I am trying to understand the proof of Tucker's theorem using Farkas lemma but there are some points that are not clear to me.
The proof I am following is in this paper at page 16.
What I do not understand is:
(1) how the system $S_1$ and $S_2$ are formulated in relation to the ones in the Farkas alternative lemma;
(2) in the following proof, Farkas lemma is applied yielding the solution vector $x^i$ but the statement is obtained from the vector $\overline{x}$. Why are we allowed to take the sum of the $x_i's$ and use the result?
Tucker's theorem:
Let $M$ be any $R^{n×n}$ skew-symmetric matrix (i.e. $L^T = −L$).
The set of solutions
$S := \{x ∈ R^n |\ Mx \succeq 0, x \succeq 0\}$
always has a non-degenerate solution of the system, i.e. a solution $\overline{x} ∈ S$ s.t. for each component $i = 1, . . . , n$ either $( M\overline{x} )_i\succ0 $ or $\overline{x}_i\succeq 0$ $ $ holds (i.e. each component satisfies at least one inequality strictly). Formally: $∃\overline{x} ∈ S$ s.t.: $\overline{x}+ M\overline{x} \succ 0$.
Proof:
Let $L$ be an arbitrary skew-symmmetric matrix, of order $n$; let $e ^i$ be the $i_{th}$ unit coordinate vector and $I$ the identity matrix of order $n$: By Farkas lemma, either the system $$S_1=\{-L^Tx \succeq\underline{0} ,\ Ix \succeq \underline{0} ,\ (-e^i)^Tx < 0\}$$
has a solution, or the system
$$\{S_2= Lv-Iz = e^i,\ v\succeq 0 ,\ z\succeq 0\}$$
has a solution $y$, with $y^T = (v^T; z^T)$ but never both.
In the first case, taking into account the equality $L^T= -L$, we have a solution $x ^i$ of $S_1$ for which $\{Lx^i \succeq\underline{0} , x^i \succeq\underline{0} , x^i_i>0\}$.
In the second case, we have a vector $v^i$ for which $\{Lv^i \succeq e^i, v^i \succeq\underline{0}\}$.
Therefore, in either case there exists a vector $x^i$ for which $Lxi + xi \succeq\underline{0}$ and the $i_th$ component of the vector on the left-hand side is positive. Thus, for the vector
$$\overline{x}:=\sum_{i=1}^nx^i$$
it holds that: $$L\overline{x}+ \overline{x}\succ \underline{0}.$$
I hope my questions were clear enough.