Disprove that x-hat is extremal I'm new to optimization and I found this problem: Disprove the statement: Give the polyhedron $P=\{x \in \mathbb{R}^n \mid Ax-By \ge b, y\le 1$}. The point $\hat{x} \in \mathbb{R}^n$ is extremal if and only if $\hat{x}$ is a feasible point in $P$. How would I begin to disprove this? If anyone knows what x-hat thing means it would also be very helpful. Thanks to anyone for the help!!
Edit:
Definition:
Let P be a polyhedron. z ∈ rec(P) \ {0} is called an extremal of P if cone{z} is an extremal ray of rec(P).
 A: I'll just point out a few facts.
Let $A=B\equiv 0$ and $b>0$, then $P=\emptyset$. In this case, if the emptyset is a polyhedron, then the sets of extremal points and feasible points coincide.
Now, let $a_i\in\mathbb R^{n+1}$ be given by $a_{i,1}=1$, $a_{i,i}=-1$ and $a_{i,j}=0$ otherwise, for $2\le i\le n+1$. Let $A=(a_{i+1,j})_{i,j}\in\mathbb R^{n+1\times n}$. For $x\in\mathbb R^n$ let $y=Ax=\sum_ix_ia_{i+1}$. Notice that if there exists $i$ with $x_i>0$, then we have $y_i<0$. Otherwise, if there exists $i$ with $x_i<0$, then we have $y_1=\sum_jx_j\le x_i<0$ because $x\le 0$. So we have $y\ge 0$ if and only if $y=0$, which we have if and only if $x=0$. So, for $B\equiv 0$ and $b\equiv 0$ we have $P=\{0\}$, which is also both the set of extremal points and feasible points, using the Wikipedia definition.
These are two examples where the statement is true, depending on the definitions. Now, we turn to an example where the statement is false.
For this purpose take $A$ from above, let $B\equiv 0$ and $b\equiv -1/(n+1)$.
Let $e_i$ be the $i$-th unit vector in $\mathbb R^{n+1}$. By construction, the set $Q=\{Ax:x\in P\}$ is the standard simplex shifted by $b$. Hence, $P$ is the convex hull of the $n$ unique corners $c_i$ given by $Ac_i=e_i+b$. The map $A:P\rightarrow Q$ is clearly a bijection that preserves feasible points and extreme points, and $0\in Q$ is clearly feasible but not extremal.
Hence, we have seen two cases where the statement holds and one where it doesn't. In general, it's not hard to see that the extremal points and the feasible points of a (general) polytope $P$ coincide if and only if $P$ has at most one point. Because, if $P$ has at least two points, take any two, consider the connecting line using that polytopes are convex, and notice that any point on this line is feasible and not extremal.
