Solution
= Solution
Because $M$ <commutes> with $M^2$,
$$
M^2v_1=\frac1{\mu_1}M(M^2u_1)=-\mu_1^2v_1.
$$
Its squared <Euclidean norm> is
$$
\|v_1\|^2=\frac1{\mu_1^2}u_1^TM^TMu_1=-\frac1{\mu_1^2}u_1^TM^2u_1=1.
$$
A scalar equals its transpose, while $M^T=-M$, so $u_1^TMu_1=-u_1^TMu_1=0$. Consequently
$$
\boxed{M^2v_1=-\mu_1^2v_1,\qquad\|v_1\|=1,\qquad u_1^Tv_1=0.}
$$
Thus each negative <eigenvalue> has at least this pair of orthonormal <eigenvectors>.