I recently encountered the Riesz Spectral Operators which roughly speaking are closed operators whose eigenvectors form a Riesz basis and I became interested in when such operators can be perturbed and still be of the same type. I must admit that my knowledge of spectral theory is a bit rusty, but I feel like this is a reasonable question.
My hypothesis is the following:
So, suppose $(\phi_n)$ is a Riesz basis for a complex hilbert space $X$. I.e. $\overline{span} \phi_n=X$ and there exist $m,M$ such that for all $N$ $$m\sum^N |\alpha_n|^2 \leq \|\sum^N a_n\phi_n \|^2 \leq M \sum^N|\alpha_n|^2.$$ Suppose further $A$ is a closed operator such that $(\phi_n)_n$ are its eigenvectors. Let us also assume $\bar \sigma_p$ totally disconnected.
Now - I wonder: can I find a (bounded) operator $B : X \to V$ where $V$is another hilbert space (possibly a subspace, or $X$ itself) such that $(B\phi_n)_n$ is a Riesz Basis for $V$ and correspond in some reasonable way to the eigenvectors of $BA$?
As I said, my knowledge of spectral theory is abit rusty, but I guess this is obviously true for $B=I$ and should also be true for $B=P_\Gamma$ (although I am not sure of this as of yet) when $\Gamma$ a jordan curve separating the point spectrum of $A$ since such projetions interact nicely with the eigenvectors. Recall $$P_\Gamma x=\frac{1}{2\pi i}\int_\Gamma R(A: \lambda ) x d\lambda.$$
Now can this be extended for a broader class $B$? What are necessary/sufficient conditions on $B$ for such statements to hold. I appreciate any links/help/references/explanations why I am daft.
EDIT: I guess the bound condition for the Riesz Basis is kind of trivial when $B$ itself is bounded. What is not trivial is then when $(B\phi_n)$is a Riesz basis and when there is a nice correspondence of the spectra.