I'm a beginner to the spectral theory of linear operators. When I was looking up materials regarding this topic, I found most of them discuss compact operators and their spectrum. I have the following two question:
Why are we specifically interested in compact operator? From what I understand, the spectrum of compact operator is discrete so that we can define eigenvalues. Are there any other insights beyond this?
How can we characterize the `spectrum' of non-compact operator? I wonder how we compute it. For instance, consider $A(\cdot)\in C(K, \mathbb{R}^n)$ (or in $L^2(K, \mathbb{R}^n)$ so that this is in a Hilbert space), where $K\subset\mathbb{R}$ is compact. How should I compute the spectrum of such operator?
Any answer or reference would be extremely helpful to me!
Some brief answers:
Re 1: "Compact operators" are of interest among linear operators for the same reason that linear operators are of interest among the set of continuous maps between vector spaces: they are interesting because they are nice objects in and of themselves that can lead to insight about more complicated objects. In a sense, compact operators are the "smallest" natural generalization of the notion of finite rank operators or of operators on a finite dimensional space. In particular, we can get much further thinking about compact operators in the same way we think of matrices that we could in the analysis of other linear operators. For example, everything (besides $0$) in the spectrum of a compact operator is actually an eigenvalue, and the spectral theorem and singular value decomposition hold for compact operator in the exact way that one "should expect".
They are also useful tools in their own right: often, differential equations (ordinary and partial) can be analyzed using compact operators, and the trace class operators play an important role in quantum mechanics.
Re 2: The spectrum of a bounded operator $A$ over a Banach space $X$ consists of all $\lambda \in \Bbb C$ for which $A - \lambda \operatorname{id}_{X}$ fails to be invertible. Unlike in the finite-dimensional case, this does not necessarily mean that $\lambda$ is an eigenvalue of $A$! In particular, $\lambda$ is an eigenvalue of $A$ when $A - \lambda \operatorname{id}_{X}$ fails to be injective, but the spectrum also includes values of $\lambda$ for which $A - \lambda \operatorname{id}_{X}$ fails to be surjective.
This distinction does not come up in the finite-dimensional case because the rank-nullity theorem assures us that if an operator on a finite-dimensional space fails to be injective, then it also fails to be surjective.
As an exercise and example, I would recommend trying to find the spectrum of the shift operator over the sequence space $\ell^2$.