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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.

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  • $\begingroup$ Apparently, the author has a previous result "regular + skew implies even-dimensional" at hand? $\endgroup$ – Hagen von Eitzen Mar 2 '18 at 7:04
  • $\begingroup$ @HagenvonEitzen Not really. Skew mapping are mentioned first at page 229, and this is mentioned at 230, and between those lines, we only definitions. $\endgroup$ – onurcanbektas Mar 2 '18 at 7:08
  • $\begingroup$ A skew-symmetric matrix of odd size has determinant zero, since $\det(A)=\det(A^t)=\det(-A)=(-1)^n \det(A)$. $\endgroup$ – Hans Lundmark Mar 2 '18 at 7:37
  • $\begingroup$ @HansLundmark yes, I do know that, but how does this helps ? $\endgroup$ – onurcanbektas Mar 2 '18 at 7:39
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    $\begingroup$ The dimension of the image space can't be odd, since in that case the determinant of $\psi_1$ would have to be zero (according to my previous comment). Which it isn't, since $\psi_1$ is regular. $\endgroup$ – Hans Lundmark Mar 2 '18 at 7:47
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A skew-symmetric linear transformation on an odd-dimensional space has determinant zero and is therefore not regular. So since $\psi_1 \colon \operatorname{Im}(\psi) \to \operatorname{Im}(\psi)$ is skew-symmetric and regular, the space $\operatorname{Im}(\psi)$ must be even-dimensional.

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While Greub is not as determinant-phobic as Axler, my impression is that he also would rather not resort to a determinantal argument if possible. What he had in mind probably goes like this: as $\psi_1^\ast\psi_1$ is self-adjoint, it has an eigenpair $(\lambda,u)$. Let $v=\psi_1(u)$. Since $\psi_1$ is skew and regular, we have $u\perp v\ne0$. Yet $\psi_1(v)=-\lambda u$ and $\psi_1^\ast\psi_1(v)=\lambda v$. Therefore $u$ and $v$ are two mutually orthogonal eigenvectors of $\psi_1^\ast\psi_1$ and they also span a two-dimensional invariant subspace of $\psi_1$. Proceed recursively, we see that $Im(\psi_1)$ is a direct sum of two-dimensional invariant subspaces, each spanned by two eigenvectors of $\psi_1^\ast\psi_1$. Hence it is even-dimensional.

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  • $\begingroup$ By the way, that is a nice proof :) Thanks a lot sir. $\endgroup$ – onurcanbektas Mar 2 '18 at 9:37

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