TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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For every square matrix over a commutative ring, $\det(A^{\mathsf T})=\det(A)$

Statement

For every $n\ge1$ , every commutative ring $R$ , and every $A\in M_n(R)$ , $\det(A^{\mathsf T})=\det(A).$

Facts & Assumptions

Given: A square matrix $A=(a_{ij})$ over a commutative ring.

[L1]

Determinant is the Leibniz sum over permutations (For $n\ge1$ , the determinant over a commutative ring by the Leibniz formula, and $|\det A|$ for a real matrix).

[L2]

Transpose satisfies $(A^{\mathsf T})_{ij}=a_{ji}$ (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[L3]

Sign is a homomorphism into $\{1,-1\}$ , so $\operatorname{sgn}(\sigma^{-1})=\operatorname{sgn}(\sigma)$ (The sign is a homomorphism $S_n\to\{+1,-1\}$ , surjective exactly when $n\ge 2$ ).

[L4]

A finite sum may be reindexed by a bijection (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

Proof

technique · direct

1.1

Substituting [L2] into [L1] gives $\det(A^{\mathsf T})=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i<n}a_{i,\sigma(i)}$ .

L1L2

2.1

Reindex the sum by $\tau=\sigma^{-1}$ and the product by $j=\sigma(i)$ . Commutativity and [L3] turn the expression into $\sum_{\tau\in S_n}\operatorname{sgn}(\tau)\prod_{j<n}a_{\tau(j),j}=\det(A)$ .

step 1.1L3L4algebra∎

Depends on

Used by

The determinant is alternating and multilinear in the rows as well as in the columns Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 83 results over 15 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

S. New, MATH 146 Linear Algebra 1 Lecture Notes, Theorem 4.23 (standard reference, not scraped)
P. Massot, Structures algébriques fondamentales, Proposition 6.4.3 (standard reference, not scraped)

For every square matrix over a commutative ring, det⁡(AT)=det⁡(A)\det(A^{\mathsf T})=\det(A)det(AT)=det(A)