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Problem:

$W$ is a complex inner product space. $A$ is a linear map defined on $W$ that satisfies: $\left \langle x,y \right \rangle=0\Rightarrow \left \langle Ax,Ay \right \rangle=0$ for any $x,y\in W$. The question of the problem is to prove that $A=\lambda B$, where $B$ is unitary.

I used the hint provided by the book, to reach the point: $A^{*}A=\lambda I$. If $\lambda=0$, then $A=0$, so $A=0.B$ where $B$ is any unitary matrix of the same size as $A$. Well, for $\lambda \neq 0$, then how do prove that there exits $B$ unitary such that $A=\lambda B$?

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    Is that $\lambda$ in $A^\star A=\lambda I$ real?2012-04-28
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    There is no reason it should be real.2012-04-28
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    @ Giuseppe Negro: yes, $\lambda$ is real and non-negative.2012-04-28

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Taking the adjoint, we can see that $\lambda=\bar\lambda$ so $\lambda$ is a real number. Let $A':=\frac{A}{\sqrt{|\lambda|}}$, then $A'^*A'=\frac{\lambda}{|\lambda|}I$. If $\lambda=1$, we are done. $\lambda$ cannot be equal to $-1$ because $A^*A$ is semi-non-negative definite.

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    Sorry, I just have one more question: if $\lambda$ is $1$, then $A=1.B$ (done). if $\lambda$ is different than $1$, then $A=\sqrt{\lambda }A^{'}$. Aren't we supposed to get $A=\lambda B$? Does this mean that this happens only when $\lambda$ is one?2012-04-28
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    Or do you mean that $A=\sqrt{\lambda }A^{'}$?2012-04-28
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    In fact I should have written $A'=B$.2012-04-28
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    here we got that $A=\sqrt{\lambda }B$, while we are required to prove that $A=\lambda B$2012-04-28
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    Oh yes, you are right, $A'$ is unitary and $A=\sqrt{|\lambda|}A'$ (there is a confusion between the $\lambda$ given in the hint and the $\lambda$ in the result)2012-04-28
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    You're right. I should have used something like $A=cB$, where $c=\sqrt{\lambda }$2012-04-28
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    Yes, in fact the most difficult part was to prove the assertion of the hint.2012-04-28