Show that $S=\{\mathbf x \in \mathbb R^n:\mathbf A \mathbf x=\mathbf b,\mathbf x \ge \mathbf 0\} \ne \emptyset$ has at least one extreme point Show that the standard polyhedron defined by $S=\{\mathbf x \in \mathbb R^n:\mathbf A \mathbf x=\mathbf b,\mathbf x \ge \mathbf 0\} \ne \emptyset $ has at least one extreme point and the set of its extreme points is a finite set.

I know that we need to find $\mathbf x \in S$ such that $$\forall \mathbf A,\mathbf B \in S, \forall \lambda \in (0,1):\mathbf x= \lambda \mathbf A+(1-\lambda)\mathbf B \implies \mathbf A=\mathbf B$$
And that shows that the set has at least one extreme point, I can find such $\mathbf x$ but unfortunately it depends on $\mathbf A$ and so it does not work for arbitrary $\mathbf A,\mathbf B \in S$.
I've seen the theorem in many sources, but could not find any proof of that.

I think that I can prove the existence at least one extreme point if $A$ is a $n \times n$ matrix.
Proof:
I will do the proof by induction.
Case $n=1$:
If $A$ is not singular and $S \neq \emptyset$ then $S=\{a\}$ (consecuence of Rouche-Frobeniu´s theorem) for some $a \in \mathbb{R}$ and then $S$ has one extreme point.
If $A$ is singular then $S=\mathbb{R}^{+} \cup \{0\}$ and then $\{0\}$ is the extreme point.
Case general:n=k
If $A$ is not singular and $S \neq \emptyset$ then $S=\{a\}$ (consecuence of Rouche-Frobeniu´s theorem) for some $a \in \mathbb{R}^{n}$ and then $S$ has one extreme point.
If $A$ is singular and $S \neq \emptyset$ then we have that $S=(a+V) \cap \{x \geq 0\}$ for some $a \in \mathbb{R}^{n}$ and a subespace $V \subseteq \mathbb{R}^{n}$.
How $A$ is singular, then exists $v=(v_{1},...,v_{n}) \in V : v \neq 0$ so exists $j \in \{1,...,n\} : v_{j} \neq 0$, and we can suppose that some index is negative (doing $\cdot (-1)$).
Then rearranging if necessary we can suppose that $a=(a_{1},...,a_{n})$, $v=(v_{1},...,v_{n})$ and $v_{n} < 0$ and $\frac{|v_{n}|}{|a_{n}|} \leq \frac{|v_{j}|}{|a_{j}|} \forall j \in \{1,...,n\}$.
Then we have that $a+\frac{v_{n}}{a_{n}}v=(a_{1}+\frac{v_{1}v_{n}}{a_{n}},...,a_{n-1}+\frac{v_{n-1}v_{n}}{a_{n}},0)$ and $a_{i}+\frac{v_{i}v_{n}}{a_{n}} \geq 0$ with $i \in \{1,...,n-1\}$, so $a+\frac{v_{n}}{a_{n}}v \in S$.
But $S \subseteq \{x \geq 0\}$ so if $a+\frac{v_{n}}{a_{n}}v=\lambda c + (1-\lambda) d$ with $\lambda \in (0,1)$ and $c,d \in S$ then $c_{n}=d_{n}=0$.
So $a+\frac{v_{n}}{a_{n}}v,c,d$ are solutions of the sistem:
$$
\begin{pmatrix}
a_{11}& \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn} \\
\end{pmatrix}
\begin{pmatrix}
x_{1} \\
\vdots \\
x_{n-1}\\
0
\end{pmatrix}
=
\begin{pmatrix}
b_{1} \\
\vdots \\
b_{n}
\end{pmatrix}
$$
and how $x_{n}=0$ we can delete the last column of our matrix and our solutions are solutions of the sistem:
$$
\begin{pmatrix}
a_{11}& \cdots & a_{1n-1} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn-1} \\
\end{pmatrix}
\begin{pmatrix}
x_{1} \\
\vdots \\
x_{n-1}
\end{pmatrix}
=
\begin{pmatrix}
b_{1} \\
\vdots \\
b_{n}
\end{pmatrix}
$$
we denote
$$A'=
\begin{pmatrix}
a_{11}& \cdots & a_{1n-1} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn-1} \\
\end{pmatrix},
X'=
\begin{pmatrix}
x_{1}\\
\vdots\\
x_{n-1}
\end{pmatrix},
B=
\begin{pmatrix}
b_{1}\\
\vdots\\
b_{n}
\end{pmatrix}
$$
then we have that $rg(A') \leq n-1$, and how $a+\frac{v_{n}}{a_{n}}v$ is solution of $A'X'=B$ we have that $rg(A')=rg(A'|B) \leq n-1$, so we can delete one row of $A'$ and $B$. We can suppose that we delete the $n$-row.
So our sistem is equivalent to the sistem:
$$
\begin{pmatrix}
a_{11}& \cdots & a_{1n-1} \\
\vdots & \ddots & \vdots \\
a_{n-11} & \cdots & a_{n-1n-1} \\
\end{pmatrix}
\begin{pmatrix}
x_{1} \\
\vdots \\
x_{n-1}
\end{pmatrix}
=
\begin{pmatrix}
b_{1} \\
\vdots \\
b_{n-1}
\end{pmatrix}
$$
and by induction hipotesis we have that this last sistem restricted to $\{x' \geq 0\}$ has at least one extreme point.
