Let $H$ be a Hilbert space, $A$ be a normal bounded operator on $H$ with spectrum $\sigma(A)=\{\lambda\in \mathbb{C}\;\;A\lambda Id \text{ is not invertible }\}$. Is $\sigma\left(\dfrac{AA^*}{2i}\right)$ the set of imaginary part of the elements of $\sigma(A)$ ? Thanks.
There is probably an elementary proof, but that's an immediate consequence of the continuous functional calculus :
$ \frac{AA^*}{2i} = f (A) $
where $f$ is the imaginary part function on $\mathbb{C}$. And when you apply a continuous function $f$ to a normal operator $A$ you have : $Spec(f(A)) = f(Spec(A))$ (you can see that by restricting to the abelian sub$C^*$algebra generated by $A$).
The spectral theorme assert that there exist a unitary operator $U$ sucht that $U^{*}AU$ is the multiplication operator by $\lambda$, $\lambda \in \mathbb{C}$. With the same unitary opeartor $U$, we have the real part (resp. the immaginary part) of $A$ mapped to $real(\lambda)$ resp. $imag(\lambda)$, with $real(A)=(1/2)(A+A^{*})$ and $imag(A)=(1/2i)(AA^{*})$.

$\begingroup$ Though I also like to consider normal operators as multiplications, in what way is this answer different from the accepted one? $\endgroup$ – András Bátkai Sep 22 '13 at 18:13
