Let $V$ be a finite dimensional inner product space, $ \phi :V \rightarrow V$ a linear operator and $\phi^*:V \rightarrow V $ its adjoint.
I wish to show:
$\phi^*$ is an inverse to $\phi$ if and only if $ \langle \phi (v), \phi (w) \rangle = \langle v,w\rangle \ \ \forall \ v,w \in V $
I understand that $\phi^* \circ\ \phi = id_V $ if and only if $ \langle \phi (v), \phi (w) \rangle = \langle v,w\rangle \ \ \forall \ v,w \in V. $
The proof I'm reading then states, by rank-nullity, it is easily deduced that $\phi \circ\ \phi^* = id_v $.
How do I show the last part?
Here are the definitions I'm working with:
An inner-product is positive definite, linear in the second variable and conjugate symmetric.
An adjoint, $\phi^*$, is a linear operator with the property: $ \forall v,w \in V, \langle \phi^*(v), w \rangle = \langle v, \phi(w) \rangle $
Note that it is generally true that for a finite dimensional space $V$ and linear operators $\psi,\psi':V \to V$, we have $$ \psi\circ \psi' = \operatorname{id}_V \implies \psi' \circ \psi = \operatorname{id}_V $$ This is sufficient to complete your proof.