Why is $\psi_1 : Im(\psi) \to Im(\psi)$ a regular skew mapping, where $\psi$ is skew.

Question

In the book of Linear Algebra by Greub, at page 230, it is stated 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.

Note: we are working on a finite real inner product space, and $\psi$ is any skew map, i.e $ \psi^* = - \psi$, where $\psi^*$ denotes the adjoint mapping.

My question is that why is $\psi_1 : Im(\psi) \to Im(\psi)$ a regular skew mapping, where $\psi$ is skew.I mean I do not see how is that possible because since $\psi$ is normal and skew mapping, it should map $Im \psi$ to $Ker \psi$, so I do not understand how there is a "induced" map with these (co)domain.

Answer 1

We denote the undelying inner product space by $V$ ( $\dim V < \infty$). Then

$V= Im ( \psi) \oplus ker ( \psi)$, hence

$Im( \psi)= Im(\psi^2)$. This shows that $\psi_1$ is surjective.

If $x \in ker( \psi_1)$ then $x \in Im ( \psi) \cap ker ( \psi)=\{0\}$ ,

thus $\psi_1$ is injective.