TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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

Statement

An endomorphism $T$ of a finite-dimensional vector space has a cyclic vector if and only if $μ_{T} = χ_{T}$ . On the zero space, $v = 0$ is cyclic and $μ_{T} = χ_{T} = 1$ .

Facts & Assumptions

Given: An endomorphism $T$ of an $n$ -dimensional vector space $V$ .

[L1]

If $m_{T, v}$ has degree $d$ , then $v, T v, \dots, T^{d - 1} v$ is a basis of $Z (v; T)$ (A vector annihilator gives a power basis and its companion matrix).

[L2]

Some vector $v$ satisfies $m_{T, v} = μ_{T}$ (Some vector has vector annihilator equal to the minimal polynomial).

[L3]

The characteristic polynomial is monic of degree $n$ , and is $1$ for $n = 0$ ( $χ_{A} (x)$ is monic of degree $n$ ; for $n \geq 1$ its $x^{n - 1}$ coefficient is $- tr (A)$ and its constant coefficient is $(- 1)^{n} \det (A)$ , while $χ_{0 \times 0} = 1$ ).

[L4]

A polynomial annihilates $T$ exactly when it is divisible by $μ_{T}$ (The annihilator ideal is nonzero and has a unique monic generator; $p (T) = 0$ if and only if $μ_{T} ∣ p$ ), and $χ_{T} (T) = 0$ (Cayley-Hamilton: every finite-dimensional endomorphism satisfies its characteristic polynomial, $χ_{T} (T) = 0$ ).

[L5]

A finite-dimensional subspace with the same dimension as its ambient space equals that space (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

Proof

technique · direct

1.1

Suppose $v$ is cyclic. A polynomial annihilates $v$ exactly when it annihilates every $q (T) v$ , because polynomial evaluations commute and these vectors span $V$ ; hence $m_{T, v} = μ_{T}$ . By [L1], $\deg μ_{T} = \dim Z (v; T) = n$ .

L1L4algebra

1.2

Conversely, suppose $μ_{T} = χ_{T}$ and choose $v$ as in [L2]. Then [L1] gives $\dim Z (v; T) = \deg m_{T, v} = \deg χ_{T} = n$ ; since $Z (v; T)$ is a subspace of $V$ , [L5] gives $Z (v; T) = V$ , so $v$ is cyclic.

L1L2L3L5choose

2.1

Fact [L4] gives $μ_{T} ∣ χ_{T}$ , while [L3] makes both monic and step 1.1 gives equal degree; therefore $μ_{T} = χ_{T}$ .

step 1.1L3L4

3.1

When $V = 0$ , [L2] chooses $v = 0$ , [L3] and [L4] give both polynomials as $1$ , and its cyclic subspace is $V$ ; thus steps 1.1-2.1 cover every case and both directions.

step 1.1step 2.1step 1.2L2L3L4∎

Depends on

Used by

The identity on F² has no cyclic vector Counterexample
A companion operator with a visible cyclic vector and equal canonical polynomials Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 120 results over 24 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 (standard reference, not scraped)