Let $A \in \mathbb{R}^{m\times n}$ and $u, b \in \mathbb R^{m}$.
I am close to proving:
$A x =b$ has a solution $\iff$ $b^{T}u \leq 0$ and $A^{T}u=0$.
Using Farkas Lemma I get to the point where
$\begin{pmatrix} A^{T} \\ -A^{T} \end{pmatrix}^{T} x = -b$ has a solution for $x\geq 0.$ $(**)$
What transformation step do I need to show that $Ax=b$ for any $x \in \mathbb R^{ n}$?
My Proof (how I got to $(**)$):
Let $b^{T}u \leq 0$ and $A^{T}u=0$ have a solution then $b^{T}u > 0 $ with $A^{T}u=0$ $(D')$has no solution,
we then rewrite $(D')$ as $-b^{T}u < 0 $ and $\begin{pmatrix} A^{T} \\ -A^{T} \end{pmatrix}u\geq 0$. By the Farkas Lemma this means that $\begin{pmatrix} A^{T} \\ -A^{T} \end{pmatrix}^{T}x=-b, x\geq 0$ has a solution. And I know that
$\begin{pmatrix} A^{T} \\ -A^{T} \end{pmatrix}^{T}=(A,-A)$
But how can I manipulate $(A,-A)x=-b, x \geq 0$ to show that
$Ax=b$ for all $x$ ?
Another question that comes to mind is the validity of the equation $(A,-A)x=-b$ as $(A,-A)\in \mathbb R^{m\times 2n}$ while $x \in \mathbb R^{n}$, so $(A,-A)x$ does not make sense, or am I overthinking this?