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.


1 Answer 1


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 $p_0,p_1,p_2,p_3$:

enter image description here

  • $\begingroup$ Thank you! What if the rectangle lies in $\mathbb{R}^n_{\geq 0}$? $\endgroup$ Commented Nov 5, 2023 at 22:32
  • $\begingroup$ @wsz_fantasy : The answer will still be no, as it is now shown. $\endgroup$ Commented Nov 6, 2023 at 3:03

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