TITLE: preservation of extreme points under linear transformation
QUESTION [5 upvotes]: Suppose $\{e_1,...,e_N\}$ is the set of all extreme points of a compact convex subset $X\subset\mathbb R^n$.   $L: \mathbb R^n\to \mathbb R^m$ is a linear transformation. $L$ is surjective but is not injective. Let $Y= L(X)$.
Would it hold that for every  $1\leq i\leq N$,   $L(e_i)$ must be an extreme point of  $Y$? Is there any   characterization on $L$ such that this property holds?
Thanks.

REPLY [3 votes]: If $L$ fails to be injective, then this will not hold.
For example, take $X \subset \Bbb R^2$ to be $\{(x,y):|x|+|y| \leq 1\}$, and take $L:\mathbb{R}^2 \to \mathbb{R}$ to be given by $L(x,y) = x$. Note that $(0,1)$ is an extreme point of $X$, but $L(0,1) = 0$ is not an extreme point of $Y$.
On the other hand, if $L$ is injective in addition to being surjective, then $L$ is an invertible linear transformation and so the statement holds.

REPLY [2 votes]: The images by $L$ of the extreme points are certainly not all extreme points of the image of the compact convex by $L$.
Take the example of a square and for $L$ the projection on one of the diagonal. The two extreme points on the diagonal are also extreme points of the image by $L$ of the square, but that is not the case for the opther two vertex of the square.
However the extreme points of the image of the compact convex is the subset of the images of the extreme points of the initial compact convex.