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?

  • 2
    $\begingroup$ Take $w = A v$ in $\langle A^\ast w | v \rangle = \langle w | A v \rangle$ $\endgroup$ – Will Jagy Oct 14 '12 at 21:04
  • $\begingroup$ $\langle Av | w \rangle= \langle v | A^*w \rangle $ for all $v,w$, this follows directly from the definition of $A^*.$ $\endgroup$ – wildildildlife Oct 14 '12 at 21:05
  • $\begingroup$ @wildildildlife - it does not. You are thinking of operators not matrices $\endgroup$ – 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 $Tv=Av$ .

  • $\begingroup$ also consider the hint by Will $\endgroup$ – Belgi Oct 14 '12 at 21:15

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.)


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