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

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Some vector has vector annihilator equal to the minimal polynomial

Statement

For every endomorphism $T$ of a finite-dimensional vector space $V$ , there is $v \in V$ such that $m_{T, v} = μ_{T}$ . If $V = 0$ , take $v = 0$ and both polynomials are $1$ .

Facts & Assumptions

Given: A finite-dimensional endomorphism $T$ .

[L1]

If $μ_{T} = \prod_{i < r} q_{i}^{e_{i}}$ is its factorisation into distinct monic irreducible powers, then $V = ⨁_{i < r} V_{i}$ with $V_{i} = \ker q_{i} (T)^{e_{i}}$ , and $T ∣_{V_{i}}$ has minimal polynomial exactly $q_{i}^{e_{i}}$ (Primary decomposition: the irreducible-power factors of $μ_{T}$ split $V$ into their invariant kernels).

[L2]

The vector annihilator is a monic divisor of the operator's minimal polynomial and detects exactly the polynomials that annihilate the vector (The vector annihilator is the unique monic generator of ${Ann}_{T} (v)$ and divides the minimal polynomial).

[L3]

The ring $F [x]$ is a unique factorisation domain (For every field $F$ , $F [x]$ is a unique factorisation domain).

Proof

technique · direct

1.1

For each nonempty primary summand $V_{i}$ in [L1], choose $v_{i} \in V_{i}$ with $q_{i} (T)^{e_{i} - 1} v_{i} \neq 0$ ; such a vector exists because otherwise $q_{i}^{e_{i} - 1}$ would annihilate $T ∣_{V_{i}}$ , contrary to its exact minimal polynomial.

L1choose

2.1

By [L2], $m_{T, v_{i}}$ divides $q_{i}^{e_{i}}$ ; [L3] makes it a power of $q_{i}$ , and step 1.1 rules out every exponent below $e_{i}$ . Hence $m_{T, v_{i}} = q_{i}^{e_{i}}$ .

step 1.1L2L3

3.1

Put $v = \sum_{i < r} v_{i}$ . Directness in [L1] gives $p (T) v = 0$ exactly when $p (T) v_{i} = 0$ for every $i$ , which by step 2.1 is exactly when every $q_{i}^{e_{i}}$ divides $p$ ; pairwise coprimality and [L3] make this equivalent to $μ_{T} ∣ p$ . Thus [L2] gives $m_{T, v} = μ_{T}$ .

step 2.1L1L2L3construct

4.1

If $V = 0$ , [L1] is the empty direct sum and the published minimal-polynomial convention gives $μ_{T} = 1$ ; taking $v = 0$ gives $m_{T, 0} = 1$ by [L2].

L1L2∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 48 results over 10 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)