Help understanding analytic proof of Farkas' lemma

I'm trying to understand the analytic proof of Farkas lemma (presented in a lecture) but I may have noted some steps wrong.

Farkas Lemma. Let $$A$$ be an $$m\times n$$ matrix with values in $$R$$ and $$b\in R^m$$. Then exactly one of the following holds

1. $$\exists x\in R^n$$ such that $$Ax=b,x\ge0$$
2. $$\exists y\in R^m$$ such that $$A^Ty\ge0,y^Tb<0$$.

Proof. The verification that both cannot hold together is straightforward. Proof that one statement must hold is, as follows.

Let $$v_1,v_2,...v_n$$ be column vectors of $$A$$ and let $$Q=\{s\in R^m:s=\sum_{i=1}^n\lambda_iv_i,\lambda_i\ge0\forall i\}$$. Suppose condition 1 doesn't hold, then we must show that 2 necessarily holds.
1 does not hold $$\Rightarrow\nexists x$$ such that $$Ax=b\Rightarrow b\notin Q$$. Also, $$0\in Q\Rightarrow Q\ne\emptyset$$.

I do not understand the steps below.

By separating hyperplane theorem (disjoint convex sets can be separated by a hyperplane) $$\exists \alpha\in R^m,\alpha\ne0$$ and $$\beta$$ such that $$\alpha^Tb>\beta~\text{ and }~\alpha^Ts<\beta\forall s\in Q\tag{*}$$ Also, $$\forall \gamma>0$$, $$\gamma v_i\in Q$$. Since $$0\in Q$$, $$\beta>0$$. $$\alpha^T(\gamma v_i)<\beta\forall\gamma>0\Rightarrow\alpha^T v_i<\frac{\beta}{\gamma}\Rightarrow\alpha^T v_i\ge0$$ Now, consider $$-\alpha=y\Rightarrow y^TA\ge0$$. From $$(*)$$, $$\alpha^Tb>\beta>0\Rightarrow y^Tb<0$$

UPDATE. Found a visualization of the proof here.

• Can you be more specific about what parts you don't understand? Do you understand the separating hyperplane theorem? Do you see how it gives you (*)? Apr 4 at 14:53
• @RobertIsrael I don't see how it gives the subsequent result. Also, fail to understand why $\beta>0$ and the next equation. Apr 4 at 15:15

By separating hyperplane theorem, we can find such a hyperplane that separates disjoint convex set. Our disjoint convex set is the point $$\{b\}$$ and $$Q$$ since we assume that $$[1]$$ doesn't hold.

The hyperplane can be written as $$\alpha^Tx=\beta$$ where $$\alpha \ne 0$$. The point $$\{b\}$$ and $$Q$$ lies on different sides of the hyperplane.

Hence, we can say that $$\alpha^Tb > \beta$$ and $$\alpha^Ts < \beta, \forall s \in Q$$.

Now, recall that we have $$0 \in Q$$, and since $$\alpha^Ts < \beta, \forall s \in Q$$, if we choose the particular $$s$$ to be $$0$$, then we have $$\alpha^T0 < \beta$$, hence we have $$\beta>0$$.

Now, if $$\gamma>0$$, we have $$\gamma v_i \in Q$$

$$\alpha^T(\gamma v_i) < \beta, \forall \gamma >0$$, hence we have $$\alpha^Tv_i <\frac{\beta}{\gamma}, \forall \gamma>0$$

There is a typo in the next line of reasoning.

Hence, by letting $$\gamma$$ to be arbitrarily large, the right conclusion should be $$\alpha^Tv_i \color{red}\le 0$$.

Now, we let $$y=-\alpha$$, then we have $$-y^Tv_i \le 0$$ or $$y^TA \ge 0$$.

From $$\alpha^Tb > \beta>0$$, we have $$-y^Tb > 0$$ or $$y^Tb<0$$.