MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $A$ is a C*-algebra and $n$ is a normal element of $A$, then we have: (By Gelfand duality for example.)

$\operatorname{spec}( |N| ) = | \operatorname{spec}(N) | := \left\{ | \lambda | ; \lambda \in \operatorname{spec}(N) \right\}$

where we define: $|n|:=(n^*n)^{1/2}$. My question is, does the converse also hold?

That is if $a\in A$ and for $r>0$:

$\left\{ r e^{it} : t \in [0, 2\pi[ \right\} \cap \operatorname{spec}(a)$ is not empty if and only if $r\in \operatorname{spec}(|a|)$

implies that $a$ is normal. (Possibly some exceptions made for the zero-element) Or bluntly speaking if the mapping $a$ to $a^*a$ does not create any "new" (or removes any "old") elements in the spectrum then $a$ is normal.

For example if $e$ is an idempotent in $A$, then $e$ is a projection if and only if $||e||=1$. Hence if $e$ is a non-projection idempotent we have $\left\{0,1\right\} = \operatorname{spec}(e) \subsetneq \operatorname{spec}(|e|)$, since $||e||>1$ and by the spectral radius $\operatorname{spec}(e^*e)$ contains an element strictly bigger that one.

Clearly if p is a projection, then p=|p|.

share|cite|improve this question
up vote 5 down vote accepted

You can take a normal operator $N$ with spectrum filling the unit disk and take its tensor product with any operator $T$ of norm less than $1$ thus effectively hiding the non-normal component's contribution to the spectrum in both $A$ and $|A|$.

share|cite|improve this answer
Replace tensor product by direct sum, and that becomes a very nice answer (better than mine). As it stands, the norm of T is rather irrelevant, and I suspect your answer is wrong unless the spectral radius of T coincides with its norm – which is of course easy to arrange, so it's not a serious objection. – Harald Hanche-Olsen Nov 5 '09 at 4:36

The shift operator seems to be a counterexample. Its spectrum is the unit circle, its absolute value is the identity operator and hence has the spectrum {1}, and it's not normal.

[Edit] Oops, no, the spectrum is the unit disk. Not a counterexample after all.

[Edit 2] But that can be fixed: Take the tensor product of the shift operator and multiplication by the identity function on L2[0,1]. The new operator has spectrum the unit disk, its absolute value has spectrum [0,1] and it's not normal.

share|cite|improve this answer
The absolute value of that is a projection, so the spectrum is {0,1}, not [0,1]. – Orr Shalit Nov 5 '09 at 3:48
@Shalit: No, the absolute value is not a projection. The multiplication operator I mentioned is after all already positive, so it is its own absolute value. You may have misread “identity operator” where I said “multiplication by the identity function”? – Harald Hanche-Olsen Nov 5 '09 at 4:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.