Permutation of Determinant Indices

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

Let $$\lambda: \mathbb{N}^* \to \mathbb{N}^*$$ be any fixed permutation on $\mathbb{N}^*$.

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

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

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

Proof

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

Let $$I: \mathbb{N}^* \to \mathbb{N}^*$$ be the identity permutation on $$\mathbb{N}^*$$.

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

From the definition of the determinant, we have:

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

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

Let $$\lambda: \mathbb{N}^* \to \mathbb{N}^*$$ be a permutation on $$\mathbb{N}^*$$ such that $$\lambda\nu = \mu$$.

Then $$\prod_{k=1}^n a_{\lambda \left({k}\right) \mu \left({k}\right)} = a_{\lambda \left({1}\right) \mu \left({1}\right)} a_{\lambda \left({2}\right) \mu \left({2}\right)} \cdots a_{\lambda \left({n}\right) \mu \left({n}\right)} = a_{1 \nu \left({1}\right)} a_{2 \nu \left({2}\right)} \cdots a_{n \nu \left({n}\right)} = \prod_{k=1}^n a_{k \nu \left({k}\right)}$$, as multiplication is commutative.

The result follows from Parity of a Permutation, that is, $$\sgn \left({\lambda}\right) \sgn \left({\nu}\right) = \sgn \left({\lambda \nu}\right)$$.


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

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

Now let $$\mu: \mathbb{N}^* \to \mathbb{N}^*$$ be a permutation on $$\mathbb{N}^*$$ such that $$\mu \nu = \lambda$$.

The result follows via a similar argument.