$\mathrm{rank}(M)=\mathrm{rank}(M^2)$ whenever $M$ is skew-symmetric

Question

On p.231 of Linear Algebra by Greub, it is stated that a real skew-symmetric matrix has the same rank as its square, i.e.,

$\mathrm{rank}(M)=\mathrm{rank}(M^2)$ whenever $M$ is real skew-symmetric.

I tried to use the fact that skew-symmetric matrix is normal and some geometric properties of normal matrices, but cannot proceed.

Any help is appreciated.

Answer 1

Since it is normal, it is diagonalizable. And it is easy to see that, for a diagonal matrix $D$, $\operatorname{rank}(D)=\operatorname{rank}(D^2)$.