I am dealing with a Fredholm operator $A$, whose kernel $K$ is non symmetrical. However when I discretise A and compute the eigenvalues of the non symmetric matrix I obtain, I observe that A only has real eigenvalues.
How can I prove it?
The kernel $K$ of the operator has the following properties:
- $K(x,x)=0$ for all $x$ of the domain $S$
- $K(x,y)\geq 0$ for all $(x,y)\in S^2$
- for all $x\in S$, the integral of $y\mapsto K(x,y)$ is always less than r<1
but:
- generally $K(x,y)$ is not equal to $K(y,x)$
- $K$ is not continuous
- $K$ is not bounded (Yet $A$ is bounded, in the sense of linear operators, and is therefore continuous too)
[EDIT]
- A is the composition of two self-adjoint operators, that do not commute.
This is not true in general and very little can be said about the spectrum of a non-self-adjoint operator by considering discretizations. Consider for example the shift operator $S : \ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$, $(Sx)_j = x_{j+1}$. This can be written as an integral operator with the Dirac measure and kernel $$k(j,k) = \begin{cases} 1 & \text{if } k = j+1, \\ 0 & \text{if } k \neq j+1. \end{cases}$$ When you truncate this into finite matrices, you get tridiagonal matrices with $0$ on the main diagonal, that is, all the eigenvalues are $0$. The spectrum of $S$ is equal to the unit circle, however.
EDIT: If you specifically want eigenvalues, restrict it to $\ell^2(\mathbb{N})$ instead. Then you have a full disk of eigenvalues.