1
$\begingroup$

$$D=X-M(X\circ P)$$ Solve for $X$

where

$D, X, P$ are vectors of size $n$.

$M$ is a matrix of size $n \times n$.

$X\circ P$ indicates element wise multiplication of $X$ and $P$ such that $(X \circ P)_i = X_i P_i$

I am aware that in general, some values of $M$ would support multiple solutions of $X$. But I suspect that in my cases there are unique solutions, so a strategy that finds any solution in the general case would suffice. Even some iterative algorithm would be acceptable.

For extra credit, is it possible to determine, based on $M$, when a unique solution exists?

$\endgroup$
2
  • 4
    $\begingroup$ If you consider the data of $P$ as "inside your matrix" $M$, you can go back to the standard theory of linear systems. Explicitly, $X\circ P$ is equal to the matrix multiplication $diag(P) X$, where $diag(P)$ is simply the diagonal matrix with entries the coefficients of $P$, so you're asking how to solve $D= (I_n - M diag(P))X$. $\endgroup$ Commented Apr 26 at 9:10
  • 1
    $\begingroup$ @AntonioLorenzin this is an awesome trick to know. I should be able to take it from here. $\endgroup$ Commented Apr 26 at 9:26

1 Answer 1

1
$\begingroup$

Rewrite your equation as $D= X - M \mathcal{D}_P X$ where $\mathcal{D}_P = \begin{pmatrix} p_1 & 0 & \ldots \\ 0 & p_2 & \ldots \\ \ldots & 0 & p_n \end{pmatrix}$

So the problem becomes a standard system of equations $D=(I-M\mathcal{D}_P)X$.

If $I-M\mathcal{D}_P$ is invertible, you find $X= (I-M\mathcal{D}_P)^{-1}D$. In this case (1 is not an eigenvalue of $M\mathcal{D}_P$) you have unique solution of the problem.

Otherwise, use standard linear algebra theorems to find solutions (if any, they will be infinite).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .