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

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

Then this would show that

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

 

[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?