Extreme rays of a polyhedral

39 Views Asked by At

I was reading the following https://faculty.coe.drexel.edu/jwalsh/eces811/linProg.pdf, and was confused at one of the lines.

They defined a polyhedral cone to be

$$\{y\in \mathbb{R}^N | Hy≤0\}.$$

Also, an extreme ray is a $\{\gamma y_∗|\gamma ≥ 0\}$ in the cone such that if

$$\{\gamma y_∗|\gamma ≥ 0\} = \mu_1\{\gamma y_1|\gamma ≥ 0\} + \mu_2\{\gamma y_2|\gamma ≥ 0\}$$ for some $\mu_1,\mu_2>0$, and if $\{\gamma y_1|\gamma ≥ 0\} , \{\gamma y_2|\gamma ≥ 0\}$ are both in the cone, then $$ \{\gamma y_∗|\gamma ≥ 0\} = \{\gamma y_1|\gamma ≥ 0\} =\{\gamma y_2|\gamma ≥ 0\}.$$

My understanding of this is an extreme ray is some direction that cannot be written as a linear combination of any other point in the cone (is that correct?).

My confusion comes in on this line

This will happen if and only if $y_∗$ obeys exactly $N −1$ linearly independent rows of $Hy ≤ 0$ with equality.

How do I interpret a cone (geometrically)? and why would solving these active constraints give extreme rays?