 Solution

Since $M$ is real, $(M^2{)}^T=(M^T{)}^2=(-M{)}^2=M^2$. For every real vector $x$,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 $M^{2}x=0$. Hence zero is an eigenvalue of $M^2$. In fact $\ker M^2=\ker M$, as the displayed norm identity shows.