I'm trying to solve the following problem:
$$(\mathbf{Ax}-\mathbf{F})\odot(\mathbf{Bx}-\mathbf{G})=\mathbf{0},$$
where $\odot$ is the Hadamard product. We also have the following restrictions:
$$\begin{align}\mathbf{Ax}-\mathbf{F}&\geq \mathbf{0}\\ \mathbf{Bx}-\mathbf{G}&\geq \mathbf{0}\end{align}.$$
I understand the logic of this problem and even i'm able to solve it if $\mathbf{x}=(x_1,x_2)'$. However, i don't know how to solve it for higher dimensions. For the 2 dimensional case i provide a minimal example:
$$\left[\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}-\begin{pmatrix}2\\2\end{pmatrix}\right] \odot \left[\begin{pmatrix}4&6\\8&10\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}-\begin{pmatrix}3\\3\end{pmatrix}\right]=\begin{pmatrix}0\\0\end{pmatrix}$$ In this case we can rewrite the problem in the following form:
$$\begin{align}&\max\left\{ x_1 + 2x_2-2,4x_1+6x_2-3\right\}\\&\max\left\{ 3x_1 + 4x_2-2,8x_1+10x_2-3\right\} \end{align},$$
or, if we express $x_2=f(x_1)$:
$$x_2=\max\left\{\frac{(2-x_1)}{2},\frac{(3-4x_1)}{6}\right\}=\max\left\{\frac{(2-3x_1)}{4},\frac{(3-8x_1)}{10}\right\}.$$
The problem is solved graphically. Solution is $(x_1,x_2)=(-2,2)$.
The question is: do you know a methodology to solve higher order problems of this type?
Thanks,
Alejandro
For ease of typing, let's use lowercase letters ($f,g$) to denote vectors instead of uppercase ($F,G$).
Then the equation can be written as $$\eqalign{ (Ax-f)\circ(Bx-g) &\ge 0 \cr\cr }$$ If either of the matrices $(A,B)$ are invertible, then there may be trivial solutions to the problem.
If $A^{-1}$ exists, then $$x = A^{-1}f$$ trivially satisfies the problem and at least one of the constraints.
Likewise $$x = B^{-1}g$$ solves the problem and one of the constraints.
In your example, I noticed that the solution that you found is of the first type. I also noticed that all of the elements of both matrices and vectors, are all positive. And that the matrices are square. Are these typical of your problem space? Because if the matrices are underdetermined (more columns than rows) then there may be another type of solution to the problem.