Permutation of Determinant Indices

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

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

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

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

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

Let $I: \N_{> 0} \to \N_{> 0}$ be the identity mapping on $\N_{> 0}$.

By Identity Mapping is Permutation, we have that $I$ is a permutation.

Let $\displaystyle \sum_\nu$ goes over all the $n!$ permutations of $\left\{{1, 2, \ldots, n}\right\}$.

From the definition of the determinant, we have:


 * $\displaystyle \det \left({\mathbf A}\right) = \sum_{\nu} \left({\operatorname{sgn} \left({I}\right) \operatorname{sgn} \left({\nu}\right) \prod_{k \mathop = 1}^n a_{k \nu \left({k}\right)}}\right)$

as $\operatorname{sgn} \left({I}\right) = 1$.

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

Then:
 * $\displaystyle \prod_{k \mathop = 1}^n a_{\lambda \left({k}\right) \mu \left({k}\right)} = a_{\lambda \left({1}\right) \mu \left({1}\right)} a_{\lambda \left({2}\right) \mu \left({2}\right)} \cdots a_{\lambda \left({n}\right) \mu \left({n}\right)} = a_{1 \nu \left({1}\right)} a_{2 \nu \left({2}\right)} \cdots a_{n \nu \left({n}\right)} = \prod_{k \mathop = 1}^n a_{k \nu \left({k}\right)}$

as multiplication is commutative.

The result follows from Parity Function is Homomorphism, that is:
 * $\operatorname{sgn} \left({\lambda}\right) \operatorname{sgn} \left({\nu}\right) = \operatorname{sgn} \left({\lambda \nu}\right)$

Next we show:
 * $\displaystyle \det \left({\mathbf A}\right) = \sum_\mu \left({\operatorname{sgn} \left({\mu}\right) \operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a_{\mu \left({k}\right) \lambda \left({k}\right)}}\right)$

Again, we start with:
 * $\displaystyle \det \left({\mathbf A}\right) = \sum_\nu \left({\operatorname{sgn} \left({I}\right) \operatorname{sgn} \left({\nu}\right) \prod_{k \mathop = 1}^n a_{k \nu \left({k}\right)}}\right)$

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

The result follows via a similar argument.