# Skew-symmetric matrix and its eigenvalues

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.

It's because $-A^2 = A^T A$ has only real nonnegative eigenvalues: if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ then $$-\lambda^2 \|v\|^2 = -v^T A^2 v = v^T A^T A v = \|A v \|^2.$$
• @Widawensen OK but there are plenty of other examples of skew-symmetric matrices with determinant $0$, including in even dimension – user356288 Jul 26 '16 at 7:54
• Ok. I've found counterexample. It has two 0 eigenvalues. $\begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ It's seems that such non-invertible matrix must have some even dimensional sub-matrix on diagonal as the zero matrix. – Widawensen Jul 26 '16 at 8:30