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I'm interested in the existence of a vector $x \in \mathbb{R}^{n, +}$ such that for a matrix $M \in \mathbb{R}^{mxn} $ and a vector $b \in \mathbb{R}^m $, $Mx=b$. I've already looked into Farkas' Lemma but that assumes a fixed $M$ and $b$. The only constraint I place on $b$ is that the elements in $b$ must follow a particular order, but not that they have specific values. For example, if $[1, 2, 3]$ is a valid vector $b$ then so is $[1, 1000, 2000]$.

As a more concrete example, consider the matrix $$M=\begin{bmatrix} 2 & .5 & 2 \\ 3 & 2 & 1 \end{bmatrix}$$ and the vector $$b_1=\begin{bmatrix}2 \\1 \end{bmatrix}$$

Then there does not exist a vector $x$ that solves the equation $Mx=b_1$ such that every element in $x$ is positive. However, the vector $$b_2=\begin{bmatrix}1.9 \\ 1\end{bmatrix} $$ has solutions $x$ where element is positive. An example is $$x=\begin{bmatrix} .025 \\ .000001 \\ .925 \end{bmatrix}$$

My question is whether there always exists an ordered $b$ such that there exists an $x$ with positive elements.

user9403
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    Could you rephrase your question? If i am understanding correctly, are you trying to solve a bunch of linear equations $Mx=b$ simultaneously? – Julian Mejia May 12 '19 at 19:31
  • No, I only want to solve one. I have a matrix $M$, and an ordered collection $b$. However, I'm free to choose the values of $b$ with the constraint that the elements of $b$ retain the same order. Put differently, I'm not given the values $b$, merely the order of the elements of $b$. – user9403 May 12 '19 at 20:27
  • I've updated the question to include an example. – user9403 May 12 '19 at 20:52

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