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.

  • 5
    $\begingroup$ A normal matrix over $\Bbb C$ is diagonalisable. $\endgroup$ – Angina Seng Aug 28 '19 at 9:02
  • $\begingroup$ $\newcommand{\rank}{\mathrm{rank}}$ A more direct proof is $\rank(M^2) = \rank(MM) = \rank(-MM') = \rank(MM') = \rank(M)$, without using any additional properties. $\endgroup$ – Zhanxiong Sep 17 '20 at 13:43

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
    $\begingroup$ (+1) Your answer is better than mine. $\endgroup$ – Fred Aug 28 '19 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


  • 1
    $\begingroup$ (+1) Your answer is better than mine. $\endgroup$ – José Carlos Santos Aug 28 '19 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.