TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Over a commutative ring, $adj (P^{- 1} A P) = P^{- 1} adj (A) P$ for every invertible $P$

Statement

Let $R$ be a commutative ring, $n \geq 1$ , $A, P \in M_{n} (R)$ , and suppose $P$ is invertible. Then

$adj (P^{- 1} A P) = P^{- 1} adj (A) P .$

Facts & Assumptions

Given: $R, n, A, P$ as in the statement, and $C : = P^{- 1} A P$ .

[F1]

Similarity over a commutative ring means $C = P^{- 1} A P$ for an invertible $P$ (Invertible square matrices and similarity over a commutative ring).

[L1]

Similar matrices have equal determinants (Similar matrices over a commutative ring have the same determinant).

[L2]

For columns $u, v$ , $\det (B + u v^{T}) = \det (B) + v^{T} adj (B) u$ (For $A \in M_{n} (R)$ and columns $u, v$ over a commutative ring, $\det (A + u v^{T}) = \det (A) + v^{T} adj (A) u$ ).

[F2]

Matrix products and transposes are given by their entry formulas (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[L3]

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

Proof

technique · direct

1.1

For arbitrary columns $u, v$ , $C + u v^{T} = P^{- 1} (A + (P u) (v^{T} P^{- 1})) P .$

F1L3algebra

2.1

Apply [L1] to step 1.1, then [L2] to both rank-one updates. Since $\det (C) = \det (A)$ , cancellation in the additive group of $R$ gives $v^{T} adj (C) u = v^{T} P^{- 1} adj (A) P u .$

step 1.1L1L2L3

3.1

For each $i, j$ , let $v$ be the column with entry $1$ at $i$ and $0$ elsewhere, and let $u$ be the analogous column at $j$ . The product formula [F2] makes $v^{T} B u = B_{i j}$ for every $B \in M_{n} (R)$ , so step 2.1 says that the $(i, j)$ entries of $adj (C)$ and $P^{- 1} adj (A) P$ are equal.

step 2.1F2

4.1

Equality of all entries proves $adj (C) = P^{- 1} adj (A) P$ , and substituting the definition of $C$ proves the statement.

step 3.1F1∎

Depends on

Used by

The adjugate of an operator on a positive-dimensional finite-dimensional vector space, defined by the adjugate matrix in any basis Definition

Dependency tree · two levels

17 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

Jaroslav Vrabel, A note on the matrix determinant lemma (standard reference, not scraped)
András Pál, Introduction to Commutative Algebra (standard reference, not scraped)

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