Why does the problem $A u = \lambda u$ bother us in Spectral Theory? Why are we interested in the inverse of $(A - \lambda I)$, and not simply of $A$?

Question

I simply don't see why it should be of our interest to check the eigenvalues of an operator and to find the resolvent of an operator.

An eigenvalue here just shows for which $\lambda$ and for which eigenfunction $u$ the equation $A u = \lambda u$ is true, nothing more. But why is this so particularly interesting for us? I really don't get it.

I mean, it just seems like an arbitrary problem, on which the whole Spectral Theory was "build" on. Like: "Look, $Au = \lambda u$ looks like an interesting problem, let's create a whole theory based on this!". Everyone can simply create such a problem, but why should $Au = \lambda u$ bother us so much to the point that we created the whole field of Spectral Theory?

Also, what I don't get is why are we bothering to calculate $(A - \lambda I)^{-1}$, and not simply $A^{-1}$?

Is it maybe because of some problems in physics that we're bothered with $Au = \lambda u$ and the resolvent of $A$? Because I really don't see a purely mathematical motivation behind that

Answer 1

The question of invertibility $(A-\lambda I)^{-1}$ is precisely the question of solvability of $Ax=\lambda x$.

The spectrum of an operator is of enormous importance to the understanding of the operator. It is the generalization of eigenvalues from linear algebra to the infinite dimensional setting. It shows how an operator may be decomposed into simpler operators. It generalizes diagonalisability. It is the mathematical foundation of quantum mechanics (ev’s of self adjoins operators are possible outcomes of measurements). And so on.

Answer 2

Jakob Elias already gave some motivation in his answer let me add some more.

First of all $A^{-1}$ is just the special case of the resolvent for $\lambda=0$.

Next, spectral theory is crucial in dynamical systems, where you are interested in how flows behave. For example if you have an ODE like $$ y'(t) = A(t) y(t) $$ where $y:\mathbb{R}\rightarrow \mathbb{R}^n$ and $A: \mathbb{R} \rightarrow Mat(n\times n, \mathbb{R})$. In general you will have a hard time solving that equation. However, if you settle for the more modest question, what happens at fixed points, you can use spectral theory to approach this.

You could of course again complain that these are just matrices. Sure, but you run into the same issue for a large class of completely integrable evolution equation (KdV, NLS, CM, KP, KP2, ...). There you use some kind of nonlinear Fourier transform, express everything in the eigenvalues of some time-dependent Schrödinger operator, evolve it there and then transform back. This proceedure allows you to define the flow of such equations even for very rough initial data.

A lot of perturbation theory is also heavily based on spectral theory. The underlying question is if you have a PDE and you make some small change in the equation how much will the solution change.

Added: If you can easily come up with such a problems, you will be soon a star among research mathematicians!