Prove vertices of a simplex are affinely independent Given that the definition of a simplex T

Here is a not-so-great solution.
The closure of $T$ is $x$ s.t. $(u_k,x)\le <!--\; \le \; -->c_k$, which is a convex polytope. Define an extreme point of a set to be a point which cannot be written as an intermediate point on a line segment between two other distinct points in the set.
It can be demonstrated that all the $v_k$ lie in the closure of $T$. I will now argue that $v_k$ are the extreme points of closure $T$, and hence closure $T$ is the convex hull of $v_k$.
Suppose $t$ is an extreme point which satisfies $C$ of the constraints $(u_k,t)=c_k$ where $C<<!--\; <\; -->n$. Another way of saying this is $Ut=c$ where $U$ is the matrix with rows $u_k$ and $c$ is the column vector with entries $c_k$. Since $U$ is $C\times <!--\; \times \; -->n$ dimensional, it has only rank $C$ and therefore it has nonempty kernel. Then we may perturb $t$ by adding a small kernel element and remain inside closure $T$ which contradicts t being an extreme point.
So the only possible extreme points are points which satisfy $n$ constraints. So the $v_k$ are the only possible extreme points. Now I check that $v_k$ are in fact extreme points. Suppose that we have a $v_k$ which is strictly interior to a line segment $[a,b]$ with $a$ and $b$ in closure $T$. Then $a$ and $b$ must satisfy the same $n$ constraints as $v_k$ or else by convexity, the line between them would not satisfy $n$ constraints. But then $v_k$ would equal $a$ and $b$. So then $v_k$ are exactly the extreme points.
A theorem of Linear Programming then tells us that the closure of $T$ is the convex hull of the $v_k$. If the $v_k$ were not affine independent, the closure of $T$ would have measure 0, which is a contradiction since $T$ is non-empty and open.

Great expert answer!