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Let $P=\{x \in \mathbb{R}^n \mid Ax \geq b, x \geq 0 \}$ be a nonempty polyhedron for matrix $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^m$.

According to Minkowski-Weyl theorem $P$ can be written as

$$ P=\text{conv}(v_1,\cdots,v_p)+ \text{cone}(d_1,\cdots,d_l) $$ for some $v_i \in \mathbb{R}^n$ and $d_j \in \mathbb{R}^n$.

Let $C=\{x \in \mathbb{R}^n \mid Ax \geq 0, x \geq 0 \}$.

Show that $\text{cone}(d_1,\cdots,d_l) \subseteq C$.

The thing that that I cannot cope with is how to connect the finite number $l$ that can be any natural number with dimension of the matrix $A$.

I tried the following:

Let $z \in \text{cone}(d_1,\cdots,d_l)$, so there exist non-negative $\mu_i$'s such that

$$ z= \sum_{i=1}^l \mu_id_i $$ where $\mu_1,\mu_2,\cdots,\mu_l \geq 0$.

We can write $z$ as the following:

$$ z= \begin{bmatrix} d_1 & d_2 & \cdots & d_l \end{bmatrix} \begin{bmatrix} \mu_1 \\ \mu_2 \\ \cdots \\ \mu_l \end{bmatrix} $$

Now, we should come up with an $m \times n$ matrix $A$ for which we have $Az \geq 0$ and $z \geq 0$ to prove the claim. But the problem is we do not have $z \geq 0$ necessarily.

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  • $\begingroup$ The statement is not true for any choice of $v_i$, so you need to chose them appropriately. $\endgroup$ – LinAlg Dec 14 '18 at 22:27
  • $\begingroup$ @ LinAlg: The statement is true because it says for some $v_i$ and $d_j$. Also, it does not say what $p$ and $l$ are. Maybe I am wrong who am using wrong number for $l$. $\endgroup$ – Sepide Dec 15 '18 at 0:22
  • $\begingroup$ a proof by contradiction is easier here: what if $z$ is not in $C$? $\endgroup$ – LinAlg Dec 15 '18 at 0:59
  • $\begingroup$ @LinAlg: How we can do that? $\endgroup$ – Sepide Dec 15 '18 at 4:13
  • $\begingroup$ if $z$ does not satisfy $Az\geq 0$ or $z\geq 0$, moving in the direction of $z$ will violate a constraint of $P$ $\endgroup$ – LinAlg Dec 15 '18 at 16:19
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We know that $P$ can be written as $$ P=\operatorname{conv}(v_1,\cdots,v_p)+ \operatorname{cone}(d_1,\cdots,d_l)=V+D. $$ The set $D$ is a cone, hence, for every $v\in V$ and $d\in D$ we have that $v+td\in P$, $\forall t\ge 0$. That is $$ A(v+td)\ge b,\quad v+td\ge 0,\quad\forall t\ge 0. $$ Now divide by $t$ and let $t\to +\infty$ \begin{align} \frac{1}{t}Av+Ad\ge\frac{1}{t}b\quad&\Rightarrow\quad Ad\ge 0,\\ \frac{1}{t}v+d\ge 0 \quad&\Rightarrow\quad d\ge 0. \end{align} Therefore, every $d\in D$ belongs to $C$.

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  • $\begingroup$ Could you help me show the reverse? I mean $ C \subseteq \text{cone}(d_1,\cdots,d_l) $. It cannot be immediately observed from your proof to show the reverse, I am having trouble of producing $x \geq 0$? $\endgroup$ – Sepide Dec 15 '18 at 18:32
  • $\begingroup$ I have proved it using different method, but you might be show it differently. I posted it as an another question: math.stackexchange.com/questions/3041877/… $\endgroup$ – Sepide Dec 15 '18 at 19:54

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