I'm studying on Rudin - Functional Analysis the part related to Banach algebra Bounded/Unbounded operator. Specifically i've studied the part of Banach algebra (where the concept of spectrum is actually mentioned...) probably in the next parts will be some more technical detail... However i had a glance to the chapter i haven't studied yet and i haven't seen a specific example of spectral analysis (the book is quite abstract, like mostly of them focused on the subject). Can you provide me an example of how a spectral analysis is performed?
At present more than the isomorphism between banach algebra and $\mathbb{C}$ (under some condition of course) i haven't seen that much... since i'm looking forward for the stuff i mentioned can you give me an example?
Not sure what you mean by "performing a spectral analysis", but here are some examples that might help you understand the ideas involved:
If $T$ is an operator on a finite dimensional Hilbert space, then its spectrum is simply the collection of all eigen-values
If $T: C[0,1] \to C[0,1]$ is the operator $$ T(f)(x) = \int_0^x f(t)dt $$ Then $T$ is injective, but not surjective (use the fundamental theorem of calculus to check this). Hence, $0$ is a spectral value, but not an eigen-value.
If $f \in C[0,1]$ (treated as a Banach algebra with the sup norm), then $$ \sigma(f) = f([0,1]) $$ because a function $g\in C[0,1]$ is invertible in $C[0,1]$ if and only if $g(t) \neq 0$ for all $t\in [0,1]$
If $x:= (\lambda_n) \in \ell^{\infty}$ (once again, a Banach algebra with the sup-norm and point-wise multiplication), then $$ \sigma(x) = \overline{\{\lambda_n\}} $$ (ie. the closure of the set $\{\lambda_n\} \subset \mathbb{C}$) Try proving this, and you should get the hang of how much of this works.