Is there a version of Gordan's Theorem in which $Ax=0$ has been replaced by $Ax=b$? That is, does the following (or something similar) hold for all $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}_+^m \setminus \{0\}$, or are there conditions on $A$ and $b$ that will make it true?

$$ \text{either} \quad \exists x \in \mathbb{R}_+^m\setminus\{0\} \centerdot Ax = b, \quad\text{or}\quad \exists y\in\mathbb{R}^n\centerdot A^\top y > 0, $$

Thanks.

Is there a version of Gordan's Theorem in which $Ax=0$ has been replaced by $Ax=b$? That is, I want a condition (possibly including conditions on $A$) for when $Ax=b$ has a solution $x \neq 0$, for $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}_+^m \setminus \{0\}$

Gordan's Theorem says that for all $A \in \mathbb{R}^{m \times n}$ we have $ \text{either} \quad \exists x \in \mathbb{R}_+^m\setminus{0} \centerdot Ax = 0, \quad\text{or}\quad \exists y\in\mathbb{R}^n\centerdot A^\top y > 0, $$

In particular, I cannot apply Farkas' Lemma to my problem because I want a condition with non-zero solutions.

Is there a version of Gordan's Theorem, stated below, that holds for the case $Ax=b$, where $b \in \mathbb{R}_+^m \setminus \{0\}$?

$$ \text{either} \quad \exists x \in \mathbb{R}_+^m\setminus\{0\} \centerdot Ax = 0, \quad\text{or}\quad \exists y\in\mathbb{R}^n\centerdot A^\top y > 0, $$

Thanks.

Is there a version of Gordan's Theorem in which $Ax=0$ has been replaced by $Ax=b$? That is, does the following (or something similar) hold for all $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}_+^m \setminus \{0\}$, or are there conditions on $A$ and $b$ that will make it true?

$$ \text{either} \quad \exists x \in \mathbb{R}_+^m\setminus\{0\} \centerdot Ax = b, \quad\text{or}\quad \exists y\in\mathbb{R}^n\centerdot A^\top y > 0, $$

Thanks.