TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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

1.1

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 $p \in F [x]$ .

L1L4algebra

2.1

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].

step 1.1L3L4

3.1

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].

L2L3algebra∎

Depends on

Used by

The minimal and characteristic polynomials have exactly the same monic irreducible factors Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 65 results over 18 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

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