in a tutorial on of our "teachers" written down this exercise and I do not understand the solution:
Given is an polyhedron in standard form $K=\{ x \in R^n | Ax =b,x \geq 0 \}$ with matrix $A\in R^{n-1xn}% and $rank(A) =n-1$. => O is not empty and has at most two vertices.
My question would be, why is P not empty and why has P at most two vertices, I actually don't understand it?
Consider $n=2$ $A =(y_1,y_2)$ so we have $K=\{ x \in R^n| (y_1,y_2)x =b,x \geq 0 \}$ and this is a line. How can a line have two vertices? There has to be a $c\in R^2$ with $c^Tx < c^Ty$ for all $y \in K$ while $y\neq x$ and because its a line that should no be possible?
A line segment has two vertices, namely the endpoints. You could also have a ray, which has one endpoint (which is a vertex), or a line, which has none.
However, the teacher is wrong in one respect: the polyhedron could be empty. For example, this is the case if all entries of $A$ are nonnegative and $b$ has some negative entries.