= Solution
Normalize $e_1$ and put $f_1=Me_1/\mu_1$. The two vectors are an <orthonormal basis> of a plane $E$ invariant under $M$, since $Me_1=\mu_1f_1$ and $Mf_1=-\mu_1e_1$. Its <orthogonal complement> is also invariant: for $w\perp E$ and $z\in E$,
$$
\langle Mw,z\rangle=-\langle w,Mz\rangle=0.
$$
Apply the <spectral theorem for real symmetric matrices> to the restriction of $M^2$ to $E^\perp$. Any remaining negative <eigenvalue> supplies another orthonormal pair by the same construction. Induction leaves only the <kernel>. Thus \b[every nonzero eigenvalue of $M^2$ has even multiplicity]. The pairs can share an eigenvalue, so this does not assert distinctness of the different $-\mu_j^2$.
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