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!

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

1 Answer 1


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.


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