Null Variable in linear programming 


Let $P=\{x\in\mathbb{R}^n|Ax=b,x\geq0\}$ be a nonempty polyhedron, and let $m$ be the dimension of the vector $b$. We call $x_j$ a null variable if $x_j=0$ whenever $x\in P$.
  (b) prove that if $x_j$ is a null variable, then there exists some $p\in\mathbb{R}^m$ for which $p'A\geq 0', p'b=0$, and such that the $j$th component of $p'A$ is positive AND (c) if $x_j$ is not a null variable, then by definition, there exists some $y\in P$ for which $y_j>0$. Use the results in parts (a) and (b) to prove that there exist $x\in P$ and $p\in\mathbb{R}^m$ such that $p'A\geq0', p'b=0, x+A'p>0$

I did prove part (a) which was fairly easy to compute simply by multiplying $p'$ on the both sides of $Ax=b$. However I am struggling solving both (b) and (c). I am assuming $p\in\mathbb{R}^m$ is a dual variable and since $x_j$ is a null variable, $j$th constraint of dual can be removed. I also think that $j$th component of $p'A$ is positive because otherwise $x\notin P$, i.e. $Ax\neq b$. But I don't clearly see it. I am completely lost on part (c) as well. Any help would be very appreciated. 

For (a) and (b), the proof is similar to that of Farka's lemma.
Hint: Consider the dual of the LP
$$\max \, \left\{e_j'x : x \in P\right\}.$$
For (c), we can classify the indices into two groups, i.e. ones with null variables and the others without. For each $j$ such that $x_j$ is a null variable, we can find a $p^j$ such that the conditions are met. Now let $p$ be the sum of these $p^j$s. We have that
$$
p'A\ge 0', \quad p'b=0, \\ \quad (A'p)_j > 0, \; \forall j: x_j \, \text{is a null variable}.
$$
Finally, for each $i$ such that $x_i$ is not a null variable, there exists $y^i \in P$ such that $y^i_i > 0$. Let y be the average of these $y^i$s. We have that
$$
y \in P \; \text{and} \; y_i > 0, \; \forall i: x_i \, \text{is not a null variable,}
$$
and that $y + A'p > 0$.
$y$ and $p$ are as desired.
Note: If there are no null variables, we can simply set $p=0$. If all variables are null, any feasible $y$ will suffice. 
