Determinant of transpose

An important fact in linear algebra is that, given a matrix A, \det A = \det {}^tA, where {}^tA is the transpose of A. Here I will prove this statement via explciit computation, and I will try to do this as cleanly as possible. We may define the determinant of A by

\det A = \sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} A_{\sigma(1), 1} \cdots A_{\sigma(n), n}.

Here S_n is the set of permutations of the set \{ 1, \dots n \}, and sgn(\sigma) is the sign of the permutation \sigma. This formula is derived from the definition of the determinant via exterior algebra. One can check by hand that this gives the familiar expressions for the determinant when n = 2, 3.

Now, since ({}^tA)_{i, j} = A_{j, i}, we have

\det {}^tA = \sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} A_{1, \sigma(1)} \cdots A_{n, \sigma(n)}

= \sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} A_{\sigma^{-1}(\sigma(1)), \sigma(1)} \cdots A_{\sigma^{-1}(\sigma(n)), \sigma(n)}.

The crucial observation here is that we may rearrange the product inside the summation so that the second indices are increasing. Let b = \sigma(a). Then the product inside the summation is

\prod_{1 \leq a \leq n} A_{\sigma^{-1}(\sigma(a)), \sigma(a)} = \prod_{1 \leq b \leq n} A_{\sigma^{-1}(b), b}

Combining this with the fact that sgn(\sigma) = sgn(\sigma^{-1}), our expression simplifies to

\sum_{\sigma \in S_n} (-1)^{sgn(\sigma^{-1})} A_{\sigma^{-1}(1), 1} \cdots A_{\sigma^{-1}(n), n}.

Noticing that the sum is the same sum if we replace all \sigma^{-1}s with \sigmas, we see that this equals \det A. So \det A = \det {}^tA. \square

I wonder if there is a more conceptual proof of this? (By “conceptual”, I mean a proof based on exterior algebra, bilinear pairings, etc…)