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

- 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 $\{x_1,\ldots, x_m\}$.
- 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:

Suppose

- We have two sets $A,B$ with $L: A\rightarrow B$, which is
**linear**. - $A,B$ are both convex hull of finite extreme points, compact and both sets have at least one extreme point.
- 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 $v_1^2 +v_2^2 =1$.