# Conic hull of a rectangle

I have a simple question that appeared in research: For a rectangle $$S :=[a_1,b_1] \times[a_2,b_2] \times \dots \times [a_n,b_n] \subset \mathbb{R}^n$$. Let $$p_0 = (a_1,a_2,\dots,a_n)$$, and define $$p_i (1 \leq i\leq n)$$ be the vector obtained by changing the $$i$$th entry of $$p_0$$ to $$b_i$$.

I think $$\operatorname{cone}(S) = \operatorname{cone}(\{ p_i\mid 0 \leq i\leq n \})$$, but I'm not sure if this is true.

Intuitively, I suspect this is true. I tried to use Farkas Lemma on this but did not reach anything promising.

I asked my professor about this and he went on talking about $$\mu$$ analysis etc, and saying something about capturing the set $$\{ A + B\triangle C\mid \|\triangle\|_\text{op} \leq 1 \}$$. However, I felt my question is very straightforward, and does not require much advanced tools

My motivation of this problem is about simultaneous stability, where all the corners are $$n \times n$$ matrix, so checking $$2^n$$ corners is computationally infeasible.

This is my first post on Mathoverflow, previously I have mostly posted on Mathematics stack exchange, but according to this post, since this is research related and might get more attention here, I decided to post here.

A counterexample is given by $$n=2$$, $$[a_1,b_1]=[-2,1]$$, $$[a_2,b_2]=[-1,2]$$. (Make a picture.)
Even if the $$n$$-box $$S$$ is required to be a subset of $$[0,\infty)^n$$, the answer will still be no. E.g., let $$n=3$$, $$p_0=(a_1,a_2,a_3)=(2,0,0)$$ and $$b_i=a_i+1$$ for $$i=1,2,3$$. Then for $$\nu:=(-1,2,2)$$ and $$i=0,1,2,3$$ we will have $$\nu\cdot p_i\le0<\nu\cdot p_*$$, where $$\cdot$$ denotes the dot product and $$p_*:=(b_1,b_2,b_3)=(3,1,1)$$. So, the vertex $$p_*$$ of this $$3$$-box is not in the conic hull of the set $$\{p_0,p_1,p_2,p_3\}$$.
For an illustration, shown below are the points $$p_0,p_1,p_2,p_3,p_*$$ and a piece of the plane $$\{x\in\mathbb R^3\colon\nu\cdot x=0\}$$ (through the points $$p_2$$ and $$p_3$$), which separates $$p_*$$ from $$p_0,p_1,p_2,p_3$$:
• Thank you! What if the rectangle lies in $\mathbb{R}^n_{\geq 0}$? Commented Nov 5, 2023 at 22:32