Conditions on boundedness of a polyhedron which makes it polytope. [closed]

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.

closed as off-topic by Namaste, Shailesh, Brian Borchers, user10354138, CesareoDec 9 '18 at 9:15

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• What does $x \ge 0$ mean? All coordinates $\ge 0$? – Paul Frost Dec 8 '18 at 23:13
• It means each element of $x$ should be greater than zero. – Saeed Dec 8 '18 at 23:21

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.

• You have shown that $Ax=0$ has trivial solution, otherwise $P$ is unbounded. Could you help me to show the other way, I mean, show if $P$ is bounded, then $Ax=0$ has trivial solution? – Saeed Dec 9 '18 at 1:02
• @Saeed: The above shows that if $P$ is bounded then only trivial solutions exist. What remains to be shown is that if there only trivial solutions then $P$ is bounded. – copper.hat Dec 9 '18 at 2:05

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.

• I understand that $x'_n$ is a bounded sequence so it has a convergent subsequence, say $x''_n$ that converges. But I do not understand why you are using $x'_n$ instead of $x''_n$? – Saeed Dec 9 '18 at 4:26
• The notation $x_{n_k}$ means a subsequence of $x_n$. (I had a typo.) – copper.hat Dec 9 '18 at 4:52