I am new to convex hulls and I have encountered the following statement:
Given a set of n points in a vector space $\{x_1 ... x_n \}$, every point $x_0$ in the convex hull they form is obtained from the expression:
$$ x_0 = \sum_{i=1}^n \alpha_i x_i$$ where $\alpha_i\geq0$ and $\sum_{i=1}^n\alpha_i=1$, with different coefficients $\alpha_i$ giving different points, always within the convex hull.
I have seen in this other answer a very clear example in $\mathbb R^2$ with only three points, for which the convex hull is just the convex combination of those three points, and corresponds to the points within the triangle formed by them. In $\mathbb R^2$, but for $n$ points, the convex hull is just the set of all points inside the polygon formed by the "outer" points of the set - enclosing the rest of "inner" points. (Formally, the smallest convex set which contains all the points in our set of $n$ points.)
If we added another point inside the triangle, the convex hull would remain the same; however, based on the expression I provide, that new point should also be included in the weighed sum (the sum goes from $0$ to $n$), in spite of lying inside such polygon and therefore not altering the convex hull. Why would such inner points be included in the sums? Are perhaps their coefficients zero (i.e. $\alpha_i=0$ precisely for those "inner" points)?
Additionally, if possible, I would like to know how the expression I give for the most general case is obtained, or at least how I could intuitively think about it to make it a bit more transparent.