I am having trouble understanding a version of a linear duality theorem which has popped up in my game theory class. It seems related to Farkas' lemma (which I understand) but I am struggling to connect the two.
Theorem. Let $A$ be an $m \times n$ matrix. Then, for any $\mathbf{b} \in \mathbb{R}^m$, exactly one of the following is true:
- There is some $\mathbf{x} \in \mathbb{R}^n$ s.t. $A\mathbf{x} \geq \mathbf{b}$
- There is some $y \in \mathbb{R}^m$ s.t. $y \geq \mathbf{0}$, $\mathbf{y}'A = \mathbf{0}$, and $\mathbf{y}'\mathbf{b}>0$.
(where for any two vectors $\mathbf{v}, \mathbf{w} \in \mathbb{R}^k$ we have that $\mathbf{v} \geq \mathbf{w} \iff v_i \geq w_i \: \forall i \in \{1,...,k\}$, and the same goes for $>,=$).
I have two questions. The first is how to relate this result to the more traditional statement of Farkas' lemma. The second is whether this theorem is even true.
For example, consider a matrix $A \in \mathbb{R}^{2 \times 2}$ whose first column has all positive entries and whose columns are all linearly dependent. Then taking any vector $\mathbf{b} \in \mathbb{R}^2$ which is not on the line spanned by $A$ (which passes through the positive quadrant and the negative quadrant), (1) clearly does not hold. However (2) does not hold either: any vector $\mathbf{y} \in \mathbb{R}^2$ which is orthogonal to the line spanned by $A$ must have a negative component.
EDIT: From @angryavian's comment, I can see why my example is not really a counterexample to the Theorem. So I suppose only my first question stands: what is the connection between this result and Farkas's lemma?