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Let $ \mathbf{b} \in \mathbb{R}_+^n$, $\mathbf{E} \in \mathbb{R}_+^{n \times m}$, $\mathbf{V} \in \mathbb{R}_+^{n \times m}$ with $\mathbf{E}^T \mathbf{1}_n = \mathbf{1}_m$ and $\mathbf{V} \mathbf{1}_m = \mathbf{1}_n$, and $\forall i, j \ \ E_{ij} \geq 0, \ \ V_{ij} \geq 0$ where $\mathbf{1}_n$ is the vector of ones of size $n$.

I have the following system of equations where I am trying to find a closed form solution for non-negative $\mathbf{p}$ (i.e., $\mathbf{p} \in \mathbb{R}_+^m$): \begin{align} \mathbf{1}_m = \mathbf{V}^T D^{-1}(\mathbf{V} \mathbf{p}) \mathbf{E} \mathbf{p} \end{align} where $D(\mathbf{x})$ is the diagonal matrix whose $(i, i)^{th}$ entry corresponds to $x_i$ and $D^{-1}(\mathbf{x})$ is the inverse of $D(\mathbf{x})$.

Is it possible to solve for non-negative $\mathbf{p}$ (i.e., $\mathbf{p} \in \mathbb{R}_+^m$) in polynomial time? What would be such an algorithm? Otherwise, can I find an approximation in polynomial time?

Denizalp
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  • Your notation isn't consistent in that $Vp$ is an $n$ by $1$ column vector and $E$ is an $n$ by $m$ vector. – Brian Borchers Jul 08 '20 at 23:23
  • Thanks for the comment, $D(\mathbf{x})$ is the diagonal matrix whose diagonal entries are those of $\mathbf{x}$. I edited the text for clarity. I hope this makes it clear. – Denizalp Jul 08 '20 at 23:35
  • It's still the case that $(Vp)Ep$ isn't properly conforming matrix multiplication. – Brian Borchers Jul 08 '20 at 23:37
  • $D^{-1}(\mathbf{V}\mathbf{p})$ is a diagonal matrix of size $n \times n$. In particular it is the inverse of the diagonal matrix whose diagonal entries correspond to $\mathbf{V}\mathbf{p}$. Is there a better way for me to express this? – Denizalp Jul 08 '20 at 23:39
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    It would be clearer if you wrote this as $\mbox{diag}(Vp)^{-1}$. It's clear enough now. – Brian Borchers Jul 09 '20 at 02:01
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    The variable to solve $p$ is in the numerator and denominator like this problem: https://math.stackexchange.com/questions/3739904/recovering-three-dimensional-vectors-after-projection-and-cross-product –  Jul 09 '20 at 13:36
  • Thanks, this might be a great starting point for me! – Denizalp Jul 11 '20 at 19:14

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