Leibniz formula for determinants


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

:egin{align}F(A)& = Fleft(sum_{k_1 = 1}^n a_{k_1}^1 (E^{k_1}, A^2, dots, A^n) ight)\& = 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 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

:egin{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)}^iend{align}

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.


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