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

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The minimal polynomial is unchanged by choosing a matrix representation or replacing a matrix by a similar one

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional vector space and let $A = [T]_{B}^{B}$ in an ordered basis $B$ . Then $T$ and $A$ have the same minimal polynomial. More generally, similar square matrices have the same minimal polynomial.

Facts & Assumptions

Given: An endomorphism $T$ , an ordered basis $B$ , its matrix $A = [T]_{B}^{B}$ , and square matrices $A, B$ with $B = P^{- 1} A P$ .

[L1]

The matrix of a composite is the product of the matrices in compatible ordered bases ( $[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C}$ ).

[L2]

Similar matrices are exactly matrix representations of one endomorphism in different ordered bases (Similarity is an equivalence relation, and two matrices represent the same endomorphism in two bases exactly when they are similar).

[L3]

The minimal polynomial is the unique monic generator of the annihilator ideal (The annihilator ideal is nonzero and has a unique monic generator; $p (T) = 0$ if and only if $μ_{T} ∣ p$ ).

[L4]

The matrix of a linear map has as its columns the coordinate columns of the images of the domain basis vectors (Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

Proof

technique · direct

1.1L1L4algebra

By induction on $k$ , [L1] gives $[T^{k}]_{B}^{B} = A^{k}$ for every $k \geq 0$ ; taking the same finite linear combination on both sides yields $[p (T)]_{B}^{B} = p (A)$ for every $p \in F [x]$ .

2.1step 1.1L3L4

A linear map is zero exactly when its matrix in a basis is zero, so step 1.1 and [L4] give $p (T) = 0$ if and only if $p (A) = 0$ . The annihilator ideals coincide, hence their unique monic generators coincide by [L3].

3.1L2L3algebra∎

If $B = P^{- 1} A P$ , induction gives $B^{k} = P^{- 1} A^{k} P$ , and therefore $p (B) = P^{- 1} p (A) P$ . Thus $p (B) = 0$ exactly when $p (A) = 0$ , so [L3] again gives $μ_{B} = μ_{A}$ . This also follows from [L2].

Depends on

Used by

Dependency tree · two levels

23 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

Keith Conrad, The Minimal Polynomial and Some Applications, §4, Theorem 4.3 (standard reference, not scraped)