 Solution

Since $M^T=-M$, $(M^2{)}^T=(M^T{)}^2=M^2$, so $M^2$ is a real symmetric matrix. Moreover,proving that it is a negative semidefinite matrix.

For odd $n$, determinant invariance under transpose givesso $\det M=0$. Hence $M$ and $M^2$ have a zero eigenvalue. Finally, if $M^{2}w=0$, then $0=w^T{M}^{2}w=-\Vert Mw{\Vert }^2$, so $Mw=0$. Thuswhere equality of these kernels includes the reverse implication, which is immediate.