$rank(M)=rank(M^2)$ whenever $M$ is skew-symmetric

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.

• A normal matrix over $\Bbb C$ is diagonalisable. – Lord Shark the Unknown Aug 28 at 9:02

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

• (+1) Your answer is better than mine. – Fred Aug 28 at 9:51

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

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

• (+1) Your answer is better than mine. – José Carlos Santos Aug 28 at 9:50