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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.

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2 Answers 2

up vote 7 down vote accepted

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 $A$).

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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 $imag(A)=(1/2i)(A-A^{*})$.

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Though I also like to consider normal operators as multiplications, in what way is this answer different from the accepted one? –  András Bátkai Sep 22 '13 at 18:13
    
I think it is more elementary . –  user36539 Sep 25 '13 at 20:45
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