For $A \in M _ { n , m } ( \mathbb { R } )$ and $b \in \mathbb { R } ^ { n }$ as in Farkas' lemma, we set $$B = \left( \begin{array} { c c c c }
0 & 0 & A & - b \\
0 & 0 & - A & b \\
- { } ^ { t } A & { } ^ { t } A & 0 & 0 \\
{ } ^ { t } b & - { } ^ { t } b & 0 & 0
\end{array} \right)$$ Let, by Tucker's theorem, $y = { } ^ { t } \left( z _ { 1 } , z _ { 2 } , x , t \right) \geq 0$ such that $B y \geq 0$ and $y + B y > 0$. Show that if $t > 0$ then for $z = z _ { 1 } - z _ { 2 }$, we have $- { } ^ { t } A z \geq 0$ and $( b \mid z ) > 0$.
For $A \in M _ { n , m } ( \mathbb { R } )$ and $b \in \mathbb { R } ^ { n }$ as in Farkas' lemma, we set
$$B = \left( \begin{array} { c c c c }
0 & 0 & A & - b \\
0 & 0 & - A & b \\
- { } ^ { t } A & { } ^ { t } A & 0 & 0 \\
{ } ^ { t } b & - { } ^ { t } b & 0 & 0
\end{array} \right)$$
Let, by Tucker's theorem, $y = { } ^ { t } \left( z _ { 1 } , z _ { 2 } , x , t \right) \geq 0$ such that $B y \geq 0$ and $y + B y > 0$.
Show that if $t > 0$ then for $z = z _ { 1 } - z _ { 2 }$, we have $- { } ^ { t } A z \geq 0$ and $( b \mid z ) > 0$.