An important fact in linear algebra is that, given a matrix , , where is the transpose of . 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 by

Here is the set of permutations of the set , and is the sign of the permutation . 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 .

Now, since , we have

The crucial observation here is that we may rearrange the product inside the summation so that the second indices are increasing. Let . Then the product inside the summation is

Combining this with the fact that , our expression simplifies to

Noticing that the sum is the same sum if we replace all s with s, we see that this equals . So .

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