Conditions on boundedness of a polyhedron which makes it polytope.

Question

Let $P=\{x \in \mathbb{R}^n \mid Ax=b, x\geq 0\}$ be a nonempty convex polyhedron (not bounded).

Show that $P$ is bounded (i.e., it is a polytope) if and only if the linear inequality $Ax=0, \,\, x\geq 0$ has trivial solution $x=0$ only.

Answer 1

If there exists some nontrivial solution $d$ to $Ax=0,x\geqslant 0$, then for any $x\in P$ we have for any $\varepsilon>0$ $$A(x+\varepsilon d)=Ax+\varepsilon Ad=Ax+0=Ax=b,$$ so that $x+\varepsilon d\in P$. It follows that $P$ is not bounded.

Answer 2

Suppose $P$ is not bounded. Then there are $x_n$ with $\|x_n\| \ge n$ such that $Ax_n =b, x_n \ge 0$.

Let $x'_n = {1 \over \|x_n\|} x_n$, and note that $A x'_n \to 0$ and $x'_n \ge 0$.

For some subsequence we have $x'_{n_k} \to x$ for some unit norm $x$. We see that $Ax = 0$ and $x \ge 0$, and so there is a non trivial solution.