Prove that $\det A = \det A^t$

Question

I am trying to use the formula $$\det A = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)a_{1,\sigma(1)}\cdots a_{n, \sigma(n)}$$

Is this correct? $$\det(A^t)= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)a_{\sigma(1), 1} \cdots a_{\sigma(n),n}$$

If it is, why?

Also, then how can I proceed to prove that the two sums are equal?

Notes: $S_n$ is a symmetric group, and $\sigma$ is a permutation.

Answer 1

Hint $$a_{\sigma(1), 1} \cdots a_{\sigma(n),n}=a_{1,\sigma^{-1}(1)}\cdots a_{n,\sigma^{-1}(n)}$$

Answer 2

\begin{eqnarray} \det A &=& \sum_\sigma \operatorname{sgn} \sigma A_{1,\sigma(1)} \cdots A_{n,\sigma(n)} \\ &=& \sum_\sigma \operatorname{sgn} \sigma A_{\sigma^{-1}(1),1} \cdots A_{\sigma^{-1}(n),n} \\ &=& \sum_\sigma \operatorname{sgn} \sigma^{-1} A_{\sigma^{-1}(1),1} \cdots A_{\sigma^{-1}(n),n} \\ &=& \sum_\sigma \operatorname{sgn} \sigma A_{\sigma(1),1} \cdots A_{\sigma(n),n} \\ &=& \det A^T \end{eqnarray}

The first step follows because $\sigma$ is a permutation, so we can rearrange the $A_{1,\sigma(1)} \cdots A_{n,\sigma(n)}$ so the second index is in the order $1,...,n$.

The second step follows because $\operatorname{sgn} \sigma = \operatorname{sgn} \sigma^{-1}$ (which follows because $\sigma \circ \sigma^{-1}$ is the identity, and $\operatorname{sgn} (ab) = \operatorname{sgn}a \operatorname{sgn} b$.

The third step follows because $\{ \sigma | \sigma \text{ is a permutation}\} = \{ \sigma^{-1} |\text{ is a permutation}\}$.