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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?

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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.

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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.

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