Leibniz formula for determinants

In algebra, the Leibniz formula expresses the determinant of a square matrix $A\; =\; (a\_\{ij\})\_\{i,j\; =\; 1,\; dots,\; n\}$ in terms of permutations of the matrix' elements. Named in honor of Gottfried Leibniz, the formula is

:$det(A)\; =\; sum\_\{sigma\; in\; S\_n\}\; sgn(sigma)\; prod\_\{i\; =\; 1\}^n\; a\_\{i,sigma(i)\}$

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:$det(A)=epsilon^\{i\_1cdots\; i\_n\}\{A\}\_\{1i\_1\}cdots\; \{A\}\_\{ni\_n\},$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 $n\; imes\; n$ matrix. Call $A^j$ the $j$-th column of $A$, i.e. $A^j\; =\; (a\_i^j)\_\{i\; =\; 1,\; dots\; ,\; n\}$, so that $A\; =\; left(A^1,\; dots,\; A^n\; ight).$

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

:

As the above sum takes into account all the possible choices of ordered $n$-tuples $left(k\_1,\; dots\; ,\; k\_n\; ight)$, 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 $left(E^\{sigma(1)\},\; dots,\; E^\{sigma(n)\}\; ight)$ so that it becomes the identity matrix; the number of columns that need to be exchanged is exactly $sgn(sigma)$. Hence, thanks to alternance, one finally gets

:

as $F(I)$ is required to be equal to $1$.

Hence the determinant can be defined as the only function

:$det\; :\; mathfrak\; M\_n\; (mathbb\; K)\; longrightarrow\; mathbb\; K$

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