## Normal operators and it’s spectrum in C*-algebras

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

spec( |N| ) = | spec(N) | := { | \lambda | ; \lambda \in spec(N) }

Where we define: |n|:=(n*n)^{1\slash 2}. My question is, does the converse also holds?

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

{ r e^{it} : t \in [0, 2\pi[ } \cap spec(a) is not empty if and only if r\in 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 {0,1}=spec(e) \subsetneq spec(|e|), since ||e||>1 and by the sprectal radius spec(e*e) contains an element strictly bigger that one.

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

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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 2009 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.

 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 2009 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 2009 at 4:34