Let $F$ be a convex bounded polytope in $\mathbb{R}^N$. Let $S$ be a linear subspace of $\mathbb{R}^N$ defined as $S=\{x \in \mathbb{R}^N: Ax = \vec{0}\}$, where $A$ is an $m \times N$ matrix, $0 \leq m \leq N$. For $m >0,$ all $m$ of the rows of $A$ are linearly independent.
Let $F_S = F \cap S$. We use $X(\mathcal{P})$ to denote the set of extreme points of any bounded polytope $\mathcal{P} \subset \mathbb{R}^N$. It is given that some of the extreme points of $F$ are in $S$, i.e. $X^* \equiv X(F) \cap S=X(F) \cap X(F_S) \neq \emptyset$. Define $aff(X^*)=$ the affine hull of $X^{*}$. I want to show that, either all of the extreme points of $F_S$ are extreme points of $F$, i.e. $X(F_S) \subseteq X(F)$, or $aff(X^*)$ forms a proper affine subspace of $S$.
My approach: Suppose not. Suppose there is an extreme point of $F_S$ which is not an extreme point of $F$ - let us call this point $x^{*}_r$ - but $aff(X^*)=S$. Therefore $x^{*}_r$ is a linear combination but not a convex combination of points in $X^*$. $x^{*}_r$ is not an extreme point of $F$, therefore it is a convex combination of extreme points of $F$ not in $S$.
Clearly, for $m \in \{0,N-1,N\}$, the claim holds.
Edit1:
Let $N-m=k$, i.e. $dim(S)=k$. We consider two cases below, in each of which we assume $aff(X^*)=S$, as stated above, and attempt to show that in that case $F \cap S$ cannot have any extreme point which is not an extreme point of $F$.
Case I. $Co(X^{*})\subseteq bd(F)$
This is the case where $F$ "rests on" $S$.
This is a special case I consider, where all the extreme points of $F$ which are in $S$, are part of the same $k$-face of $F$, call it $K$. By assumption, $aff(X^*)=S$. Therefore, in this case it is clear that $F \cap S = K$. But $K$ is, by definition, the convex hull of the extreme points of $K$, which are also extreme points of $F$. Therefore, $F \cap S$ cannot have any extreme points which are not extreme points of $F$.
Case II: $Co(X^{*})\nsubseteq bd(F)$
This is the case where $S$ "cuts through" $F$.
Define $S^{+}=\{x \in \mathbb{R}^N: Ax \geq \vec{0}\}$ and $S^{-}=\{x \in \mathbb{R}^N: Ax \leq \vec{0}\}$, replacing rows of $A$ with their negatives if necessary, to ensure $F^{+} \equiv F \cap S^{+} \neq \emptyset$ and $F^{+} \neq F$. Clearly, if such a redefinition of $A$ is not possible, it means $F \subseteq S^{+}$ or $F \subseteq S^{-}$ and original question already said $F \cap S \neq \emptyset$, $\therefore F \cap S$ is on the boundary of $F$ and we are back to Case I. Therefore WLOG, for Case II, we assume such a redefinition of $A$ is possible.
Clearly, $F \cap S = F^{+} \cap S=K^{+}$, a $k$-face of $F^{+}$. Clearly $X^{*} \subseteq K^{+}$, and $K^{+} \subset aff(X^{*})$.
Obviously, what I now need to show is that $K^{+}$ cannot have any extreme point which was not an extreme point of the original polytope $F$.
After that I'm not able to proceed. Any help is most appreciated. Thanks in advance.
Edit 2: In an earlier version of the question I mistakenly referred to $aff(X^*)$ as the span of $X^*$. Thank you user diracdeltafunk for pointing this out subsequently. diracdeltafunk's answer to this question is based on that earlier (incorrect) version of the question.
Edit 3: I am not very conversant with projections, but this answer makes me think maybe we can try using projections? Maybe try proving something like $X(F_S) \subseteq X(F) \Leftrightarrow X(P(F_S)) \subseteq X(P(F))$ for all projections $P:\mathbb{R}^N \rightarrow \mathbb{R}^{(N-1)}$, then use this iteratively (for smaller and smaller values of $N$) and finally use the fact that the claim is always true if $S$ is a line - to complete the proof?
