Determinant of transpose | Arithmetic variety

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 $\sigma$ s, 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…)

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