TITLE: Determinant of skew-hermitian matrix
QUESTION [6 upvotes]: Given a skew-hermitian matrix $A \in \mathbb{C}^{N\times N}$, then $A = -A^H = -(A^*)^{T}.$
We can also say that $A^T = (-(A^*)^T)^T = -A^*.$  Thus, when computing the determinant we get
$$
\det(A) = \det(A^T) = \det(-A^*) = (-1)^N\det(A^*) = (-1)^N(\det(A))^*.
$$
For $N$ even, we have $\det(A) = (\det(A))^*$, and we can conclude that $\det(A)$ is real.  
For $N$ odd, we have $\det(A) = -(\det(A))^*$.  
Is the only conclusion that $\det(A) = 0$ for odd $N$?

REPLY [3 votes]: Let $y = det(A)$. For $N$ odd, we have $y = -\bar y$. Therefore, $y$ is imaginary (or can be $0$, too).