# Normal operator + only real eigenvalues implies self-adjoint operator?

Let say we are in a complex vector space, is there an example of a normal operator with only real eigenvalues(or without eigenvalues) that is not a self-adjoint operator? Cause of the spectral theorem it is impossible for the finite dimensional case. I have no idea in the infinite case. I would appreciate any help! Thanks!

• Consider the multiplication by your favorite bounded complex valued function that doesn't take any particular value on a set of positive measure in $L^2$. Feb 15, 2013 at 15:34
• @fedja What does it mean, not to take any particular value on a set? Feb 15, 2013 at 15:48

To elaborate on fedja's comment: Let $$(X,\mu)$$ be a measure space, let $$h$$ be a bounded measurable complex-valued function on $$X$$, and let $$T$$ be the multiplication operator on $$L^2(X,\mu)$$ defined by $$Tf = hf$$. Show that $$T$$ is normal, and is self-adjoint iff $$h$$ is real-valued almost everywhere. Now show that $$\lambda$$ is an eigenvalue of $$T$$ iff $$\mu(\{h= \lambda\}) > 0$$. Taking as an example $$X = [0,1]$$ with Lebesgue measure, you should be able to use this to construct a normal, non-self-adjoint operator with only real eigenvalues, or with no eigenvalues at all.
The correct statement would be if a normal operator has only real spectrum (i.e. $$\sigma(T) \subset \mathbb{R}$$) then it is self-adjoint. But in infinite dimensions, an operator's spectrum may be more than just its eigenvalues.