# Showing $\langle A^*Av|v \rangle = \langle Av | Av\rangle$

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?

• Take $w = A v$ in $\langle A^\ast w | v \rangle = \langle w | A v \rangle$ – Will Jagy Oct 14 '12 at 21:04
• $\langle Av | w \rangle= \langle v | A^*w \rangle$ for all $v,w$, this follows directly from the definition of $A^*.$ – wildildildlife Oct 14 '12 at 21:05
• @wildildildlife - it does not. You are thinking of operators not matrices – Belgi Oct 14 '12 at 21:19

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 $Tv=Av$ .
In the world of matrices, $$\langle u,Av\rangle=u^*(Av)=u^*(A^*)^*v=(A^*u)^*v=\langle A^*u,\rangle$$ (using $(A^*)^*=A$ and $(AB)^*=B^*A^*$).