If a is in a Banach algebra (with identity 1) then the spectrum of a is a set consisting of [itex]\lambda \in \mathbb{C}[/itex] such that [itex](a-\lambda 1)[/itex] is not invertible. That is, there does not exist [itex](a-\lambda 1)^{-1} \in A[/itex] such that [itex](a-\lambda 1)^{-1} (a-\lambda) = (a-\lambda)(a-\lambda 1)^{-1} \neq 1[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

So the spectrum of an element of a unital Banach algebra is a set of complex numbers satisfying a certain property.

My question is: Does it work the other way?

What if a is invertible, that is, if [itex]a\in A^{-1}[/itex], then what is the spectrum of [itex]a^{-1}[/itex]?

Would the spectrum of [itex]a^{-1}[/itex] be the set of all (inverse) complex numbers [itex]\lambda^{-1} \in \mathbb{C}[/itex] such that [itex](a-\lambda 1)^{-1}[/itex] is NOT invertible?

To prove this, all I would have to do is show that there does not exist an element [itex]b \in A[/itex] such that

[tex](a-\lambda 1)^{-1}b = b(a - \lambda 1)^{-1} = 1[/itex]

Then this would show that

[tex]\sigma(a^{-1}) = \{\lambda^{-1}\in\mathbb{C}\,:\,(a-\lambda 1)^{-1}\mbox{ is not invertible }\}[/tex]

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# Homework Help: Spectrum problem

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