Estimate eigenvectors of symmetric matrix with almost vanishing diagonal Is there a way to approximate the eigenvectors of a symmetric matrix with almost vanishing diagonal elements, i.e. with the block matrix form,
\begin{equation}
M=\left( \begin{array}{cc} 
\alpha\epsilon _1  & A \\  
A ^T  & \alpha\epsilon _2 
\end{array} \right)
\end{equation}
with $\alpha \ll 1$? I'd like the result to be to first order in $\alpha$. I thought that maybe I could first take it to block diagonal form using a an satz for the unitary transformation as I've seen done in another similar case (the seesaw mechanism for studying neutrino masses in particle physics), but didn't lead anywhere.
 A: Note that below I replaced $A^{T}$ by $A^{\ast }$, the adjoint of $A$.
In case $A$ is real they coincide but it has advantages to use a complex
formalism. The problem is a standard perturbation situation in quantum
mechanics and is treated in nearly all textbooks. Thus we have a
Hamiltonian
\begin{equation*}
H_{\alpha }=\left(
\begin{array}{cc}
\alpha \varepsilon _{1} & A \\
A^{\ast } & \alpha \varepsilon _{2}%
\end{array}
\right) =H_{0}+\alpha V,\;H_{0}=\left(
\begin{array}{cc}
0 & A \\
A^{\ast } & 0
\end{array}
\right) ,\;V=\left(
\begin{array}{cc}
\varepsilon _{1} & 0 \\
0 & \varepsilon _{2}
\end{array}
\right) ,
\end{equation*}
where $V$ is small. The parameter $\alpha $ is introduced for convenience in
tracking orders in a perturbation expansion. The idea is to express various
quantities in terms of the eigenvalues and eigenvectors of $H_{0}$ which are
usually known. Suppose that $\mathbf{f}$ is an eigenvector of $H_{0}$ at the
eigenvalue $\lambda $,
\begin{equation*}
H_{0}\mathbf{f}=\lambda \mathbf{f},
\end{equation*}
or
\begin{equation*}
\left(
\begin{array}{cc}
0 & A \\
A^{\ast } & 0%
\end{array}%
\right) \left(
\begin{array}{c}
f_{1} \\
f_{2}
\end{array}
\right) =\lambda \left(
\begin{array}{c}
f_{1} \\
f_{2}
\end{array}
\right) .
\end{equation*}
A little calculation shows that
\begin{eqnarray*}
AA^{\ast }f_{1} &=&\lambda _{{}}^{2}f_{1} \\
A^{\ast }Af_{2} &=&\lambda ^{2}f_{2}
\end{eqnarray*}
Note that some of the eigenvalues may be degenerate (have multiplicity $>1$
). Then we denote the eigenprojector corresponding to $\lambda $ by $
P_{\lambda }$.
An elegant method to treat the perturbation problem is by Kato (T. Kato,
Perturbation Theory for Linear Operators). Thus let
\begin{equation*}
R_{\alpha }(z)=[z-H_{\alpha }]^{-1}
\end{equation*}
be the resolvent of $H_{\alpha }$ and
\begin{equation*}
R_{0}(z)=[z-H_{0}]^{-1}
\end{equation*}
It is analytic outside the spectrum of $H_{\alpha }$ which consists of the
finite set of eigenvalues of $H_{\alpha }$ (which are real). We can write
\begin{equation*}
R_{\alpha }(z)=R_{0}(z)[1-\alpha VR_{0}(z)]^{-1}
=R_{0}(z)[1-{\alpha}VR_{0}(z)]+\alpha ^{2}\{VR_{0}(z)\}^{2}-\alpha ^{3}\{VR_{0}(z)\}^{3}+\cdots
],
\end{equation*}
which is a series expansion in $\alpha $. Let now $\Gamma $ be a contour
that wraps around the eigenvalue $\lambda $ of $H_{0}$ (with associated
eigenprojector $P_{\lambda }$) but avoids its other eigenvalues. Then
\begin{equation*}
P_{a}=\frac{1}{2\pi i}\int_{\Gamma }dzR_{\alpha }(z)
\end{equation*}
is a projector and equals the sum of all eigenprojectors of $H_{\alpha }$
inside $\Gamma $,
\begin{equation*}
P_{a}=\sum_{n}P_{\alpha }^{(n)}
\end{equation*}
whereas ($\lambda _{\alpha }^{(n)}$ is the eigenvalue associated with $%
P_{\alpha }^{(n)}$)
\begin{equation*}
\frac{1}{2\pi i}\int_{\Gamma }dzzR_{\alpha }(z)=\sum_{n}\lambda _{\alpha
}^{(n)}P_{\alpha }^{(n)}.
\end{equation*}
Introducing the perturbation expansion, above, we have
\begin{eqnarray*}
P_{\alpha } &=&\frac{1}{2\pi i}\int_{\Gamma }dz\{R_{0}(z)-\alpha
R_{0}(z)VR_{0}(z)\}+\mathcal{O}(\alpha ^{2}) \\
&=&P_{\lambda }-\alpha \frac{1}{2\pi i}\int_{\Gamma }dzR_{0}(z)VR_{0}(z)+
\mathcal{O}(\alpha ^{2}) \\
&=&P_{\lambda }+\alpha P^{(1)}+\mathcal{O}(\alpha ^{2}).
\end{eqnarray*}
In case the eigenvalue problem for $H_{0}$ is solvable we can then calculate
$P^{(1)}$ and, to first order in $\alpha $, the $\lambda _{\alpha }^{(n)}$.
Note that the original eigenvalue $\lambda $ can split up. Note further that
truncation after the first order in $\alpha $ can lead to a result that is
no longer a projector. 
A full exposition, including the mathematical details, can be found in
Kato's book.
A: Here is an intuitive approach:
Note that $\det(A-\lambda I) = \det(A-\lambda I)^T = \det(A^T-\lambda I)$ and so if $\lambda$ is an eigenvalue of $A$ then it is also an eigenvalue of $A^T$. Now, let $v,w \neq 0$ be such that $A^Tv = \lambda v$ and $Aw = \lambda w$, so we have
$$\begin{pmatrix} \epsilon_1  & A \\ A ^T  & \epsilon_2 \end{pmatrix}\begin{pmatrix} v \\ w \end{pmatrix} =\begin{pmatrix} \epsilon_1 v+ A w \\ A^T v+ \epsilon_2 w \end{pmatrix} =\begin{pmatrix} \epsilon_1 w+ \lambda v \\ \lambda w+ \epsilon_2 v \end{pmatrix} %\approx \lambda \begin{pmatrix} v \\ w \end{pmatrix}$$
Now let $\mu_{i}$ be the largest eigenvalue of $\epsilon_i$ and assume $\mu_i \ll \lambda$, then we have 
$$\|\epsilon_i u \| \leq \mu_{i} \|u\| \ll \lambda \|u\|,$$ 
it follows that for any $x,y$ with $\|x\| = \|y\|$, we have
$$\|\epsilon_i x + \lambda y\| \leq \mu_{i} \|x\|+\lambda \|y\| \approx \|\lambda y\|.$$
and thus it seems reasonable to assume that
$$\epsilon_1 v+ \lambda w \approx \lambda w, \quad \text{ and } \quad  \lambda v+ \epsilon_2 w \approx \lambda v.$$
From which follows that 
$$\begin{pmatrix} \epsilon_1  & A \\ A ^T  & \epsilon_2 \end{pmatrix}\begin{pmatrix} v \\ w \end{pmatrix} \approx \lambda \begin{pmatrix} v \\ w \end{pmatrix}.$$
