5

Given a closed convex cone $C \subset \mathbb{R}^n$ and a matrix $M \in \mathbb{R}^{m\times n}$, is the set $S = \{Mx\mid x \in C\}$ also a closed convex cone?

Firstly, $S$ must be a convex cone. But how about the closeness? I conjecture that $S$ must be closed if $C$ is a closed convex cone. However, I fail to come up with a rigorous proof (so maybe the claim is false). Note that for a more general $C$ (i.e., $C$ is not closed convex cone), the claim is not necessarily true.

Any hint or counterexample is really appreciated.

PSPACEhard
  • 10,381

1 Answers1

8

No, this is not true. You could take the rotated Lorentz cone $$C = \{(x,y,z) \in \mathbb{R}^3: y^2 \le x z; x \ge 0; z \ge 0\}$$ and project it onto the $y$-$z$-plane. The result is an open half-plane together with the origin.

Some conditions ensuring that the image is closed, can be found in this paper.

copper.hat
  • 178,207
gerw
  • 33,373
  • Sorry, I forgot to add $x \ge 0$, $z \ge 0$. – gerw Sep 30 '15 at 08:36
  • Yes, because $K^*$ is always a closed, convex cone, regardless of $K$. – gerw Sep 30 '15 at 13:03
  • @NP-hard: Thanks. I completely missed the $z$ on the $y^2 \le x z$. Sounds of deletion... – copper.hat Sep 30 '15 at 15:21
  • Delightful counterexample. – copper.hat Sep 30 '15 at 15:37
  • 3
    It was not so straightforward to see it is the open half-plane together with the origin. In order to see it, one can project into $x-y$ plane as well. The full projection area is the union of all the regions plotted here: https://www.desmos.com/calculator/uzd0tfqqt1 – R. W. Prado May 20 '22 at 18:04