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Given a set $X \subset R^n$ with $m$ points. We can find it's Convex Hull and together with set of extreme points $E(X)$. And none of any points are linear multiplier of each other.

Under a linear projection of $f: R^n \to R^{n-1}$, we can find extreme points of $Y:=\{f(x)| x \in X\}$, denote $E(Y)$. which would again be extreme points of $X$, since linear projection preserves convexity.
$PI((E(Y))) \cap E(X) \neq \emptyset$, the pre-image of extreme points of $Y$ contains a subset of extreme points of $X$

There are ${m \choose n }$ such linear projection defined by the data points. Namely picking $n$ data points, looking at the affine subspace they span, and then projecting orthogonally onto that affine subspace

Would the union of ${m \choose n}$ linear projection and find extreme points in the lower dimension be sufficient to recover all the extreme points in the original set?

would this hold $E(X)\subseteq \cup_{i \in {m \choose n}}PI(E(Y_i))$ ?

For example, if in 2d space(n=2) and 10 points (m=10). Any two points can define a line, we would have ${10 \choose 2}$ lines defined by the data. If we project the 10 points to any line, the convex hull would form a line segment, we would detect 2 extreme points. If we project towards all the lines and collect all the extreme points there would be $2*{10 \choose 2}$ points, which contains duplicates of course. but if we deduplicate them, would all the extreme points of the 2d space being captured by these procedures?

peng yu
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  • what do you mean basis? the triangle would be a perfect example of capturing all projection.since all the sides of the triangle is a line defined by the data, if you project data to the sides, it captures all extreme points – peng yu Jul 02 '20 at 16:07
  • Do you mean $f(X)$ instead of $f(x)$? – diracdeltafunk Jul 02 '20 at 16:12
  • thanks, it is, i have made some explicit notation ${f(x)| x \in X}$ – peng yu Jul 02 '20 at 16:25
  • Thanks for rewording the question! However, there's still an issue: the preimage of an extreme point under an orthogonal projection need not consist of extreme points. For example, consider the unit square $X=[0,1]\times[0,1]$ in $\mathbb{R}^2$. The extreme points $(0,0)$ and $(1,0)$ span the affine subspace ${(x,0) : x \in \mathbb{R}}$ (aka the $x$-axis). The image of $X$ under the orthogonal projection to the $x$-axis is $[0,1]$. The extreme points of $[0,1]$ are $0$ and $1$, but the preimage of $0$ is ${(0,y):y\in [0,1]}$, which contains non-extreme points of $X$! – diracdeltafunk Jul 02 '20 at 16:38
  • wouldn't (0, 1) be an extreme point of X? – peng yu Jul 02 '20 at 16:41
  • Yes, but your claim that the inverse image of an extreme point of $Y$ will be a subset of $E(X)$ is false. Do you mean to ask something like "is every extreme point of $X$ an element of the inverse image of some extreme point of some $f(X)$"? – diracdeltafunk Jul 02 '20 at 16:43
  • ah yeah, that is what I mean :) sorry my rigorous language is pretty limited, i have changed that into $PI(E(Y)) \cap E(X) \neq \emptyset$ – peng yu Jul 02 '20 at 16:45
  • Ok, thanks! I've edited my answer. – diracdeltafunk Jul 02 '20 at 17:37

1 Answers1

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There are $\binom{m}{n-1}$ such linear projection defined by the data points.

I don't understand how: do you mean that you are picking some data points, looking at the affine subspace they span, and projecting orthogonally onto that affine subspace? If, so $n-1$ points will define an $(n-2)$-dimensional affine subspace, so you probably want to choose $n$ data points to project onto an $(n-1)$-dimensional space.

Apologies if this isn't what you meant — your question is pretty imprecise as stated so this was my best attempt at interpreting it. Also, what do you mean by an "extreme point of $X$", since $X$ is supposed to be finite (hence not convex unless $\lvert X \rvert \leq 1$)? Finally, how is an extreme point of $f(X)$ supposed to be the same as an extreme point of $X$, since these live in different spaces? I think this question is answerable but needs to be clarified first. I'll edit this answer once the question is clear and precise :)

Edit The question has been clarified, and the answer is no. Indeed, this fails immediately in $\mathbb{R}^2$ (and in all higher dimensions):

Let $X$ be the convex hull of $p_1 = (0,0)$, $p_2 = (0,1)$, $p_3 = (1,3)$, and $p_4 = (1,2)$ (these four points are the extreme points of $X$). Then $p_2$ is not sent to an extreme point of $f(X)$ for any orthogonal projection $f$ onto the affine span of any two of the $p_i$'s.

diracdeltafunk
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  • so you probably want to choose data points to project onto an (−1)-dimensional space.

    yeah you are right, sorry for the confusion, let me clear myself a bit

     – peng yu Jul 02 '20 at 16:15
  • thanks! interesting! i'm wondering how to construct linear projections that suits the conditions. – peng yu Jul 02 '20 at 17:41
  • I should note that there is always at least one orthogonal projection which sends a given extreme point to an extreme point, the issue is that such an orthogonal projection need not be onto any hyperplane parallel to the affine span of $n$ elements of $X$ (the condition that the hyperplane be the affine span of points of $X$ is very restrictive, because this only gives you at most $\binom{m}{n}$ normal directions! the odds that one of the hyperplanes that works has one of these normal directions is therefore low). – diracdeltafunk Jul 02 '20 at 17:45
  • ah, i forget another constraint, such that no two points are linear multiplier of each others. – peng yu Jul 02 '20 at 17:54
  • Just add $(2,0)$ to all of the points in my example. Then this additional constraint will be satisfied. – diracdeltafunk Jul 02 '20 at 17:57
  • ah right right... is there any analytic proof? a special case is sound, but my goal is to get exactly why makes it impossible – peng yu Jul 02 '20 at 17:58
  • Yes, you can try to pinpoint the issue with analysis! This is easiest to do in $\mathbb{R}^2$. Write down the formulas for the orthogonal projections onto the affine spans of two elements of $X$. Write down what it would mean for a given extreme point $e$ to map to an extreme point under one of these orthogonal projections (you should get an inequality involving a dot product). In the end, the constraint is that we need there to exist $i \neq j$ such that $p_i - p_j$ lies in a cone bounded by the orthogonal directions to the edges incident to $e$. Let me know if you need more details! – diracdeltafunk Jul 02 '20 at 18:04
  • thanks, i'll try that, btw, i have a very similar question, in case you are interested https://math.stackexchange.com/questions/3742073/would-any-two-consecutive-extreme-points-on-a-convex-hull-can-be-linearly-projec – peng yu Jul 02 '20 at 19:05