Representation of a Face of a Polytope In Ziegler's book "Lectures on Polytopes" it is proven that each face $F$ of a polytope $$P=\mathrm{conv}(V)=\{x\in\mathbb{R}^n:~Ax\leq\textbf{1}\}$$ (assuming $0\in\text{int}(P)$) can be expressed as
$$
F=\mathrm{conv}(V')
$$
where $V'\subseteq V$ is a subset of the vertices of $P$. How to prove $F$ can also be expressed without loss of generality as
$$
F=\{x\in\mathbb{R}^n:~A''x\leq\textbf{1},~A'x=\textbf{1}\}
$$
for some partition of the matrix
$$
A=\begin{bmatrix}A'\\A''\end{bmatrix}?
$$
From the definition of face I know every set of the form $\{x\in\mathbb{R}^n:~A''x\leq\textbf{1},~A'x=\textbf{1}\}$ is indeed a face. My question is about the converse statement.
 A: If $F$ is $\DeclareMathOperator{\conv}{conv}\conv(V')$,
let $A'$ be the rows $r$ of $A$ such that $r x = 1$ for all $x \in V'$,
and $A''$ be the other rows of $A$.
Then any point $x$ in $F$ is $\sum_{v \in V'} \alpha_v v$ where $\sum \alpha_v = 1$, so $A' x = \sum_{v \in V'} \alpha_v A' x = 1$.
And of course $A'' x \leq 1$ since $x \in P$.
On the other hand, any point $y$ satisfying $A' y = 1$ and $A'' y \leq 1$ must be in $P$, so it is a convex combination of points in $V$; $y = \sum_{v \in V} \alpha_v v$.
Any of the points $v \in V \setminus V'$ have $A' v < 1$, and so $\alpha_v$ must be zero; otherwise $A' y$ would be less than one.
Thus $y$ is in $F$.
We do need some arguments to establish that any vertices $v \in V$
not in the face $F$ have $A' v < 1$ (by which I mean that at least one entry is less than one.) I would argue

*

*Each facet of $P$ corresponds to a row $r$ of $A$, with a point $x\in P$ in the facet iff $rx=1$;

*So if $A′v=1$, then $v$ is contained in all the facets which contains the face $F$, and so $v$ must be in $F$.

Fleshing out the details seems like it could be annoying, but perhaps you already have theorems establishing the necessary facts (for instance, that a face is the intersection of the facets containing it).
