[proofplan]
The proof extracts the coefficient of order $\lambda$ from the eigenvalue equation
\begin{align*}
H(\lambda)\psi(\lambda)=E(\lambda)\psi(\lambda).
\end{align*}
Because the eigenvector branch is analytic in the graph norm, the first correction vector lies in $D(H_0)$ and the perturbation term $V\psi_0$ is meaningful. Projecting the first-order equation onto the unperturbed eigenspace kills the range of $H_0-E_0I$ and leaves exactly the finite-dimensional eigenvalue equation for the compression $P_0VP_0$. The final multiplicity statement uses the analytic type-A isolated-cluster hypothesis through the standard finite-dimensional analytic reduction of the spectral cluster.
[/proofplan]
[step:Show that the compressed perturbation is a well-defined finite-dimensional operator]
Since $E_0$ is an eigenvalue of $H_0$, every vector in $\mathcal E_0$ belongs to $D(H_0)$. Since $D(H_0) \subset D(V)$, every vector in $\mathcal E_0$ belongs to $D(V)$. Thus for each $u \in \mathcal E_0$, the vector $Vu \in \mathcal H$ is defined, and therefore $P_0Vu \in \mathcal E_0$ is defined.
The compressed perturbation is the map
\begin{align*}
P_0VP_0: \mathcal E_0 \to \mathcal E_0
\end{align*}
given by
\begin{align*}
u \mapsto P_0Vu.
\end{align*}
This proves that $P_0VP_0$ is well-defined on $\mathcal E_0$. Since $\mathcal E_0$ is finite-dimensional, this is a finite-dimensional linear operator. Moreover, for $u,v \in \mathcal E_0$, symmetry of $V$ gives
\begin{align*}
(P_0VP_0u,v)_{\mathcal H} = (Vu,v)_{\mathcal H}.
\end{align*}
Since $V$ is symmetric on $D(V)$ and $u,v \in D(V)$,
\begin{align*}
(Vu,v)_{\mathcal H} = (u,Vv)_{\mathcal H}.
\end{align*}
Because $u \in \mathcal E_0$ and $P_0$ is the [orthogonal projection](/theorems/437) onto $\mathcal E_0$,
\begin{align*}
(u,Vv)_{\mathcal H} = (u,P_0Vv)_{\mathcal H} = (u,P_0VP_0v)_{\mathcal H}.
\end{align*}
Hence $P_0VP_0$ is symmetric on the finite-dimensional [Hilbert space](/page/Hilbert%20Space) $\mathcal E_0$.
[/step]
[step:Differentiate the eigenvalue equation at the unperturbed eigenvalue]
Because $\psi$ is analytic as a map into $D(H_0)$ equipped with the graph norm, there exists a vector $\psi_1 \in D(H_0)$ such that
\begin{align*}
\psi(\lambda) = \psi_0 + \lambda \psi_1 + O(\lambda^2)
\end{align*}
in the graph norm of $H_0$ as $\lambda \to 0$. In particular,
\begin{align*}
H_0\psi(\lambda) = H_0\psi_0 + \lambda H_0\psi_1 + O(\lambda^2)
\end{align*}
in $\mathcal H$.
Since $D(H_0) \subset D(V)$ and $V$ is $H_0$-bounded, convergence in the graph norm of $H_0$ implies convergence after applying $V$. Hence
\begin{align*}
V\psi(\lambda) = V\psi_0 + O(\lambda)
\end{align*}
in $\mathcal H$.
Substituting the expansions into
\begin{align*}
H(\lambda)\psi(\lambda) = E(\lambda)\psi(\lambda)
\end{align*}
gives, on the left-hand side,
\begin{align*}
H(\lambda)\psi(\lambda) = H_0\psi_0 + \lambda H_0\psi_1 + \lambda V\psi_0 + O(\lambda^2).
\end{align*}
On the right-hand side,
\begin{align*}
E(\lambda)\psi(\lambda) = E_0\psi_0 + \lambda E_0\psi_1 + \lambda E_1\psi_0 + O(\lambda^2).
\end{align*}
The zeroth-order terms agree because $\psi_0 \in \mathcal E_0$, so $H_0\psi_0=E_0\psi_0$. Equating the coefficients of $\lambda$ in $\mathcal H$ gives
\begin{align*}
H_0\psi_1 + V\psi_0 = E_0\psi_1 + E_1\psi_0.
\end{align*}
Equivalently,
\begin{align*}
(H_0 - E_0I)\psi_1 = (E_1I - V)\psi_0.
\end{align*}
[guided]
The point of using graph-norm analyticity is that it lets us differentiate the eigenvalue equation without leaving the domain of the unbounded operator $H_0$. Since $\psi$ is analytic as a map into $D(H_0)$ with the graph norm, there is a vector $\psi_1 \in D(H_0)$ such that
\begin{align*}
\psi(\lambda) = \psi_0 + \lambda \psi_1 + O(\lambda^2)
\end{align*}
in the graph norm of $H_0$. The graph norm controls both the vector and its image under $H_0$, so this also gives
\begin{align*}
H_0\psi(\lambda) = H_0\psi_0 + \lambda H_0\psi_1 + O(\lambda^2)
\end{align*}
in $\mathcal H$.
We also need to expand the perturbation term. The hypothesis that $V$ is $H_0$-bounded means that $V$ is continuous from $D(H_0)$ with the graph norm into $\mathcal H$. Therefore the graph-norm expansion of $\psi(\lambda)$ implies
\begin{align*}
V\psi(\lambda) = V\psi_0 + O(\lambda)
\end{align*}
in $\mathcal H$. Multiplying by the explicit factor $\lambda$ in $H(\lambda)=H_0+\lambda V$ gives
\begin{align*}
\lambda V\psi(\lambda) = \lambda V\psi_0 + O(\lambda^2).
\end{align*}
Now substitute into the eigenvalue equation. The left-hand side becomes
\begin{align*}
H(\lambda)\psi(\lambda) = H_0\psi_0 + \lambda H_0\psi_1 + \lambda V\psi_0 + O(\lambda^2).
\end{align*}
The eigenvalue expansion and eigenvector expansion give the right-hand side:
\begin{align*}
E(\lambda)\psi(\lambda) = E_0\psi_0 + \lambda E_0\psi_1 + \lambda E_1\psi_0 + O(\lambda^2).
\end{align*}
At order $1$, both sides agree because $\psi_0 \in \ker(H_0-E_0I)$, which means $H_0\psi_0=E_0\psi_0$. Comparing the order-$\lambda$ terms gives
\begin{align*}
H_0\psi_1 + V\psi_0 = E_0\psi_1 + E_1\psi_0.
\end{align*}
Moving the $E_0\psi_1$ term to the left and the $V\psi_0$ term to the right yields the first-order perturbation equation
\begin{align*}
(H_0 - E_0I)\psi_1 = (E_1I - V)\psi_0.
\end{align*}
This equation is the analytic version of “matching the coefficient of $\lambda$,” with the domain issues handled by graph-norm analyticity and $H_0$-boundedness of $V$.
[/guided]
[/step]
[step:Project the first-order equation onto the degenerate eigenspace]
We apply $P_0$ to the first-order equation
\begin{align*}
(H_0 - E_0I)\psi_1 = (E_1I - V)\psi_0.
\end{align*}
Since $\psi_1 \in D(H_0)$ and $H_0$ is self-adjoint, the vector $(H_0-E_0I)\psi_1$ is orthogonal to $\ker(H_0-E_0I)=\mathcal E_0$. Indeed, for every $u \in \mathcal E_0$,
\begin{align*}
((H_0-E_0I)\psi_1,u)_{\mathcal H} = (\psi_1,(H_0-E_0I)u)_{\mathcal H} = 0.
\end{align*}
Therefore
\begin{align*}
P_0(H_0-E_0I)\psi_1 = 0.
\end{align*}
Applying $P_0$ to the right-hand side gives
\begin{align*}
0 = E_1P_0\psi_0 - P_0V\psi_0.
\end{align*}
Since $\psi_0 \in \mathcal E_0$, we have $P_0\psi_0=\psi_0$ and $P_0V\psi_0=P_0VP_0\psi_0$. Hence
\begin{align*}
P_0VP_0\psi_0 = E_1\psi_0.
\end{align*}
Thus $E_1$ is an eigenvalue of $P_0VP_0$ with eigenvector $\psi_0$.
[/step]
[step:Identify all possible first-order shifts with the spectrum of the compression]
Let
\begin{align*}
m := \dim \mathcal E_0.
\end{align*}
The type-A analyticity and isolated finite multiplicity of $E_0$ are used here through the standard [analytic perturbation theorem](/theorems/6967) for isolated self-adjoint eigenvalue clusters: after shrinking $\lambda_0>0$ if necessary, the cluster issuing from $E_0$ is represented by $m$ analytic eigenvalue branches and by analytic eigenvector branches whose values at $\lambda=0$ may be chosen to form a basis of $\mathcal E_0$. The hypotheses needed for this theorem are exactly the type-A analyticity of $H(\lambda)$, self-adjointness for real $\lambda$ sufficiently close to $0$, and isolation with finite multiplicity of $E_0$.
Thus choose analytic eigenvalue-eigenvector branches
\begin{align*}
E_j: (-\lambda_0,\lambda_0) \to \mathbb R
\end{align*}
and
\begin{align*}
\psi_j: (-\lambda_0,\lambda_0) \to D(H_0)
\end{align*}
for $j \in \{1,\dots,m\}$, counted with algebraic multiplicity inside the cluster, such that the vectors $\psi_j(0)$ form a basis of $\mathcal E_0$. Write
\begin{align*}
E_j(\lambda) = E_0 + \lambda E_{j,1} + O(\lambda^2)
\end{align*}
as $\lambda \to 0$. Applying the preceding step to each branch gives
\begin{align*}
P_0VP_0\psi_j(0) = E_{j,1}\psi_j(0)
\end{align*}
for every $j \in \{1,\dots,m\}$.
Since $\{\psi_j(0) : 1 \leq j \leq m\}$ is a basis of $\mathcal E_0$, the displayed equations diagonalize the operator $P_0VP_0:\mathcal E_0 \to \mathcal E_0$ in that basis. Therefore the multiset of first-order coefficients $\{E_{j,1}:1\leq j\leq m\}$ is exactly the multiset of eigenvalues of $P_0VP_0$, counted with multiplicity. In particular, every first-order coefficient is an eigenvalue of the compression, and every eigenvalue of the compression occurs as the first-order coefficient of one of the analytic branches.
[guided]
The last point is a multiplicity argument, not just a pointwise implication. The previous step proves the following for any analytic branch: if
\begin{align*}
E_j(\lambda) = E_0 + \lambda E_{j,1} + O(\lambda^2)
\end{align*}
and if $\psi_j(0) \in \mathcal E_0$ is its initial eigenvector, then
\begin{align*}
P_0VP_0\psi_j(0) = E_{j,1}\psi_j(0).
\end{align*}
This proves that each branch derivative $E_{j,1}$ is an eigenvalue of the compression, but it does not by itself prove that all eigenvalues of the compression occur.
The missing ingredient is supplied by the analytic isolated-cluster hypothesis. The standard analytic perturbation theorem for isolated self-adjoint type-A families says that, after possibly shrinking $\lambda_0>0$, the spectral cluster issuing from the isolated eigenvalue $E_0$ admits exactly $m=\dim\mathcal E_0$ analytic eigenvalue branches, counted with multiplicity, and corresponding analytic eigenvector branches whose initial values can be chosen to form a basis of $\mathcal E_0$. Its hypotheses are satisfied here because $H(\lambda)$ is assumed analytic of type A near $0$, the operators are self-adjoint for real $\lambda$ sufficiently close to $0$, and $E_0$ is isolated with finite-dimensional eigenspace.
Choose such branches and write their initial vectors as $u_j:=\psi_j(0)$ for $j \in \{1,\dots,m\}$. Then $\{u_1,\dots,u_m\}$ is a basis of $\mathcal E_0$, and the branchwise first-order equation gives
\begin{align*}
P_0VP_0u_j = E_{j,1}u_j
\end{align*}
for every $j \in \{1,\dots,m\}$. Thus the matrix of $P_0VP_0$ in the basis $\{u_1,\dots,u_m\}$ is diagonal with diagonal entries $E_{1,1},\dots,E_{m,1}$. Hence the eigenvalues of $P_0VP_0$ counted with multiplicity are exactly the first-order coefficients of the analytic eigenvalue branches. This proves both inclusions: every branch derivative is an eigenvalue of the compression, and every eigenvalue of the compression is realized by some branch derivative.
[/guided]
Therefore the possible first-order shifts of the degenerate eigenvalue $E_0$ are precisely the eigenvalues of the finite-dimensional compressed perturbation $P_0VP_0:\mathcal E_0 \to \mathcal E_0$.
[/step]
