I'm looking over the proof of the singular-value decomposition and a prelim to the proof is that $A^{*}A$ has non-negative eigenvalues, where $ A^* = \overline{A}^T$. We proved this in class doing as follows:
Let $\lambda \in \mathbb{C} $ be an eigenvalue for $A^*A$ corresponding to an eigenvector of norm 1. Then
$$ \lambda = \lambda\langle v | v \rangle = \langle \lambda v | v \rangle $$ $$ \lambda = \langle A^*Av|v \rangle = \langle Av | Av\rangle \ge 0 $$
I'm having trouble seeing how the equivalence in the second equation works. I'm not having any luck trying to derive it from the inner-product space axioms either. How is that equivalence derived?
Hint 1: In orthonormal basis $B$ it holds that $[T]_{B}^{*}=[T^{*}]_{B}$
Hint 2: By definition $\langle Tu,v\rangle=\langle u,T^{*}v\rangle$
Hint 3: You can consider $T:V\to V$ defined by $Tv=Av$ .