v; T)$ , so [L2] supplies $p \in F [x]$ with $S v = p (T) v$ .

2.1step 1.1L2L3algebra

For every $k \geq 0$ , commutation gives $S (T^{k} v) = T^{k} S v = T^{k} p (T) v = p (T) T^{k} v$ ; the power basis in [L2] therefore makes $S = p (T)$ on all of $V$ .

3.1step 2.1L3∎

Conversely every $p (T)$ commutes with $T$ term by term in [L3]. This proves both directions, including the zero space where the only endomorphism is the zero map.

Depends on

Used by

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Dependency tree · two levels

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Sources

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