# When is there a solution to these coupled eigenvalue equations?

I am trying to find the fixed point of a dynamical system, which requires solving two coupled eigenvalue-like equations. These equations are, in general, overconstrained. I'd like to have a simple criterion for whether or not there will be a solution.

Assume that there are two given matrices, $A$ and $B$. I want to find a diagonal matrix $X$ and a nonzero vector $y$ such that the following two conditions are satisfied: \begin{align} X A y &= y \\ (A - X B)y &= 2y \end{align} (All matrices are of size $n \times n$, and $y$ is of size $n$.) Please feel free to suggest any reasonable simplifying assumptions if it helps solve the problem.

Here is a hand-wavy argument to suggest that the problem is slightly overconstrained. I have $2n$ free variables, and the two conditions above impose $2n$ constraints, suggesting that there is generally a solution. However, they have the obvious solution $y=0$, which I am excluding, so it's almost as if there are $2n+1$ constraints. Therefore, it seems like solutions to this problem should be possible if one extra condition is satisfied by $A$ and $B$, which would make one of the constraints degenerate. (Yes, this is extremely wishy-washy and possibly wrong.)

I posted a related question here in the hope that it would give me some insight into this problem, but so far I am still stuck. Thanks in advance for any ideas!

• maybe you can require the vector $y$ to satisfy $\|y\|=1$; that will exclude the $y=0$ solution... – Suvrit Oct 10 '13 at 13:53
• Of course. There are many ways to impose the $f \ne 0$ condition, but the main point is that it makes the system slightly overconstrained. – sasquires Oct 10 '13 at 19:43
• Actually, what I was hoping with $\|y\|=1$ is flip it around into a fixed-point iteration so that we can invoke Brouwer's theorem to guarantee existence of $X$ and $y$... – Suvrit Oct 10 '13 at 19:54

I tried the $n=2$ case, using Maple to find a "plex" Groebner basis for the equations. The result is rather complicated, and too large to show here. The first member of the basis, for example, is $y_1$ times an irreducible polynomial of $143$ terms and total degree $8$ in the entries of $A$ and $B$, which must be $0$ in order to have a solution with $y_1 \ne 0$. So I don't think you'll get a "simple" criterion.
EDIT: You want the $2n \times n$ matrix $$\pmatrix{X A - I\cr A - X B - 2 I\cr}$$ to have rank $< n$. That is equivalent to all its ${2n} \choose n$ $n \times n$ submatrices (obtained by selecting $n$ of the rows) having determinant $0$. Each of those determinants is a polynomial ...
• It's occurs to me that you can construct a scary-looking polynomial in all eight variables just using simple ingredients like $\det(A)$ or $\det(B-I)$ or something, and we could be seeing a complicated result that has a simple origin. But that still doesn't bring us any closer to a solution. – sasquires Oct 11 '13 at 14:54