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Every alternating multilinear $F$ satisfies $F(A)=F(I)\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_i a_{\sigma(i),i}$

Statement

Let $n\ge1$ , let $R$ be a commutative ring, and let $F:M_n(R)\to R$ be alternating and column-multilinear. Then for $A=(a_{ri})$ , $F(A)=F(I_n)\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i<n}a_{\sigma(i),i}.$

Facts & Assumptions

Given: A matrix $A=(a_{ri})\in M_n(R)$ and an alternating column-multilinear function $F:M_n(R)\to R$ .

[L1]

Column multilinearity expands $F$ in each column (Column-multilinear, alternating, normalized and antisymmetric functions on square matrices over a commutative ring).

[L2]

Alternating multilinear functions are antisymmetric under swaps (Every alternating multilinear matrix function is antisymmetric under a column swap).

[L3]

The sign of a permutation is $(-1)^{\operatorname{inv}(\sigma)}$ (Inversions, inversion number, the sign $\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}$ , and even and odd permutations).

[L4]

Sign is a homomorphism on $S_n$ (The sign is a homomorphism $S_n\to\{+1,-1\}$ , surjective exactly when $n\ge 2$ ).

[L5]

Finite sums and products over finite index sets are defined in commutative monoids (A finite sum in a commutative monoid indexed by an arbitrary finite set, The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

[L6]

Every finite permutation is a product of transpositions (Every finite permutation is a product of transpositions, so the transpositions generate $S_n$ ).

[L7]

Every transposition factorisation of a permutation has parity prescribed by its inversion sign (Every transposition factorisation of $\sigma$ has parity $(-1)^{\operatorname{inv}(\sigma)}$ ).

Proof

technique · direct

1.1

For $r<n$ , let $e_r$ be the column with entry $1$ in row $r$ and $0$ elsewhere. Since column $i$ is $\sum_{r<n}a_{ri}e_r$ , repeated multilinearity expands $F(A)$ as the finite sum over tuples $(r_i)_{i<n}$ of $\bigl(\prod_{i<n}a_{r_i i}\bigr)F(e_{r_0},\ldots,e_{r_{n-1}})$ .

L1L5algebra

2.1

If a tuple repeats a row index, its $F$ -value is zero by alternation. The surviving tuples use every element of $n$ once, so they are precisely the permutations $r_i=\sigma(i)$ .

step 1.1L1

3.1

Factor $\sigma$ into transpositions. Repeated antisymmetry changes $F(I_n)$ by one minus sign per transposition. Each transposition has sign $-1$ , so [L4] identifies the product of these signs with $\operatorname{sgn}(\sigma)$ ; equivalently [L3] and [L7] identify it with the factorisation-independent inversion parity. Substitution into step 2.1 gives the formula.

step 2.1L2L3L4L6L7algebra∎

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Sources

S. New, MATH 146 Linear Algebra 1 Lecture Notes, Theorem 4.20 (standard reference, not scraped)
P. Massot, Structures algébriques fondamentales, Definition 6.4.1 (standard reference, not scraped)

Every alternating multilinear FFF satisfies F(A)=F(I)∑σ∈Snsgn⁡(σ)∏iaσ(i),iF(A)=F(I)\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_i a_{\sigma(i),i}F(A)=F(I)∑σ∈Sn​​sgn(σ)∏i​aσ(i),i​