Farkas' Lemma and Theorem of Alternatives Let's define the DUAL CONE of a subset Y of $\mathbb{R}^n$ as follows: $$Y^*=\{x\in\mathbb{R}^n\mid \langle x,y\rangle \ge 0\quad\forall y\in Y\}.$$ If $Y$ is a finite subset of $\mathbb{R}^n$, i.e. if there are $v_1,\dots,v_m\in\mathbb{R}^n$, with $m\in\mathbb{Z}^+$ such that $Y=\{v_1,\dots,v_m\}$, then I will write $(v_1,\dots,v_m)^*$ to denote the dual cone of $\{v_1,\dots,v_m\}$.
Now I'd like to prove that $((v_1,\dots,v_m)^*)^*=\{\sum_{i=1}^m\lambda_iv_i\mid \lambda_i\in\mathbb{R}^\ge\quad\forall i=1,\dots, m\}$ by using Theorem of Alternatives. The set on the right hand side of the equality is called the CONVEX CONIC HULL of $\{v_1,\dots,v_m\}$ and it is the set of all conic combinations (linear combinations with nonnegative scalars) of the vectors $v_1,\dots, v_m$.
Theorem of Alternatives
The Theorem of Alternatives states the following:

Let $A$ be a $m\times n$ matrix, $B$ a $p\times n$ matrix, $C$ a $q\times n$-matrix with $A\ne 0$. Let's consider the system
$$S_1:\; \begin{cases}
Ax<0\\
Bx\le 0\\
Cx=0
\end{cases}$$
The system $S_1$ has no solutions in $\mathbb{R}^n$ IF AND ONLY IF $\exists \theta\in(\mathbb{R}^\ge)^m, \lambda\in(\mathbb{R}^\ge)^p,\mu\in\mathbb{R}^q$ such that $A^T\theta+B^T\lambda+C^T\mu=0\quad$ (system $S_2$).

My idea is to consider the matrix $M$ having as rows the vectors $v_1,\dots, v_m$ and to notice that $(v_1,\dots,v_m)^*=\{x\in\mathbb{R}^n\mid Mx\ge 0\}$ but I don't know how to continue. Can anyone help me, please?
 A: From your reply to my comment, I gather that the result you're looking for is
$$
\big(v_1,v_2,\dots,v_m\big)^{**}=\big\{\sum_\limits{i=1}^m\lambda_iv_i\,\big| 
\,\lambda_i\in\mathbb{R}^\ge\ \forall i=1,\dots, m\big\}\ ,
$$
or, equivalently, since you've already shown that $\ \big(v_1,v_2,\dots,v_m\big)^*=\big\{x\in\mathbb{R}^n\,\big|\,Mx\ge0\,\big\}\ $,
$$
\big\{x\in\mathbb{R}^n\,\big|\,Mx\ge0\,\big\}^*=\big\{\sum_\limits{i=1}^m\lambda_iv_i\,\big| 
\,\lambda_i\in\mathbb{R}^\ge\ \forall i=1,\dots, m\big\}\ .
$$
If $\ y\in\big\{\sum_\limits{i=1}^m\lambda_iv_i\,\big| 
\,\lambda_i\in\mathbb{R}^\ge\ \forall i=1,\dots, m\big\}\ $, then $\ y^T=\lambda^TM\ $ for some $\ \lambda\in\mathbb{R}^m\ $ with $\ \lambda\ge0\ $. So if $\ x\ $ is any vector in $\ \big\{x\in\mathbb{R}^n\,\big|\,Mx\ge0\,\big\}\ $, then $\ \langle y,x\rangle=\lambda^TMx\ge0\ $ because both $\ \lambda\ge0\ $ and $\ Mx\ge0\ $. Therefore $\ y\in$$\big\{x\in\mathbb{R}^n\,\big|\,Mx\ge0\,\big\}^* $, and it follows that
$$
\big\{\sum_\limits{i=1}^m\lambda_iv_i\,\big| 
\,\lambda_i\in\mathbb{R}^\ge\ \forall i=1,\dots, m\big\}\subseteq\big\{x\in\mathbb{R}^n\,\big|\,Mx\ge0\,\big\}^*
$$
Now let $\ a\in\big\{x\in\mathbb{R}^n\,\big|\,Mx\ge0\,\big\}^*\ $. Then for every $\ x\in\mathbb{R}^n\ $ with $\ Mx\ge0\ $ we must have $\ 0\le\langle a,x\rangle=a^Tx\ $. This means that the system of inequalities
\begin{align}
a^Tx&<0\\
Mx&\ge0
\end{align}
can have no solution.  Can you finish it off from here?  I think the easiest way is to appeal to the form of Farkas's lemma as it's given in Wikipedia.  In this connection, there appears to be something missing from the system $\big(S_2\big)$  as given in your statement of the theorem of alternatives (I'm guessing that $\ \theta\ $ should be required to be non-zero).  As given, the system $\big(S_2\big)$ always has the solution $\ \theta=0, \lambda=0, \mu=0\ $.
