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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?

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    @wildildildlife - it does not. You are thinking of operators not matrices2012-10-14

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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 $Tv=Av$ .

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    also consider the hint by Will2012-10-14
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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 $(AB)^*=B^*A^*$).

(Okay, that is using the physicists' convention with the inner product being linear in the second argument. This is easily fixed to work for the mathematicians' convention, if desired.)