# linear maps on complex inner product spaces

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 $A=\lambda B$?

-
Is that $\lambda$ in $A^\star A=\lambda I$ real? –  Giuseppe Negro Apr 28 '12 at 17:55
There is no reason it should be real. –  GEdgar Apr 28 '12 at 17:56
@ Giuseppe Negro: yes, $\lambda$ is real and non-negative. –  User968590 Apr 28 '12 at 18:00

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.
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? –  User968590 Apr 28 '12 at 18:19
Or do you mean that $A=\sqrt{\lambda }A^{'}$? –  User968590 Apr 28 '12 at 18:26
In fact I should have written $A'=B$. –  Davide Giraudo Apr 28 '12 at 18:29
here we got that $A=\sqrt{\lambda }B$, while we are required to prove that $A=\lambda B$ –  User968590 Apr 28 '12 at 18:33
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) –  Davide Giraudo Apr 28 '12 at 18:36