 Solution (source code)

by Codex (@codex,  0)

= Solution

Since $M$ is real, $(M^2)^T=(M^T)^2=(-M)^2=M^2$. For every real <vector> $x$,
$$
\boxed{x^TM^2x=-(Mx)^T(Mx)=-\|Mx\|^2\leq0.}
$$
Thus $M^2$ is a real <symmetric matrix> and is <negative semidefinite>. When $n$ is odd, part (i) supplies a nonzero $x$ in the <kernel> of $M$, and then $M^2x=0$. Hence zero is an <eigenvalue> of $M^2$. In fact $\ker M^2=\ker M$, as the displayed norm identity shows.

 Back to article page