Determinant, same for transpose?

Question

$$\det(A) = \det(A^\top)$$

Wiki confirms this. How can I see this this is true? I tested out a few examples and it seems to hold, however unsure the exact reasoning.

For instance:

$$\begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = ad - bc$$

Vectors $(a,b)$ and $(c,d)$ define the parallelogram, but how would the transpose define a similar shape with same area?

What is the transpose of $\left[\begin{smallmatrix} a & b \ c & d \end{smallmatrix}\right]$, and what does the determinant formula say about its determinant? — Arthur, Jan 10 '21 at 22:42

score 1 · Answer 1 · answered Jan 10 '21 at 22:44

Use the Leibniz definition of determinants, that $$\det(A) = \sum\limits_{\sigma\in S_n}\text{sgn}(\sigma)\prod\limits_{i=1}^na_{\sigma(i),i}$$

Recognize that the transpose of a permutation matrix is simply the inverse of the permutation matrix and so has the same sign.

Your sum then consists of exactly the same terms before and after the transposition with the same signs as originally had, just perhaps appearing in a slightly different order.

Determinant, same for transpose?

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