# Show sum of two polyhedral cones is a polyhedral cone?

Polyhedral cone is defined as $$C= \{x \in \mathbb{R}^n \mid Ax \geq 0\}$$. Let $$C_1$$ and $$C_2$$ be two polyhedral cones in $$\mathbb{R}^n$$. Show that $$C_1+C_2$$ is also a polyhedral cone.

My try: Let $$x + y \in C_1 + C_2$$ we need to show there is a matrix $$D$$ such that $$D(x+y) \geq 0$$. To show that we need to use the facts that $$x \in C_1 \rightarrow Ax \geq 0$$ and $$y \in C_2 \rightarrow By \geq 0$$ for some appropriate $$A$$ and $$B$$.

My question: How can we use $$A$$ and $$B$$ to connect them to $$D$$.

## 1 Answer

First, $$z\in C_1+C_2$$ if and only if there are $$x,y$$ such that $$x+y=z$$, $$Ax\ge0$$, $$Bx\ge0$$. That is, $$z$$ is the third component of vectors in the cone $$\left\{ (x,y,z) \in (\mathbb R^n)^3: \pmatrix{I & I & -I \\ -I&-I&I \\ A & 0&0\\0&B&0} \pmatrix{x\\y\\z}\ge \pmatrix{ 0\\0\\0\\0} \right\}.$$ Now use Fourier-Motzkin elimination to compute the projection onto the $$z$$-component. It follows that $$C_1+C_2$$ is a polyhedral cone.