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 "https://math.stackexchange.com/questions/282036/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 "https://math.stackexchange.com/questions/1591561/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 hull of finite extreme points, 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 $B$.

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

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

Moreover, if $A,B$ are convex hull of infinite extreme points, will I get the same answer?

This part comes from an example:

The extreme points of $A$ can be $SO(2)$ , i.e. a set of $2×2$ rotation matrices with $0≤θ≤2π$, to make an one-to-one map. The extreme points of $B$ is a set of rank one, $2×2$ positive semidefinite matrices with trace one. So $B$ can be built by $vv^T$ where $v = [v_1 v_2]^T$ with $v_1^2 +v_2^2 =1$.


1 Answer 1


So, you should try to map the vertices of a tetrahedron linearly onto the vertices of a square. For example by a projection $\mathbb R^3 \to \mathbb R^2$.

  • $\begingroup$ I see. It is surjective but not injective in your example and this is for finite extreme points. Therefore it is not injective for the case of infinite extreme points certainly. $\endgroup$ Sep 11, 2016 at 0:49

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