What is the dual cone of a convex polyhedron? A convex polyhedron is defined as $P=\{x \in \mathbb{R}^n \mid Ax \geq b\}$. On the other hand, the dual cone of any set $S$ is defined as $S^{*}=\{ y \in \mathbb{R}^n \mid x^{\top}y \geq 0 \,\,\,\,\,\forall x \in S\}$. 
Question: what is the dual cone of $P$? or how can we characterize $P^{*}$ in terms of $A$ and $b$.
$$
P^{*}=\{ y \in \mathbb{R}^n \mid x^{\top}y \geq 0 \,\,\,\,\,\forall x \in P\}
$$
 A: Here is an optimization take, using (strong) Lagrange duality. Let $ C = \{x: Ax \ge b\} $ and let $\delta_C$ denote the convex analysis indicator of $C$, that is, $\delta_C(x) = 0$ if $x \in C$ and $=\infty$ if $x \notin C$. 
Consider the optimization problem:
\begin{align*}
 \phi(y) &= \inf_{Ax \ge b} x^T y \\
&= \inf_{x \in \mathbb R^n} x^T y + \delta_C(x), \\ 
&\stackrel{(a)}{=} \inf_{x \in \mathbb R^n} \,\sup_{\lambda \ge 0} \big[ x^Ty  + \lambda^T (b-Ax) \big] \\
&\stackrel{(b)}{=} \sup_{\lambda \ge 0}\inf_{x \in \mathbb R^n}\big[ x^Ty  + \lambda^T (b-Ax) \big] \\
&= \sup_{\lambda \ge 0}\big[ \lambda^T b + \inf_{x \in \mathbb R^n} x^T(y - A^T \lambda) \big] \\
&\stackrel{(c)}{=} \sup_{\substack{\lambda \ge 0, \\ A^T \lambda = y}} \lambda^T b
\end{align*}


*

*Step (a) is the common way of encoding inequlity constraints as part of the objective function. Note that
$$
\sup_{\lambda \ge 0} \lambda^T (b-Ax)  = 
\begin{cases}
 0 & b - Ax \le 0 \\
 \infty & \text{otherwise}
\end{cases} = \delta_C(x)
$$
for the $C$ defined above. (If one of the coordinates of $b-Ax$ is positive, you can send the corresponding coordinate of $\lambda$ to $\infty$.)

*Step (b) is the strong duality. 

*Step (c) is obtained by noting that unless $y-A^T\lambda = 0$, the $\inf$ over $x$ evaluates to $-\infty$, hence these values of $\lambda$ can be dropped when considering the outer $\sup$ optimization.


The dual cone is characterized by $\phi(y) \ge 0$. The above dual expression for $\phi$ says that $y$ belongs to the dual cone if and only if there exists $\lambda \ge 0$ such that $y = A^T\lambda$ and $\lambda^T b \ge 0$. That is 
$$P^* = \{ A^T \lambda \mid \lambda \ge 0, \;\lambda^T b \ge 0\}.$$
