Determinant of transpose intuitive proof

Question

We are using Artin's Algebra book for our Linear Algebra course. In Artin, det(A^T) = det(A) is proved using elementary matrices and invertibility. All of us feel that there should be a 'deeper' or a more fundamental or a more intuitive proof without using elementary matrices or invertibility. The one our prof came up with used linear transformations between tensor algebras, wedges and exterior algebras which we do not understand. Are there any other proofs for det(A^T) = det(A) ? Edit: also, is there a geometric proof? For the 2*2 case at least?

Answer 1

The determinant of a matrix does not change when you compute it via cofactor expansion along column or row. Thus expanding along a row in $A$ is equivalent to expanding along a column in $A^t$. I'm not sure if this is what you meant by "using invertibility".

Answer 2

If $A$ is diagonalizable, so that there exists an invertible matrix $C$ such that $D=CAC^{-1}$ is diagonal, then $D=D^t=(CAC^{-1})^t=C^{-t}A^tC^t$. Since $D$ and $D^t$ have the same determinant, simply because the two matrices are in fact equal, it follows at once from this that $A$ and $A$ and $A^t$ have the same determinant.

As diagonalizable matrices are dense in $M_n(\mathbb C)$, and the map $A\in M_n(\mathbb C)\mapsto \det(A)-\det(A^t)\in\mathbb C$ is continuous, we see at once that this function is in fact constant. What you want follows from this.