TITLE: $rank(M)=rank(M^2)$ whenever $M$ is skew-symmetric
QUESTION [4 upvotes]: 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., 

$rank(M)=rank(M^2)$ 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.

REPLY [7 votes]: Let $M$ be a real skew-symmetric $n \times n$ - matrix, $(\cdot, \cdot)$ the usual inner product on $ \mathbb R^n$ and $|| \cdot ||$ the induced norm.
Let $x \in ker(M^2)$, then
$$ 0= (M^2x,x)=(Mx, M^Tx)=(Mx, - Mx)=-||Mx||^2.$$
This gives $ker(M^2) \subseteq ker(M).$ The other inclusion $ker(M) \subseteq ker(M^2)$ is clear. Thus
$$ker(M^2) = ker(M).$$
Now invoke the rank - nullity theorem to get
$$rank(M)=rank(M^2).$$
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