Permutation of Determinant Indices

Theorem
Let $\mathbf A = \left[{a}\right]_{n}$ be a square matrix of order $n$ over a field.

Let $\lambda: \N_{> 0} \to \N_{> 0}$ be any fixed permutation on $\N_{> 0}$.

Let $\det \left({\mathbf A}\right)$ be the determinant of $\mathbf A$.

Let $\left({S_n, \circ}\right)$ be the symmetric group of $n$ letters.

Then:
 * $\displaystyle \det \left({\mathbf A}\right) = \sum_{\mu\in S_n} \left({ \operatorname{sgn} \left({\mu}\right) \operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a_{\lambda \left({k}\right), \mu \left({k}\right)}}\right)$
 * $\displaystyle \det \left({\mathbf A}\right) = \sum_{\mu\in S_n} \left({ \operatorname{sgn} \left({\mu}\right) \operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a_{\mu \left({k}\right), \lambda \left({k}\right)}}\right)$

where:
 * the summation $\displaystyle \sum_\mu$ goes over all the $n!$ permutations of $\left\{{1, 2, \ldots, n}\right\}$
 * $\operatorname{sgn} \left({\mu}\right)$ is the sign of the permutation $\mu$.

Proof
First it is shown that:
 * $\displaystyle \det \left({\mathbf A}\right) = \sum_\mu \left({\operatorname{sgn} \left({\mu}\right) \operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a_{\lambda \left({k}\right), \mu \left({k}\right)}}\right)$

Let $\nu: \N_{> 0} \to \N_{> 0}$ be a permutation on $\N_{> 0}$ such that $\lambda \circ \nu = \mu$.

The product can be rearranged as:
 * $\displaystyle \prod_{k \mathop = 1}^n a_{\lambda \left({k}\right), \mu \left({k}\right)} = a_{\lambda \left({1}\right), \mu \left({1}\right)} a_{\lambda \left({2}\right), \mu \left({2}\right)} \cdots a_{\lambda \left({n}\right), \mu \left({n}\right)} = a_{1, \nu \left({1}\right)} a_{2, \nu \left({2}\right)} \cdots a_{n, \nu \left({n}\right)} = \prod_{k \mathop = 1}^n a_{k, \nu \left({k}\right)}$

from Field Axiom $M 2$: commutativity of product.

By Parity Function is Homomorphism:
 * $\operatorname{sgn} \left({\mu}\right) \operatorname{sgn} \left({\lambda}\right) = \operatorname{sgn} \left({\lambda}\right) \operatorname{sgn} \left({\nu}\right) \operatorname{sgn} \left({\lambda}\right) = \operatorname{sgn}^2 \left({\lambda}\right) \operatorname{sgn} \left({\nu}\right) = \operatorname{sgn} \left({\nu}\right)$

and so:
 * $\displaystyle \det \left({\mathbf A}\right) = \sum_{\nu \mathop \in S_n} \left({\operatorname{sgn} \left({\nu}\right) \prod_{k \mathop = 1}^n a_{k, \nu \left({k}\right)}}\right)$

which is the usual definition for the determinant.

Next it is to be shown:
 * $\displaystyle \det \left({\mathbf A}\right) = \sum_{\mu \mathop \in S_n} \left({\operatorname{sgn} \left({\mu}\right) \operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a_{\mu \left({k}\right), \lambda \left({k}\right)}}\right)$

Let $\nu: \N_{> 0} \to \N_{> 0}$ be a permutation on $\N_{> 0}$ such that $\mu \circ \nu = \lambda$.

The result follows via a similar argument.