Intuition regarding convex hulls and convex combinations I know that the following definitions of convex hull are equivalent: 
$\textbf{Definition 1: }$The convex hull of $k$ points in $\mathbb{R^n}$ is the smallest convex set containting the $k$ points.
$\textbf{Definition 2: }$The convex hull of $k$ points in $\mathbb{R^n}$ is the set $C$ of all convex combinations of the $k$ points.
Here we take convex combination to be a linear combination $c_1 a_1 + \ldots + c_k a_k$ where $c_i \geq 0$ and $\sum c_i =1$ Could someone geometrically explain why the restrictions on $c_i$ force $C$ to be the convex hull of the $k$ points. For just two points it is clear to me but for the case of three points or more I find it hard to grasp intuitively. Why do these algebraic constraints on the coefficients of the combinations produce figures like triangles, squares, tetrahedrons, etc.
Thanks.
 A: 
For just two points it is clear to me

Think iteratively (assuming $c_j \lt 1$ for the sake of the following argument, which doesn't lose generality since cases where one of the $c_j=1$ correspond to actual points in the given set, which are known to be included in the convex hull anyway): $$\;\displaystyle a =c_1 a_1 + \ldots + c_k a_k = c_1 a_1 + (1-c_1) \left(\frac{c_2}{1-c_1}a_2+ \cdots + \frac{c_k}{1-c_1}a_k\right)=c_1 a_1+(1-c_1)a'$$
Therefore $a$ is on the segment between $a_1$ and $a'$ where $\displaystyle a'=\frac{c_2}{1-c_1}a_2+ \cdots + \frac{c_k}{1-c_1}a_k\,$. Note that $\displaystyle \frac{c_j}{1-c_1} \ge 0$ and $\displaystyle \sum_{j=2}^k \frac{c_j}{1-c_1}=1\,$, so $a'$ is a convex combination of the $k-1$ points $a_2, \cdots,a_k$. Repeat the process to show that $a'=\lambda a_2 + (1-\lambda)a''\,$ so $a'$ is on the segment between $a_2$ and $a''$ where $a''$ is a convex combination of $a_3, \cdots,a_k$, etc. In the end you'll get to a segment ending at $a_{k}\,$.

[ EDIT ]   Following up on a comment, as far as the geometric intuition goes, it may help to look at it from the end backwards:


*

*you start at point $\,A_k=a_k\,$, which obviously belongs to the convex hull

*then $\displaystyle\,A_{k-1}=\frac{c_{k-1} a_{k-1} + c_k a_k}{c_{k-1}+c_k}\,$ takes you to a point $A_{k-1}$ on segment $a_{k-1}a_k\,$, which again obviously belongs to the convex hull

*next, $\displaystyle\,A_{k-2}=\frac{c_{k-2} a_{k-2} + c_{k-1} a_{k-1} + c_k a_k}{c_{k-2}+c_{k-1}+c_k}=\frac{c_{k-2} a_{k-2} +(c_{k-1}+c_k)A_{k-1}}{c_{k-2}+c_{k-1}+c_k}\,$ takes you to a point $A_{k-2}$ on segment $a_{k-2}A_{k-1}\,$, which is inside the triangle $a_{k-2}a_{k-1}{a_k}$ i.e. the convex hull of points $a_{k-2}, a_{k-1}, a_k\,$, therefore inside the convex hull of all points again

*after $\,k\,$ steps you end up at $\,A_1=a\,$, and the entire path from $\,a_k\,$ to $\,a\,$ consists of line segments contained in the convex hull
