Are these two definitions of the effective Hamiltonian equivalent? Consider a Hamiltonian $H$ on a Hilbert space $\mathcal H_A \oplus\mathcal H_B$ and let $P$ (and $Q$) denote the projection operator onto $A$ (and $B$). There are two common definitions of an effective Hamiltonian $H_\textrm{eff}(E)$ on subspace $A$:


*

*The Schroedinger equation can be written as
$$ \left( \begin{array}{cc} PHP & PHQ\\
QHP & QHQ \end{array} \right) \left( \begin{array}{c} \psi_A \\ \psi_B\end{array} \right) = E \left( \begin{array}{c} \psi_A \\ \psi_B\end{array} \right)$$
One can straightforwardly eliminate $\psi_B$ to obtain
$$ H_\textrm{eff}(E) |\psi_A\rangle = E  |\psi_A\rangle \qquad \textrm{with } \;  \boxed{H_\textrm{eff}(E) := P \left( H + H Q \frac{1}{E-QHQ} Q H \right) P}$$

*In the Green's function/resolvent approach, one simply defines
$$ \boxed{\frac{1}{E-H_\textrm{eff}(E)} := P \frac{1}{E-H} P }$$
(The motivation for this has to do with the fact that the left-hand side has the same poles as the right-hand side would have even without the $P$ projectors, hence $H_\textrm{eff}(E)$ has the same eigenvalues as $H$ for eigenvectors which have a non-zero support on $A$.)


My question is simply: how can I see whether these two definitions are equivalent?
(Note that this is not a duplicate! A very similar question was asked before, but there the answer that was accepted in fact did not answer the exact question that I am asking. There is also another question which has to do with deriving the effective Hamiltonian, which is not what I am asking here.)
 A: I'll prove $(2\implies 1)$; the reverse implication can be found just by reading the proof bottom to top. To get equation (1) from equation (2), we need to figure out how to invert $P\frac{1}{E-H}P$. Once we've figured that out, the rest of the proof follows straightforwardly.
For any operator of the form
$$
\hat{O}=\left(\begin{array}{cc}\hat{A} & \hat{B}\\\hat{C}&\hat{D}\end{array}\right)
$$
it is a theorem from linear algebra that
$$
\hat{O}^{-1}=\left(\begin{array}{cc}(\hat{A}-\hat{B}\hat{D}^{-1}\hat{C})^{-1} & \square\\\square&\square\end{array}\right)
$$
where the $\square$s denote elements we don't care about.
For the specific case of $\hat{O}=E-H=\left(\begin{array}{cc}E-PHP&-PHQ\\-QHP&E-QHQ\end{array}\right)$, we have that
$$
\hat{O}^{-1}=\frac{1}{E-H}=\left(\begin{array}{cc}([E-PHP]-PHQ[E-QHQ]^{-1}QHP)^{-1} & \square\\\square&\square\end{array}\right)
$$
Thus, we have
$$
P\frac{1}{E-H}P=([E-PHP]-PHQ[E-QHQ]^{-1}QHP)^{-1}
$$
therefore,
$$
\frac{1}{E-H_{eff}}=([E-PHP]-PHQ[E-QHQ]^{-1}QHP)^{-1}
$$
We can now easily take the inverse of both sides! As promised, we very quickly get to the final answer:
\begin{array}{rcl}
E-H_{eff}&=&[E-PHP]-PHQ[E-QHQ]^{-1}QHP\\
H_{eff}&=&PHP+PHQ[E-QHQ]^{-1}QHP\\
\end{array}
