Before entering my problem, let me review some related results:

Suppose $\mathcal{S}$ is a convex hull of finite points: $\mathcal{S}= \operatorname{conv}(x_1,x_2,\ldots,x_m)$, then

  1. By "Convex hull of extreme points" , we know $\mathcal{S}$ is the convex hull of its extreme points, which is the subset of $\{x_1,\ldots, x_m\}$.
  2. By "preservation of extreme points under linear transformation", if the transformation is linear, let $A$ = {the extreme points of the image of a compact convex set}, then the preimage of $A$ must be the extreme points of that compact convex set. But the inverse in not true. The inverse is true only under such transformation is injective and surjective.

My question is:


  1. We have two sets $A,B$ with $L: A\rightarrow B$, which is linear.
  2. $A,B$ are both convex, compact and both sets have at least one extreme point.
  3. Suppose $L$ maps extreme points in $A$ one-to-one and onto to extreme points in $B$.

So obviously, $A$ and $B$ are convex hull of their extreme points.

Question: Is $L$ a one-to-one and onto linear map for $A\rightarrow B$?

If not, could you please provide an counter-example?

  • $\begingroup$ Are $A$ and $B$ generic convex sets, or are they (or at least one of them), like $\mathcal{S}$, the convex hull of a finite number of points? $\endgroup$ – Fimpellizieri Sep 10 '16 at 22:53
  • $\begingroup$ @Fimpellizieri Actually $A,B$ are convex hull of infinitely number of points. For example, $A=SO(2)$, ie $2\times 2$ rotation matrices with $0\leq \theta \leq 2\pi$. $B$ is a set of rank $1$, $2\times 2$ positive semidefinite matrices with trace one. So $B$ can be built by $vv^T$ with $v_1^2+v_2^2=1$. $\endgroup$ – sleeve chen Sep 10 '16 at 23:16
  • $\begingroup$ @Fimpellizieri Does your question make any subtle difference in the answer? $\endgroup$ – sleeve chen Sep 10 '16 at 23:17
  • $\begingroup$ It is better if the answer is general. Since I have no idea if it makes any crucial difference if number of extreme points is finite, or infinite. $\endgroup$ – sleeve chen Sep 10 '16 at 23:21
  • $\begingroup$ I was thinking about simple cases (polytopes) first, looks more tractable. I don't have an answer yet. $\endgroup$ – Fimpellizieri Sep 10 '16 at 23:48

I think the assertion is false.

Let $A \subset \mathbb{R}^3$ be a tetrahedron and let $L$ be a projection onto a two-dimensional subspace, such that $B = L\,A$ is a quadrangle.


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