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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.

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

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    $\begingroup$ (+1) Your answer is better than mine. $\endgroup$ – Fred Aug 28 '19 at 9:51
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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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    $\begingroup$ (+1) Your answer is better than mine. $\endgroup$ – José Carlos Santos Aug 28 '19 at 9:50

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