In the book of Linear Algebra by Greub, at page 230, it is claimed that
More general, it will now be shown that the rank of a skew transformation is always even. Since every skew mapping is normal (see sec. S.5) the image space is the orthogonal complement of the kernel. Consequently, the induced transformation $\psi_1 : Im(\psi) \to Im(\psi)$ is regular. Since $\psi_1$ is again skew, it follows that the dimension of $Im(\psi)$ must be even. It follows from this result that the rank of a skew-symmetric matrix is always even.
However, I cannot understand how does the fact that the dimension of $im(\psi)$ should be even follow from his argument.
Note that, I have seen this answer, but if works on a complex space, and I'm particularly interested in understanding the Greub's argument.