Consider the vector equality
$$x = A[v]_+ + B[-v]_+$$
where $[v]_+$ is the elementwise rectifier function, i.e.
$$[x]_+ = \begin{cases} x &\mbox{if } x > 0, \\ 0 & \mbox{otherwise}. \end{cases} $$
Question: Under what conditions on A and B does there always exist a solution $v$ for every $x$, i. e. when does
$$\forall x\in\mathbb{R}^N \,\,\exists v\in\mathbb{R}^M\quad x = A[v]_+ + B[-v]_+$$
hold?
For simplicity consider $N = M$ and $A, B$ full rank. In the case $N = M = 1$ we simply need $A*B < 0$. The case $N = M = 2$ is considerably harder: A trivial solution is $B = -A$, but from numerical tests I see that there exist many non-trivial solutions (I set A = Id, and draw B randomly from an iid Gaussian distribution, then about every third draw returns a valid B). I tested some intuitions (like eigenvalues of $AB^\top$) but none of them proofed sufficient (or even necessary) for a solution. Unfortunately, one cannot simply compute the optimal $v$ analytically (in terms of MSE), and then test for equality, so I am a bit stuck on how to handle that problem. Any hints are greatly appreciated!