$\mathrm{rank}(M)=\mathrm{rank}(M^2)$ 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.

$\newcommand{\rank}{\mathrm{rank}}$ A more direct proof is $\rank(M^2) = \rank(MM) = \rank(-MM') = \rank(MM') = \rank(M)$, without using any additional properties. — Zhanxiong, Sep 17 '20 at 13:43

score 7 · Accepted Answer · answered Aug 28 '19 at 09:06

7

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

answered Aug 28 '19 at 09:06

José Carlos Santos

427,504

score 7 · Answer 2 · edited Jan 27 '24 at 12:51

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

$$\mathrm{rank}(M)=\mathrm{rank}(M^2).$$

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

2 Answers2