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I have the following vector equation: $$ {\bf Ax} + {\bf b} + {\bf Cx}^{ \circ - 1} = {\bf 0}_n $$ Where ${\bf x}$ is a vector of unknown variables. ${\bf b},{\bf x}, {\bf x}^{\circ - 1}, {\bf 0}_n \in \mathbb{R}^{n}$; ${\bf A},{\bf C} \in \mathbb{R}^{n \times n}$ and $\bf C$ is a diagonal matrix. ${\bf x}^{\circ - 1}$ is the Hadamard (pointwise) product inverse of ${\bf x}$, that means ${\bf x} \circ {\bf x}^{\circ - 1} = {\bf 1} \in \mathbb{R}^{n}$. I thought about converting the equation into a quadratic vector equation and ended with something like this: $$ {\bf x} \circ \left({\bf Ax}\right) + {\bf Bx} + {\bf c} = {\bf 0}_n $$ Where ${\bf B} \in \mathbb{R}^{n \times n}$ is a diagonal matrix whose diagonal entries are the same of ${\bf b}$ and ${\bf c} \in \mathbb{R}^{n}$ is a vector whose entries are equal to the diagonal entries of ${\bf C}$. I have not progressed any further than that. Is there any way of analytically solving this equation? whether by doing some kind of "completing the square" or any other method.

NOTE: If necessary it can be assumed that the entries of ${\bf A}$ and ${\bf C}$ are non-negative, and that ${\bf A}$ and ${\bf C}$ are invertible.

EDIT: Another equivalent equation is: $$ \left( {\bf X}^T{\bf AX} + {\bf X}^T{\bf B} + {\bf C} \right) {\bf 1}_n = {\bf 0}_n $$ Where ${\bf 1}_n \in \mathbb{R}^{n}$ is a vector of $1$'s and ${\bf X}$ is a diagonal matrix whose diagonal entries are the same of ${\bf x}$.

EDIT2: Second equivalent equation is: $$ \left({\bf Ax}\right) \circ \left({\bf Ax}\right) + {\bf ABx} + {\bf Ac} = {\bf 0}_n $$ But still, no progress have been made.

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  • $\begingroup$ I am tempted to vote to close, but I cannot find a reason. Unfortunately "asking for something akin to magic" is not offered as a choice. $\endgroup$ Commented Mar 20, 2017 at 0:49
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    $\begingroup$ I am not supposing an analytical solution exists, that is in fact my question. For me an answer like "the equation can be solved using X" is as valid as "no closed solution exists because of Y" or "your question cannot be answered because what is known about that type of equation is Z". I am sorry if I am asking for something akin to magic, but I am not that knowledgeable about linear algebra. $\endgroup$ Commented Mar 20, 2017 at 2:27
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    $\begingroup$ You could look into solving a system of quadratic equations (for edit2); there are a few solvers (e.g., Bertini) for numerically solving systems of polynomial equations. (I am not sure that all your formulations are equivalent though! e.g., in edit2, if $A=0$, then $x=0$ provides a solution, whereas this is not the case for the other versions). $\endgroup$
    – Suvrit
    Commented Mar 20, 2017 at 3:57
  • $\begingroup$ @Suvrit you are right. The equation of "edit2" is obtained by doing a linear combination of the scalar equations of the original vector equation. I suspect the equations are equivalent as long as A is non-singular. As for the numerical solvers, I will consider it. It seems like an analytic solution is a no-go, but I do not want to lose hope. $\endgroup$ Commented Mar 20, 2017 at 4:15
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    $\begingroup$ As I said earlier, I do not much about linear algebra. When I reached this equation for the first time I thought it would be possible to solve it, but I failed and ended up asking the question here. As I started to read more about the topic I realised that's probably not the case. $\endgroup$ Commented Mar 20, 2017 at 21:22

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