most recent 30 from http://mathoverflow.net 2013-05-19T22:55:49Z http://mathoverflow.net/feeds/question/109448 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109448/imaginary-part-of-a-spectrum loup blanc 2012-10-12T10:29:22Z 2012-10-12T18:33:10Z

<p>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.</p>

http://mathoverflow.net/questions/109448/imaginary-part-of-a-spectrum/109449#109449 Simon Henry 2012-10-12T10:39:30Z 2012-10-12T10:46:10Z

<p>There is probably an elementary proof, but that's an immediate consequence of the continuous functional calculus : </p> <p>$ \frac{A-A^*}{2i} = f (A) $ </p> <p>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 $A$).</p>