Here is an example of $\min \max $ that should be converted to a linear program.
$$\min_{x \in \mathbb{R}^n} \,\,\,\max_{i = 1, \cdots, m} a_i^{\top}x \tag{1}$$
We know the standard primal linear program is defined as follows:
$$
\min \,\,\,\,\,\,c^{\top}x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\, (P)\\
s.t. \,\,\,\, Ax = b\,\, , x \geq 0
$$
Also, the dual of $(P)$ is defined as
$$
\max_{y \in \mathbb{R}^n} \,\,\,\,\,\,\,\,b^{\top}y \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\, (D)\\
s.t. \,\,\,\, A^{\top}y \leq c \,\,\,\,\,\
$$
My try: I tried to convert $(1)$ to $(D)$, then it would be easy to convert $(D)$ to $(P)$.
Let $t= \max_{i = 1, \cdots, m} a_i^{\top}x$, then
$$
\min_{x \in \mathbb{R}^n} t \\
t \geq a_i^{\top}x \,\,\,\,\,\,\forall i = 1, \cdots, m
$$
Now we can write
$$
\max_{x \in \mathbb{R}^n} -t \\
a_i^{\top}x-t \leq 0 \,\,\,\,\,\,\forall i = 1, \cdots, m
$$
or in the matrix form
$$
\max_{x \in \mathbb{R}^n} \begin{bmatrix}
\textbf{0}\\
-1
\end{bmatrix}^{\top}
\begin{bmatrix}
x\\
t
\end{bmatrix} \\
\begin{bmatrix}
a_1^{\top} & -1 \\
a_2^{\top} & -1 \\
\vdots\\
a_m^{\top} & -1
\end{bmatrix}
\begin{bmatrix}
x\\
t
\end{bmatrix} \leq \textbf{0} \tag{2}
$$
Question: I cannot convert $(2)$ to $(P)$ because $c= \textbf{0}$. What am I missing and mistaken?
Please complete my answer and do not provide other solutions
Best Answer
The original program (1) as written is always at least feasible no matter the choice of $a_1,\ldots, a_m$, with $x \in \mathbb{R}^n = 0$ a feasible solution, with an objective value of 0. Furthermore, either the optimum objective value of (1) is either 0 or $-\infty$. [Make sure you see why.]
Thus your linear program (2) has either an optimum objective value of either 0 or $\infty$ i.e., unboundedness.
Thus (2)'s dual which is in the form of (P) either has an an optimum objective value of either 0 or is infeasible. Which in fact is precisely what you get w $c=0$.