TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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A vector annihilator gives a power basis and its companion matrix

Statement

Let $m_{T, v} = x^{d} + a_{d - 1} x^{d - 1} + \dots + a_{0}$ . Then $(v, T v, \dots, T^{d - 1} v)$ is an ordered basis of $Z (v; T)$ . In this basis the restriction $T ∣_{Z (v; T)}$ has the companion matrix with ones on the subdiagonal and last column $(- a_{0}, - a_{1}, \dots, - a_{d - 1})^{T}$ . If $d = 0$ , then $v = 0$ , $Z (v; T) = 0$ , and both the basis and matrix are empty.

Facts & Assumptions

Given: An endomorphism $T$ , a vector $v$ , and its monic vector annihilator $m = m_{T, v}$ of degree $d$ .

[L1]

For an endomorphism of a finite-dimensional vector space, ${Ann}_{T} (v)$ has a unique monic generator $m_{T, v}$ and $p (T) v = 0$ exactly when $m_{T, v} ∣ p$ (The vector annihilator is the unique monic generator of ${Ann}_{T} (v)$ and divides the minimal polynomial).

[L2]

Division by monic $m$ writes each $p \in F [x]$ uniquely as $p = q m + r$ with $\deg r < d$ or $r = 0$ (Division algorithm for polynomials over a field).

[L3]

The cyclic subspace is $Z (v; T) = {p (T) v : p \in F [x]}$ (Cyclic subspaces, cyclic vectors, and vector annihilators).

[L4]

Matrix columns are the coordinates of the images of ordered basis vectors (Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

Proof

technique · direct

1.1L1L2L3

By [L2], $p (T) v = r (T) v$ because $m (T) v = 0$ ; [L3] therefore shows that $v, T v, \dots, T^{d - 1} v$ span $Z (v; T)$ .

1.2L1algebra

A linear relation among those powers gives a polynomial $r$ of degree below $d$ with $r (T) v = 0$ ; [L1] says $m ∣ r$ , so $r = 0$ . Thus the list is independent and hence a basis.

1.3L1L4algebra

The first $d - 1$ basis vectors shift to the next ones, while $m (T) v = 0$ gives $T^{d} v = - a_{0} v - \dots - a_{d - 1} T^{d - 1} v$ ; [L4] yields the stated companion matrix.

2.1L1L3∎

If $d = 0$ , monicity makes $m = 1$ , so [L1] gives $v = 0$ and [L3] gives the zero cyclic subspace; the empty basis and matrix then establish the endpoint case.

Depends on

Used by

Dependency tree · two levels

16 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.

Sources

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