The question in title arises because I am trying to prove that a polytope is the convex hull of its vertices, i.e., $\mathcal{P}=conv(V)$. Here is how far I have got.
Convex hull of a finite set of vectors is a polytope. So for $v_1,…,v_k$, $\mathcal{Q}=conv(v_1,…v_k)$ is a polytope and I can show that for this polytope, $\mathcal{Q}$, $v_i$ must be vertices for all $i$. So I now know that if I take convex hull of a set of vectors, I get a polytope whose vertices are those vectors. But I require that vertices determine a polytope uniquely or else I cannot complete my proof. Is this true and if not, how do I prove that a polytope is convex hull of its vertices?
Best Answer
Yes, in general this is how one defines a convex polytope. In general, a convex polytope is defined as the convex hull of a finite set (of vertices), page 14:
So maybe you should provide the definition of convex polytopes that you are using. There another way to characterize a convex polytope $P$. Which is by means of a finite set of inequalities. Some times this is refered as the H-V theorem were H stands for half-spaces and V for vertices. Sometimes also as the Fundamental Theorem of Polytopes:
Polyhedral set is defined in the same text as