CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Similar matrices over a commutative ring have the same determinant

Statement

Let $n \geq 1$ . If $A, B \in M_{n} (R)$ are similar over a commutative ring, then $\det (B) = \det (A)$ .

Facts & Assumptions

Given: An invertible $P$ with $B = P^{- 1} A P$ .

[L1]

Similarity over $R$ means $B = P^{- 1} A P$ for an invertible $P$ (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, $\det (A B) = \det (A) \det (B)$ ).

[L3]

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

[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, $\det (B) = \det (P^{- 1}) \det (A) \det (P)$ .

L1L2L3L4

2.1

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

step 1.1L3algebra∎

Depends on

Used by

Over a commutative ring, adj(P⁻¹AP)=P⁻¹adj(A)P for every invertible P Theorem

Dependency tree · two levels

15 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

S. New, MATH 146 Linear Algebra 1 Lecture Notes, Ch. 4 (standard reference, not scraped)