Let $A\in \mathbb R^{m,n}$ then there exists and $x \in \mathbb R^n x \neq0 ;x \ge 0; Ax=0$ if and only if there exists no vector $y$ such that $y^tA \gt 0$
I do know Farkas Lemma but I dont know how to apply it to that specific case.
Would appreciate any help, hints.
The version of Farkas Lemma I know is the following with $A,x,y$ as above :
There exists and $x \ge 0$ such that $Ax=b$ if and only if there does not exist a vector $y$ such that $y^TA \ge 0$ and $b^ty\lt 0$
Edit: $x\ge 0 $ means $x_i\ge 0$ for every $i=1….n$
Best Answer
First note that all alternative theorems are very very special cases of separation theorem. So you can solve them using separation theorem. I myself have not memorized either of them, since they all saying separating two sets (usually cones) in different languages.
But here you want solve problem using Farkas Lemma. I'll give you hint:
Left to Right : If there exist $ x \in \mathbb R^n x \neq0 ;x \ge 0; Ax=0$ and there is $y \in \Bbb R^n$ such that $y^t A > 0 $ then $ 0 = y^t Ax > 0$ which is contradiction.
Right to Left: Assume the inequality system $ y^t A > 0 $ has no solution! Thus for all scalar $s > 0 $ the inequality system $y^t A - s e \ge 0$ has no solution, where $e= (1,1,...,1)$ therefore setting $z=\begin{pmatrix} y \\ -s \end{pmatrix} $ and $ \bar{A} =\begin{pmatrix} A\\ e \end{pmatrix}$ and $ b^t :=(0,0,0...,0,1)$, one can observe that in newly defined symbols we have the system $$ z^t \bar{A} \ge 0 , \quad b^t z <0 $$ has no solution, Now just apply farkas lemma to new system!