A property of eigenvectors of real antisymmetric block matrices Consider the block matrix
$$M=\left[\begin{matrix}0 & A \\ -A^T & 0\end{matrix}\right]$$
with $A$ a real matrix.
An eigenvector of $M$ obeys
$$\left[\begin{matrix}0 & A \\ -A^T & 0\end{matrix}\right] \left[\begin{matrix} v_1 \\ v_2\end{matrix}\right]=\lambda \left[\begin{matrix} v_1 \\ v_2\end{matrix}\right]$$
where the eigenvector is split into blocks just like the matrix.
$\lambda$ is of course purely imaginary, because $M$ is antisymmetric (anti-Hermitian). The eigenvector $[v_1,v_2]^T$ is of course only defined up to a factor, but it appears that this factor can be chosen such that all $v_1$ are purely real, and all $v_2$ purely imaginary.
This is consistent with the eigenvalue equation: it tells us that if $v_1$ is real, $v_2$ must be imaginary, and if $v_2$ is imaginary, $v_1$ must be real. But this is a circular line of reasoning.
How can we prove that $v_1$ is purely real and $v_2$ purely imaginary?
 A: One method is to find an explicit form for these eigenvectors.  In particular: suppose $A$ has singular value decomposition $A = U \Sigma V^T$.  Suppose $u_j$ are the columns of $U$ and $v_j$ are the columns of $V$.  We note that for every $j$,
$$
Av_j = \sigma_j u_j, \qquad A^Tu_j = \sigma_j v_j
$$
With that, we can note that
$$
\pmatrix{0&A\\-A^T&0} \pmatrix{v_j\\\pm i u_j} = \pmatrix{\pm i A u_j\\-A^Tv_j} = (\pm i \sigma_j)\pmatrix{v_j\\\pm i u_j}
$$
Thus, every vector of the form 
$\pmatrix{v_j\\\pm i u_j}$
is an eigenvector.
Since there are $n$ of these which are linearly independent (over $\Bbb C$), we have a complete basis of eigenvectors.
A: You can write
$$
\eqalign{
  & \left( {\matrix{
   0 & A  \cr 
   { - A^{\,T} } & 0  \cr 
 } } \right)\left( {\matrix{
   {v_{\,1} }  \cr 
   {v_{\,2} }  \cr 
 } } \right) = \left( {\matrix{
   0 & A  \cr 
   {i^{\,2} A^{\,T} } & 0  \cr 
 } } \right)\left( {\matrix{
   {v_{\,1} }  \cr 
   {v_{\,2} }  \cr 
 } } \right) = i\mu \left( {\matrix{
   {v_{\,1} }  \cr 
   {v_{\,2} }  \cr 
 } } \right)  \cr 
  & \quad \quad  \Downarrow   \cr 
  & \left( {\matrix{
   0 & { - i\,A}  \cr 
   {i\,A^{\,T} } & 0  \cr 
 } } \right)\left( {\matrix{
   {v_{\,1} }  \cr 
   {v_{\,2} }  \cr 
 } } \right) = \mu \left( {\matrix{
   {v_{\,1} }  \cr 
   {v_{\,2} }  \cr 
 } } \right) \cr} 
$$
Then, depending on $A,\mu$, the components of $v_1,v_2$ can be mixed, alternatively real and imaginary (e.g. with $A=I$).
In the general case, one vector must be real and the other imaginary. 
