I have a matrix equation of the form:

$$ A^{-1} = B + A \circ C $$

where $\circ$ denotes the Hadamard product (i.e., $(A\circ C)_{ij} = A_{ij}B_{ij}$). How can I determine if a solution for $A$ exists, if it's unique, and ultimately solve for $A$?

Edit: I'm only interested in the case of real valued matrices.


You can rewrite it without the inverse as $$A(B+A\circ C)=I. \tag{*}$$ Since this is quadratic in $A$, I can imagine that without other constraints there will be several real solutions (up to $2^n$).

Do you have maybe other structural properties on $B$ and $C$? Such as symmetry, low rank, or positiveness of their elements?

For instance (shameless self-advertising), I have studied in a wide class of matrix equations that includes (*) under constraints on the positivity of their entries --- for this equation, if I am not mixing up things, these constraints would correspond to $C\leq 0$, $B^{-1}\geq 0$ (elementwise).

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  • $\begingroup$ I require that $A$ be positive semidefinite. I have a vague hope that this is enough to make the solution unique, but if not I suspect that any solution would be equally useful for my application (a metric learning algorithm). I glanced through the paper you link. It's not obvious to me how to rewrite this equation into a suitable form, and I don't think my $B$ and $C$ obey any elementwise positive constraints. $\endgroup$ – Mike Izbicki Oct 16 '13 at 19:39
  • $\begingroup$ Then probably the theory there doesn't apply to your case. The constraint that $A$ is SPD looks unusual -- typically in matrix eqns when one looks for symmetric solutions the equation itself has some "natural" symmetry, but this is not your case. $\endgroup$ – Federico Poloni Oct 16 '13 at 21:31
  • $\begingroup$ If you are fine with any solution, a simple experiment could be running a fixed point iteration such as $X_{k+1}=(B+X_k\circ C)^{-1}$ (or Newton's method, if you want to be more sophisticated) and see if it leads you somewhere. $\endgroup$ – Federico Poloni Oct 16 '13 at 21:33
  • $\begingroup$ I've tried hammering my equation into a form suitable for your method, but unfortunately I don't think I can get it there. I did manage to remove the Hadamard product though, so I've asked another question here: mathoverflow.net/questions/145225/… . $\endgroup$ – Mike Izbicki Oct 19 '13 at 0:44
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    $\begingroup$ @DaeyoungLim It's possible that there is one, but I don't know of any result that applies to that case. Matrix equations with Hadamard products are not very commonly studied, as far as I know. $\endgroup$ – Federico Poloni Aug 25 '16 at 17:07

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