Solving Coupled Eigenvalue equations I wish to solve the following set of coupled eigenvalue equations. How should I do it?
For real matrices $A$,$B$,$D$ and vectors $x \in R^m$, $y \in R^n$
$$
A x + B y = \lambda x
$$
$$
B^T x + D y = \mu y
$$
where, $A$ and $D$ are symmetric.
Background:
I am trying to solve the following optimization problem:
$$
\min x^T A x + y^T D y + x^T By
$$
$$
\textrm{such that } \, x^T x = 1 , y^T y = 1
$$
This leads to the above eigenvalue problem. $\lambda$ and $\mu$ are the Langrangian parameters for the constraints.
EDIT: I need to actually find numerical solutions for these equations given $A$, $B$, $D$
 A: If you write $I^m$ for an $m \times m$ identity matrix, you can make a big matrix equation out of it $$\left(\begin {array}{c  c}A-\lambda I^m & B\\B^T & D-\mu I^n \end {array}\right)\left(\begin {array}{c}x \\ y \end {array}\right)=0$$    which is a classic eigenvalue problem.  Chapter 11 of Numerical Recipes (free for the obsolete versions) or any numerical analysis text can help you.
A: This is not a complete solution... but maybe this will help. 
Write $u = \begin{pmatrix}x\\y\end{pmatrix}$. Then the objective can be written as $u^TQu$ where $Q = \begin{pmatrix}A & B/2\\B^T/2 & D\end{pmatrix}$. The constraints can be written as $u^TM_1u = 1$ and $u^TM_2u = 1$.
Now, for $\alpha\in(0,1)$, let $M = \alpha M_1 + (1-\alpha)M_2$. Form a constraint as $u^TMu = 1$. It is clear that the minimization with this constraint alone will give a lower bound on the required minimum value. This minimization can be done easily. Let $R = \sqrt{M}$, as $M$ is p.d., and $v = Ru$. Then, the constraint is simply $v^Tv = 1$ and the objective can also be written as $v^TQ'v$, where $Q' = R^{-1}QR^{-1}$. The solution to this is the minimum eigenvalue of $Q'$, which can be obtained by Ritz iterations etc.
Hence the supremum of these lower bounds give a lower bound for for the given minimization problem. The interesting thing to prove is that the supremum of these lower bounds for various $\alpha\in(0,1)$ is the required minimum. This can be shown easily when $B=0$, but I haven't been able to prove it in the general case.
A: Append x and y 
form z belonging to R (m+n) by appending y to x
So, now, you will get 
m * (m+n) matrix on the left hand side through 1st equation and n*(m+n) matrix on the left hand side through 2nd equation. 
your variable is of size m+n.
hence, append the two equations...
now, you get a square matrix on the LHS of size m+n *m+n -- solve
A: Perhaps adding the two equations in order to reeduce it to one equation with eqienvalue (lambda + mu) and Eqienvector x+y is the trick. Then find the inverse using a calculator or standard methods. But there is probably a trick involving the fact that the matrices satisfies the properties they do. 
