Tucker's theorem from Farkas lemma

Question

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.

Answer 1

Below is a rewording of the argument that's hopefully more transparent than the one in the paper. I'll use the symbols $\ge $ and $>$ for componentwise inequality. Recall that Farkas' lemma may be stated as follows:

Suppose that $A$ is an $m\times n$ matrix, and that $b \in \mathbb{R}^m$. Then either there is some $x\ge 0$ such that $Ax = b$, or there is some $p$ such that $p^\top A \ge 0$ and $p^\top b <0$.

If $L$ is a $k\times k$ matrix, choose $A$ to the be $k\times 2k$ matrix $A=[\,\,-L \,\,|\,\, I\,\,]$, and take $b = -e_1$ for specificity. Then Farkas' lemma tells us that exactly one of the following situations holds:

(a) There is some $x\ge 0$ such that $[\,\,-L \,\,|\,\, I\,\,] \,x = b$. In this case, write $x$ as a concatenation of its first $k$ coordinates and last $k$ coordinates, $x = (v,z)$. Notice that $v, z \ge 0$ since $x\ge 0$, and that we have $-Lv + Iz = -e_1 \iff Lv -z = e_1 \iff Lv = e_1 + z$. This last expression shows us that $Lv > 0$, and adding this to $v\ge 0$, we obtain $Lv+v >0$. So $v$ is our desired vector.

(b) There is some $p$ such that $p^\top [\,\,-L \,\,|\,\, I\,\,]\ge 0$ and $p^\top (-e_1) < 0$. Rewrite the constraints to give $[-p^\top L \,\,|\,\, p^\top\,\,] \ge 0$ and $p_1 >0$. Using skew-symmetry, the first constraint holds iff $Lp \ge 0$ and $p\ge 0$. Now, the constraints $p\ge 0$ and $p_1 >0$ imply that $p >0$, so adding inequalities, we see that $Lp + p>0$, and thus $p$ is our desired vector.

In either case, there's a vector $w\ge 0$ such that $Lw \ge 0$ and $Lw+w >0$, so we're done.