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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|.

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2 Answers 2

up vote 4 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 $|A|$.

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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.

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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

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