I'm trying to solve the following problem:
Let $V$ and $W$ be two finite dimensional vector spaces over $\mathbb{C}$ with inner products $p$ and $q$ respectively. Let $\psi: V \rightarrow W$ be a surjective linear transformation and suppose there is a linear transformation $\phi:W \rightarrow V$ such that $q(\psi(v),w) = p(v,\phi(w))$ for all $v \in V$, $w \in W$.
Show that $\phi$ is injective and that $(Ker(\psi))^{\perp} = Im(\phi)$
I have already proved that $\phi$ is injective and in order to show that $(Ker(\psi))^{\perp} = Im(\phi)$ I have proved that $Im(\phi) \subset (Ker(\psi))^{\perp}$ as follows:
Let $w \in Im(\phi)$, then there exists $v \in V$ such that $v=\phi(w)$. Let $u \in Ker(\psi)$, then:
$0 = q(\psi (u), w) = p(u, \phi(w)) = p(u, v)$. Since $p(u, v) = 0$ then $p(u, v) = \overline{p(u,v)} = p(v,u) = 0$ and therefore, since $u$ was arbitrary $v \in Ker(\psi)^{\perp}$.
What I need to prove know is that $Ker(\psi)^{\perp} \subset Im(\phi)$ but I'm stuck in there. I know that if $v \in Ker(\psi)^{\perp}$ then $p(v,u)=0$ for all $u \in Ker(\psi)$, but I don't know how to could I show that $v= \phi(w)$ for some $w$.
Firstly, we will show that $ (Im(\phi))^{\perp}\subset Ker(\psi) $. If this result is true, it can be got by simply use the fact that $ A\subset B $ $ \Rightarrow $ $ B^{\perp}\subset A^{\perp} $ and $ (A^{\perp})^{\perp}=A $ in finite dimensional linear space. If $ w\in(Im(\phi))^{\perp} $, then for any $ v\in V $, $$ 0=p(w,\phi(v))=q(\psi(w),v). $$ This directly implies that $ \psi(w)=0 $, i.e. $ w\in Ker(\psi) $.