I have the discrete Laplace operator on an infinite Hilbert space with an orthonormal basis $\psi_x$ ($\forall x \in \mathbb Z$), given by $\Delta \psi_x=\psi_{x-1}+\psi_{x+1}$. If I introduce a finite rank self adjoint perturbation $A$ to $\Delta$, then I want to prove that the continuous spectrum does not change, this perturbation only introduces finitely many new eigenvalues to the existing spectrum. How could I go about it?
I know that $\Delta$ has the continuous spectrum given by all $\lambda = 2 \cos k$, where $k$ is the momentum of the state whose eigenvalue we're finding: $\psi_k = \sum_x e^{ikx} \psi_x$, then the solution to $(\Delta - I \lambda )\psi = 0$ is $\psi = \psi_k$, thus getting the expression for $\lambda$.
Does a similar reasoning work to show that the continuous spectrum is equivalent even after a finite rank perturbation? How can I see it?
More generally, I am aware that there exists a theorem which guarantees the spectral stability of compact infinite operators under a finite rank perturbation. But I wanted to know the exact statement of the theorem (and a possible reference?)
PS: Sorry for the sloppy language, physicist by trade.
