Spectrum of Invertible Elements in Unital Banach Algebra: A Proof

[Oxymoron]

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

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 a\in A^{-1}, then what is the spectrum of a^{-1}?

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

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

(a-\lambda 1)^{-1}b = b(a - \lambda 1)^{-1} = 1[/itex]<br /> <br /> Then this would show that <br /> <br /> \sigma(a^{-1}) = \{\lambda^{-1}\in\mathbb{C}\,:\,(a-\lambda 1)^{-1}\mbox{ is not invertible }\}

 

[Oxymoron]

PS. Is the last line of the above post:

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

equivalent to saying:

\sigma(a^{-1}) = \{\lambda^{-1}\in\mathbb{C}\,:\,\lambda\in\sigma(a)\}

??

 

[matt grime]

If a is invertible then a-t= a(1-a^{-1}t). Which gives you the answer.

I think you have to many things going on there, too many double negatives, and why define the spectrum as the set of lambda^(-1)'s? just make it the set of mu's and show mu is in the spec of a inverse if and only if one over mu is in the spec of a. Your choice of presentation makes it more complicated than it needs to be.

 

[Oxymoron]

Posted by Matt Grime

If a is invertible then a-t= a(1-a^{-1}t).

Hmm, I don't see how this gives me the answer. I am assuming your "t" is my lambda?

Posted by Matt Grime

I think you have to many things going on there, too many double negatives, and why define the spectrum as the set of lambda^(-1)'s? just make it the set of mu's and show mu is in the spec of a inverse if and only if one over mu is in the spec of a. Your choice of presentation makes it more complicated than it needs to be.

You know what, I was getting the same idea. So your saying that I should define a new set of complex numbers: \{\mu\in\mathbb{C}\} and show that \mu_i \in\sigma(a^{-1}) if and only if \mu^{-1} \in \sigma(a)?

 

[Oxymoron]

To show \sigma(a^{-1}) = \{\lambda^{-1}\,:\,\lambda\in\sigma(a)\} could I just show that (a-\lambda 1)^{-1} \in A^{-1}?

That is, show that if I multiply (a-\lambda 1) by a^{-1} then I get an invertible element? Is that what you where trying to point out?