Permutation of Determinant Indices

Theorem

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

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

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

Then:

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

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

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

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:

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

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

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

Then:

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

as multiplication is commutative.

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

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

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

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

The result follows via a similar argument.

$\blacksquare$