asymmetric Dual of the asymmetric Dual is Primal? It is known that in Linear Programming the Dual of the Dual is the Primal.
In wikipedia I saw that apart from the symmetric Dual I knew, there is also the asymmetric one:
https://en.wikipedia.org/wiki/Dual_linear_program#Vector_formulations
I tried to show that the "Dual of the Dual is the Primal" would still hold in this case but I didn't manage to.
My question is: Does the "Dual of the Dual is the Primal" still holds for the asymmetric case? And if yes, could you please provide a comprehensive proof ?
 A: Okey. I managed.
The asymmetric duality says:
Primal 
max ${c^Tx}$
s.t. $Ax ≤ b$
produces the folowing
Dual
min $b^Ty$
s.t. $A^Ty = c$
$y ≥ 0$
I want to prove that I can go back to Primal from the Dual applying only the knowledge of this asymmetric transformation.
First I need to bring the Dual in the 'standard' form as in Primal:
$$
max\; -b^Ty\\
s.t.\quad 
\begin{align}
   A^Ty & \le c\\
   -A^Ty & \le -c\\
   -y & \le 0\\
\end{align}\\
$$
$$
\implies
\begin{align}
max\; -b^Ty\\
s.t. \qquad \begin{bmatrix}A^T\\-A^T\\-I\end{bmatrix}
\begin{bmatrix}y\\y\\y\end{bmatrix} =
\begin{bmatrix}c\\-c\\0\end{bmatrix}
\end{align}
$$
Where $I$ is the identity matrix.
Applying the assymetric Dual transformation we get:
$$
min\; \begin{bmatrix}c^T & -c^T & 0\end{bmatrix}w\\
s.t.
\begin{align}
\quad \begin{bmatrix}A & -A & -I\end{bmatrix}w & = -b\\
w & \ge 0
\end{align}\\
$$
let $w = \begin{bmatrix}u & v & k\end{bmatrix}^T$. Then:
$$
min\; c^Tu -c^Tv\\
s.t.
\begin{align}
\quad Au -Av -Ik & = -b\\
u,v,k & \ge 0
\end{align}\\
$$
reforming:
$$
max\; c^T(v - u)\\
s.t.
\begin{align}
\quad A(v - u) & = b - Ik\\
u,v,k & \ge 0
\end{align}\\
$$
let $x = v-u$
$$
max\; c^Tx\\
s.t.
\begin{align}
\quad Ax & = b - Ik\\
k & \ge 0
\end{align}\\
$$
We no longer have guarantee for the sign of $x$.
Since $k \ge 0$ the Right Hand Side will be greater if we omit the last term:
$$
max\; c^Tx\\
s.t.
\begin{align}
\quad Ax \le b \\
\end{align}\\
$$
Which is exactly the Primal Problem.
