I checked some examples and I always received that skew-symmetric matrix of even dimension has only pure imaginary eigenvalues.
For example:
$\begin{bmatrix}
0 & 2 & 3 & 1 \\
-2 & 0 & 1 & 4 \\
-3 & -1 & 0 & 1 \\
-1 & -4 & -1 & 0
\end{bmatrix}$
Eigenvalues:
$( 0.000, 5.406i)
( 0.000,-5.406i)
( 0.000, 1.665i)
( 0.000,-1.665i)$
How can be explained such property?
Additionally why skew-symmetric of even dimension has non-zero determinant in opposition to odd dimensional skew-symmetric matrices ?
(I'm not considering here zero matrices)
Interesting is also fact that probably every matrix (of even dimension) can be decomposed into symetrical part which has only real eigenvalues and skew-symmetrical which has only pure imaginary values what makes interesting analogy to complex numbers and their two parts, but I don't know whether there are importatnt consequences of this fact.
Best Answer
It's because $-A^2 = A^T A$ has only real nonnegative eigenvalues: if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ then $$-\lambda^2 \|v\|^2 = -v^T A^2 v = v^T A^T A v = \|A v \|^2.$$
Skew-symmetric matrices do not have to have nonzero determinant, the zero matrix is a counterexample.