Let $G$ be an operator with discrete spectrum on Hilbert space $H$ such that $\ker G$ is different from $\{0\}$.
Let $P$ be the orthogonal projection onto $\ker G$, and let $G_{0} = G+P$.
My question is:
Are these conditions sufficient to say that $0\in\rho(G_{0})$? If the answer is negative, the addition of the condition of $G$ is normal can guarantee that $0\in\rho(G_{0})$?
In case $G$ is self-adjoint $P$, and hence $Q=1-P$, reduces $G$ so $${G_0}=QGQ+P.$$ Then ${G_0}f=0$ implies that $QGQf=0$ and $Pf=0$ separately. Since $QGQ$ has a vanishing kernel on the subspace projected upon by $Q$, also $Qf=0$ so $f=0$ and hence $0$ is in the resolvent set of $G_0$.