Show that the determinant of a 4x4 skew-symmetric matrix is non-negative

Question

I've tried expanding along rows and columns with general variables which only brought a mess and didn't get me anywhere. The mess decreased a little when I used the fact that the determinant of an odd order skew-symmetric matrix is $0$ however it still didn't help me towards solving the problem.
I'm rather unsure with how to proceed with answering the question without using a ton of messy algebra.

Answer 1

If you are not familiar with spectral theorem the easiest way to do it is just to use combinatorial definition of determinant. Since you have anti-simmerty and zeros on the main diagonal it not that bad:

$$det\begin{pmatrix} 0 & a & b & c \\ -a & 0 & d & e\\ -b & -d & 0 & f\\ -c & -e & -f & 0\\ \end{pmatrix} = a^2f^2 + b^2e^2+c^2d^2 - 2(af\cdot be) - 2 (be \cdot cd) + 2 (af \cdot cd ) = (af -be + cd) ^ 2 \ge 0 $$

Another approach: If you know that a skew symmetric matrix has a non-zero eigenvalue, then it is pure imaginary and has the form $\lambda i$ and there is another eigenvalue in the form $-\lambda i$. So if all eigenvalues are non-zero, then the determinant is positive since $det = \prod \lambda_j i (-\lambda_j i) = \prod \lambda_j^2 > 0$.