Spectra of T and T* when T is a bounded linear operator

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The discussion focuses on the spectra of a bounded linear operator T and its adjoint T* in a Hilbert space. It is established that while the spectra are equal for finite rank operators, this does not hold for arbitrary operators in infinite dimensions. However, if T is self-adjoint, the spectra of T and T* are indeed equal. Additionally, for normal operators, it is confirmed that the spectral radii of T and T* are the same due to a general theorem regarding normal elements in C*-algebras. Overall, the relationship between the spectra and spectral radii of T and T* varies based on the properties of the operator.
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Hi,

If T is a bounded linear operator on a Hilbert space, what can we say about the spectra of T and T* (\sigma(T)=\{\lambda:T-\lambda I is not invertible})?
 
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I'm asking because they are equal for finite rank operators and I hope the relation is equally nice in infinite dimensions. :)
 
They're not related for arbitrary operators in \infty-dimensional HS. If T is self-adjoint, the 2 spectra are equal.
 
Is it true at least for normal operators that the spectral radii are the same?
 
@mathwonk: I am not sure whether your question is meant to help the topic starter in the right direction, but since there hasn't been an answer yet, I'll give one:

Yes, then they are the same. A general theorem says that for a normal element x in a C*-algebra, its spectral radius equals its norm. Since x and x* have the same norm, they have the same spectral radius.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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