Consider the equation $$\pmb{c}+\text{sign}(G\pmb{c}) = L$$


$\pmb{c}$ is a $n\times1$ matrix.


$G$ is a $n\times n$ matrix which is also positive definite.


matrices $G$ and $c$ are real.


$L$ is a $n\times 1$ matrix whose entries are from the set $\{-1,1\}$.


Can this equation be solved for the matrix $\pmb{c}$?
Suppose there is no signum function there, then the solution is $$\pmb{c} = (G+I\_n)^{-1}L$$
and as $G$ is positive definite, there exists a unique solution. But with the presence of the $\text{sign}$ function, the problem doesn't seem to belong to linear algebra. I request for a reference to any subject or book for this type of equations. Does this belong to linear programming?(I don't know anything about it, so I am hoping it is related).


PS: please tag appropriately.


PS 2: $\text{sign}([a\_{i,j}]\_{m\times n}) = [\text{sign}(a\_{i,j})]\_{m\times n}$.



An otherwise linear matrix equation with the presence of a signum function : reference request
You can solve the problem via mixed integer linear programming as follows. Let $\epsilon>0$ be a small constant tolerance. For $i \in \{1,\dots,n\}$, let $[\ell\_i,u\_i]$ be constant lower and upper bounds on $(Gc)\_i$, and let binary decision variables $z^-\_i$ and $z^+\_i$ indicate whether $(Gc)\_i<0$ or $(Gc)\_i>0$, respectively, so $\text{sign}(G\pmb{c})=\pmb{z^+}-\pmb{z^-}$. The constraints are:
\begin{align}
c\_i - z^-\_i + z^+\_i &= L\_i &&\text{for all $i$} \tag1\\
z^-\_i + z^+\_i &\le 1 &&\text{for all $i$} \tag2\\
\ell\_i z^-\_i + \epsilon z^+\_i \le (Gc)\_i &\le -\epsilon z^-\_i + u\_i z^+\_i &&\text{for all $i$} \tag3
\end{align}
Constraints $(2)$ and $(3)$ together enforce three cases:
\begin{align} 
(z^-\_i,z^+\_i)=(1,0) &\implies (Gc)\_i\le -\epsilon < 0 \\
(z^-\_i,z^+\_i)=(0,0) &\implies (Gc)\_i = 0 \\
(z^-\_i,z^+\_i)=(0,1) &\implies (Gc)\_i\ge \epsilon > 0
\end{align}


