Compactness and being finite-rank operators Suppose $T$ is a selfadjoint unbounded operator with compact resolvent on a given separable Hilbert space $H$, and $P$ a projection onto $\ker T$. Define $T_1$ to be $T+P$. Is it true that $|T_1|^{-1}$ is compact, $P$ is finite-rank operator and $|T_1|=|T|+P$?
If the answers are "not" what else should we assume about $T$ to get positive answers?
I am looking for some minimal condition of such perturbations. If we know that $P$ is compact then the last question has a positive answer. Does this condition follow from the compactness of the resolvent of $T$?
 A: Yes to all your questions.
If $T$ is self-adjoint and has a compact resolvent then:


*

*$\sigma(T)$ is discrete with only accumulation point being $\infty$.

*There is a Hilbert basis of $H$ consisting of eigenvectors of $T$.

*Each eigenspace of $T$ is finite dimensional.


As an example, point 3. follows since if an eigenspace were infinite dimensional then $(z-T)^{-1}$ would also have an infinite dimensional eigenspace (to eigenvalue $\neq0$) contradicting compactness.
You may then write
$$T=\sum_{\lambda\in\sigma(T)} \lambda \,P_\lambda$$
where $P_\lambda$ is the projection onto the eigenspace of $\lambda$ (which is finite dimensional). Note that $|T|=\sum_{\lambda\in\sigma(T)}|\lambda|\,P_\lambda$ and that $T+P_0= \sum_{\lambda\in\sigma(T), \lambda\neq 0} \,\lambda\,P_\lambda+ 1\,P_0$, of which the absolute value is $|T+P_0|= \sum_{\lambda\neq 0}|\lambda|\,P_\lambda + 1\,P_0 = |T|+P_0$. Further you can see that
$$(T+P_0)^{-1} = \sum_{\lambda\neq0} \lambda^{-1}\,P_\lambda + P_0$$
which is globally defined (due to $0$ not being an accumulation point of $\sigma(T)$) and compact by the $P_\lambda$ all being finite dimensional and $\sigma(T)$ having no accumulation points other than $\infty$.
