I've read the following exercise in my book: Let $A\in\mathbb R^{m\times n},b\in\mathbb R^m,c\in\mathbb R^n$.
Then exactly one holds:
$Ax=0,c^t\cdot x=1$ with $x\geq0$ has a solution
$A^ty\geq c$ has a solution
I've tried to prove it but having some troubles with the solution.
My attempt: Suppose 1. and 2. are right. Then $$y^t\cdot x\leq(A^ty)^t\cdot x=y^t\cdot Ax=0$$This is a contradiction. So both aren't solvable together.
So suppose 1. hasn't got any solution. How can you show 2. has a solution?
Thanks for helping!
Edit: I want to use Farkas' Lemma.
The statement is the direct application of the Farkas' lemma for some handy block system. Let $$\tag{1}\tilde{A}=\begin{bmatrix}A\\c^T\end{bmatrix}, \quad \tilde{b}=\begin{bmatrix}0_{m\times 1}\\1_{1\times 1}\end{bmatrix}. $$
Farkas' lemma says that one of the following is true:
There is an $x\geq 0$ such that $\tilde{A}x=\tilde{b}$. From (1), this is equivalent to the existence of $x\geq 0$ such that $$ Ax=0,\quad c^Tx=1. $$
There is a $\tilde{y}$ such that $\tilde{A}^T\tilde{y}\geq 0$ and $\tilde{b}^T\tilde{y}<0$. Partition $\tilde{y}$ as $\tilde{y}^T=[y^T,-\mu]$. Using (1), this alternative is equivalent to the existence of a $y$ and $\mu$ such that $$ A^Ty-\mu c\geq 0, \quad -\mu<0. $$ It follows that $A^Ty\geq \mu c$ and $\mu>0$. Hence setting $z=y/\mu$ gives $A^Tz\geq c$.