Question

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 $\sigma\left(\dfrac{A-A^*}{2i}\right)$ the set of imaginary part of the elements of $\sigma(A)$ ? Thanks.

Answer 1

There is probably an elementary proof, but that's an immediate consequence of the continuous functional calculus :

$ \frac{A-A^*}{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 $A$).