I am reading a paper where the sentence appears: "The bounded operator $T^{\frac{1}{2}}$ has the same eigenvectors as T, and its eigenvalues are the complex square roots of those of T."
The sentence is obvious if $T$ is positive, compact, and self-adjoint, by the spectral representation of $T$.
I wonder how we can know the sentence holds when the operator $T$ is a normal operator and may have some negative eigenvalues or even some imaginary eigenvalues. In this case, we cannot even use the spectral mapping theorem because $f(x)=x^{1/2}$ may not be continuous on the spectrum of $T$.
To whom it might be of any interest, the paper is here, and the sentence appears in page 14. And, actually here $T=zI-Q$ and $T$ is just normal but not positive nor compact, nor self-adjoint. (By the way, I feel that the exposition in this part is somewhat insufficient or incorrect.)
Also, Rudin's functional analysis book does not seem to cover the related material very well. Some recommended reference, if any, would be appreciated.