Given a linear program, we have the primal as follows:
\begin{array}{lll}
\max: & c^Tx\\
\text{s.t.} & Ax \leq b\\
& x\geq 0\\
\end{array}
And we also have the dual as follows:
\begin{array}{lll}
\min: & y^Tb\\
\text{s.t.} & y^TA \geq c\\
& y \geq 0\\
\end{array}
Now, the question is (a)to design a single linear feasibility formulation (i.e. Find $z$ such that $Bz \leq k$) that exactly captures all the optimal solutions of both primal and dual. (b)show that $(x^*,y^*)$ is feasible in the formulation if and only if $x^*$ is an optimal solution of the primal and $y^*$ is an optimal solution of the dual.
Here's what I got so far: The single linear feasibility formulation is the following:
$\begin{pmatrix}A & 0 \\ 0 & -A^T \end{pmatrix}$$\begin{pmatrix}x \\ y\end{pmatrix} \leq \begin{pmatrix}b \\ -c^T\end{pmatrix}$, where $A \in \mathbb{R}^{m\times n},x\in\mathbb{R}^{n\times 1}, y\in\mathbb{R}^{m\times 1}$.
From this formulation, it should capture all the optimal solution of both primal and dual. Now,for part (b), the converse direction is trivial because if $x^*$ is an optimal, then it means $Ax^* \leq b$ and $y^*$ is an optimal solution of the dual means $-A^Ty^* \leq -c^T$, so $z=(x^*,y^*)$ is feasible in the formulation.
However, I am stuck in the forward direction, which is if $(x^*,y^*)$ is feasible in the formulation, then $x^*$ is an optimal solution of the primal and $y^*$ is an optimal solution of the dual. Can I get some help for that? Do I have to use the strong duality theorem in order to prove that?