= Solution
Choose $e_1$ real, as permitted by the <spectral theorem for real symmetric matrices>. Matrix powers commute, so
$$
M^2(Me_1)=M(M^2e_1)=-\mu_1^2Me_1.
$$
The proposed partner $e_2=Me_1$ is nonzero because
$$
\|e_2\|^2=-e_1^TM^2e_1=\mu_1^2\|e_1\|^2>0.
$$
Also the real scalar $e_1^TMe_1$ equals its own transpose $e_1^TM^Te_1=-e_1^TMe_1$, so it vanishes. Therefore \b[$e_1$ and $e_2$ are orthogonal eigenvectors with the same eigenvalue]. This is the basic pairing in the <paired eigenspaces of the square of a real skew-symmetric matrix>.
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