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A general polyhedral cone $\mathcal{P} \subseteq \mathbb{R}^n$ can be represented as either $\mathcal{P} = \{x \in \mathbb{R}^n : Ax \geq 0 \}$ or $\mathcal{P} = \{V x : x \in \mathbb{R}_+^k , V \in \mathbb{R}^{n \times k} \}$.

I am trying to do show the dual of $\mathcal{P}$, $\mathcal{P}^*$, is a polyhedral set.

I start by writing $\mathcal{P}^* = \{ y \in \mathbb{R}^n : y^T V x \geq 0$ $\forall x \in \mathbb{R}_+^k \}$. (1)

A polyhedral set is a set of the form $\{x \in \mathbb{R}^n : Ax \leq b \}$ (2).

Any ideas how to go from (1) to (2). I also know that an image of a polyhedral set under a linear map is also polyhedral.

PS: Similar questions like these have very advanced solution methods; I am looking for a much simpler way.

rims
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2 Answers2

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It's quite non-trivial to prove that your two definitions of polyhedral cone are equivalent. However, granted that, let $\mathcal{P}$ be a polyhedral cone satisfying your first definition, then $$ \begin{aligned} \mathcal{P}^* =& \{y\in\mathbb{R}^n:y^TVx\ge 0, \forall x \in \mathbb{R}_+^k\} \\ =& \{y\in\Bbb R^n:x^T\left(V^Ty\right)\ge0, \forall x\in \Bbb R_+^k\} \\ =& \{y\in\Bbb R^n:V^Ty\ge0\} \end{aligned} $$ so that $\mathcal{P}^*$ meets your first definition of polyhedral cone.

nalzok
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Angina Seng
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By metric geometry duality most often is obtained via spherical reciprocation.

So just insert a sphere into the cone such that it is tangent to both, the base and the lateral shell, then obviously those contact points to both surface parts will define the (then inscribed) dual cone.

--- rk

Dr. Richard Klitzing
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