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

The reverse has been proven in How to show polyhedral cone of nonnegative vectors contains finitely generated cone?

My try:

$C$ can be written as the following:

$$ Ax \geq 0, x \geq 0 \,\,\, \Rightarrow A' = \begin{bmatrix} A \\ I \end{bmatrix} x \geq 0 $$

Hence, $C=\{x' \in \mathbb{R}^{2n} \mid A'x' \geq 0 \}$ which is a polyhedral cone and using the theorem saying that every polyhedral cone is a finitely generated cone we are done.

I want to show this using rigorous prove, and not using that theorem.


We know that $$ P=\text{conv}(v_1,\cdots,v_p)+ \text{cone}(d_1,\cdots,d_l)=V+D. $$ Take $x\in C$ $\Leftrightarrow$ $Ax\ge 0$, $x\ge 0$, and $p\in P$ then $$ A(p+tx)=Ap+tAx\ge b,\quad p+tx\ge 0,\quad\forall t\ge 0, $$ hence $p+tx\in P$, $\forall t\ge 0$. From the representation above $$ p+tx=v_t+d_t\quad\Leftrightarrow\quad \frac{1}{t}p+x=\frac{1}{t}v_t+\frac{1}{t}d_t. $$ Let $t\to+\infty$ then $\frac{p}{t}\to 0$ and $\frac{v_t}{t}\to 0$ ($V$ is bounded), hence $$ \frac{1}{t}d_t\to x. $$ Since $\frac{d_t}{t}\in D$ and $D$ is closed, we must have $x\in D$.

  • $\begingroup$ If we assume that $V$ is bounded, then $v_t/t \rightarrow 0$ makes sence. But what is the reason $p_t/t \rightarrow 0$? Because $p_t$ already contains $d_t$ which is not bounded. $\endgroup$ – Sepide Dec 16 '18 at 23:13
  • $\begingroup$ @Sepide $V$ is bounded as a convex hull of finitely many points, it is a fact, not an assumption. $p$ is a fixed point, it does not depend on $t$. $\endgroup$ – A.Γ. Dec 17 '18 at 5:52

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