TITLE: Every skew-symmetric matrix has a non-negative determinant
QUESTION [6 upvotes]: Let $A$ be a skew-symmetric $n\times n$-matrix over the real numbers. Show that $\det A$ is nonnegative.

I'm breaking this up into the even case and odd case (if $A$ is an $n\times n$ skew-symmetric matrix).
So when $n$ is odd, we have:
$\det(A)=\det(A^T)=\det(-A)=(-1)^n\det(A)\Rightarrow \det(A)=-\det(A)\Rightarrow \det(A) = 0$
So $\det(A)$ is non-negative when $n$ is odd.
When $n$ is even, we have:
$\det(A)=\det(A^T)=\det(-A)=(-1)^n\det(A)\Rightarrow \det(A)=\det(A)$
But why can't $\det(A)$ be negative in this case?

REPLY [2 votes]: See Lemma 5.6 (and the following discussion) in Keith Conrad's Bilinear Forms.