I'm trying to prove that:
$A$ positive definite Hermitian $\iff$ $\exists P \in GL(E)$ s.t. $A=P^*P$.
$GL(E)$ is the set of all invertible matrices.
Proof:
$\Leftarrow:$
Suppose $\exists P \in GL(E)$ s.t. $A=P^*P$
To show: $A=A^*$. We know $A=P^*P.$ And $A^*=\overline{(P^*P)^T} = \overline{P^TP^{*T}} = P^*P \Rightarrow A=A^* \Rightarrow A$ is Hermitian $\Rightarrow \lambda \in \mathbb{R}$(eigenvalues) $\Rightarrow x^TAx =0 \iff x=0 \Rightarrow A$ is positive definite Hermitian.
$\Rightarrow:$
How can I show this? And is my proof so far correct?
You're missing a transpose
$$ \overline{(P^* P)^T} = \overline{P^T P^{*T}}\\ = \overline{P^T} \overline{P^{*T}}\\ = P^* P $$.
You didn't finish the proof for positivity. You only gave the equivalent condition that you need to prove.
$$ x^* A x = x^* P^* P x = \mid Px \mid^2 == 0\\ $$
implies $Px = 0$ which then implies $x=0$ because $P \in GL(E)$.
For the other direction, you know that $A = Q^{-1} \Lambda Q$ with $\Lambda$ a diagonal matrix with the positive eigenvalues. $Q$ being the change of basis. $\Lambda$ has a positive square root because you can take positive square roots of all the diagonal entries. Call that $D$. It is real, diagonal and invertible. Then $A=Q^{-1} D^* D Q$. Now use what you know about $Q^{-1}$ and $Q^*$.