Leibniz formula for determinants

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In algebra, the Leibniz formula expresses the determinant of a square matrix in terms of permutations of the matrix' elements. Named in honor of Gottfried Leibniz, the formula is

$\det(A) = \sum_{\sigma \in S_n} \sgn(\sigma) \prod_{i = 1}^n a_{\sigma(i)}^i$

for an n×n matrix, where sgn is the sign function of permutations in the permutation group S_n, which returns +1 and –1 for even and odd permutations, respectively.

Another common notation used for the formula is in terms of the Levi-Civita symbol and makes use of the Einstein summation notation, where it becomes

$\det(A)=\epsilon^{i_1\cdots i_n}{A^{1}}_{i_1}\cdots {A^{n}}_{i_n},$

which may be more familiar to physicists.

In the sequel, a proof of the equivalence of this formula to the conventional definition of the determinant in terms of expansion by minors is given.

Theorem. There exists exactly one function

$F : \mathfrak M_n (\mathbb K) \longrightarrow \mathbb K$

which is alternate multilinear w.r.t. columns and such that $F (I) = 1$ .

Proof.

Let $F$ be such a function, and let $A = (a_i^j)_{i = 1, \dots, n}^{j = 1, \dots , n}$ be an $n \times n$ matrix. Call $A j$ the $j$ -th column of $A$ , i.e. $A^j = (a_i^j)_{i = 1, \dots , n}$ , so that $A = \left(A^1, \dots, A^n\right)$ .

Also, let $E k$ denote the $k$ -th column vector of the identity matrix.

Now one writes each of the $A j$ 's in terms of the $E k$ , i.e.

$A^j = \sum_{k = 1}^n a_k^j E^k$ .

As $F$ is multilinear, one has

$\begin{align} F(A)& = F\left(\sum_{k_1 = 1}^n a_{k_1}^1 (E^{k_1}, A^2, \dots, A^n)\right)\\ & = \cdots\\ & = \sum_{k_1, \dots, k_n = 1}^n \prod_{i = 1}^n a_{k_i}^i F((E^{k_1}, \dots, E^{k_n})). \end{align}$

As the above sum takes into account all the possible choices of ordered $n$ -tuples $\left(k_1, \dots , k_n\right)$ , it can be expressed in terms of permutations as

$\sum_{\sigma \in \mathfrak S_n} \prod_{i = 1}^n a_{\sigma(i)}^i F((E^{\sigma(1)}, \dots , E^{\sigma(n)})).$

Now one rearranges the columns of $\left(E^{\sigma(1)}, \dots, E^{\sigma(n)}\right)$ so that it becomes the identity matrix; the number of columns that need to be exchanged is exactly $sgn(σ)$ . Hence, thanks to alternance, one finally gets

$\begin{align} F(A)& = \sum_{\sigma \in \mathfrak S_n} \sgn(\sigma) \prod_{i = 1}^n a_{\sigma(i)}^i F(I)\\ & = \sum_{\sigma \in \mathfrak S_n} \sgn(\sigma) \prod_{i = 1}^n a_{\sigma(i)}^i \end{align}$