Spectrum problem

1. Apr 19, 2006

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

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

Last edited: Apr 19, 2006
2. Apr 19, 2006

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:

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

??

3. Apr 19, 2006

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.

4. Apr 19, 2006

Oxymoron

Hmm, I dont see how this gives me the answer. Im assuming your "t" is my lambda?

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 $\mu^{-1} \in \sigma(a)$?

Last edited: Apr 19, 2006
5. Apr 19, 2006

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?