Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Similar matrices over a commutative ring have the same determinant

Statement

Let n≥1n\ge1. If A,B∈Mn(R)A,B\in M_n(R) are similar over a commutative ring, then det⁡(B)=det⁡(A)\det(B)=\det(A).

Facts & Assumptions

Given: An invertible PP with B=P−1APB=P^{-1}AP.

[L1]

Similarity over RR means B=P−1APB=P^{-1}AP for an invertible PP (Invertible square matrices and similarity over a commutative ring).

[L2]

Determinants multiply over a commutative ring (For same-sized finite square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B)\det(AB)=\det(A)\det(B)).

[L3]

det⁡(P−1)=det⁡(P)−1\det(P^{-1})=\det(P)^{-1} (If AA is invertible over a commutative ring, then det⁡(A−1)=det⁡(A)−1\det(A^{-1})=\det(A)^{-1}).

[L4]

Matrix multiplication is associative (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products).

Proof

technique · direct
1.1

By [L1], [L2] and associativity, det⁡(B)=det⁡(P−1)det⁡(A)det⁡(P)\det(B)=\det(P^{-1})\det(A)\det(P).

L1L2L3L4
2.1

Substitute [L3] and commute the scalar factors in RR: det⁡(P)−1det⁡(P)=1\det(P)^{-1}\det(P)=1, leaving det⁡(B)=det⁡(A)\det(B)=\det(A).

step 1.1L3algebra∎

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