Geometric n-simplex is convex

Given the definition of a geometric n-simplex:

\displaystyle\sigma^n=\{x=\sum_{i=0}^{n}t_i a^i \mid t_i\geq 0\ \text{and }\sum_{i=0}^{n}=1\}\subseteq\mathbb{R}^n

where \{a^0,\dots, a^n\} are geometrically independent, we can show that the n-simplex is convex (i.e. given any two points, the line connecting them lies in the simplex).

Write x=\sum_{i=0}^n t_i a^i, y=\sum_{i=0}^n s_i a^i.

Consider the line from x to y: \{ty+(1-t)x\mid 0\leq t\leq 1\}.

\begin{aligned}    ty+(1-t)x&=t\sum_{i=0}^n s_i a^i+(1-t)\sum_{i=0}^n t_i a^i\\    &=\sum_{i=0}^n (s_i t+t_i-tt_i)a_i\\    s_it+t_i-tt_i&=s_i t+t_i (1-t)\\    &\geq 0(0)+(0)(1-1)\\    &=0\\    \sum_{i=0}^n s_i t+t_i-tt_i &=t\sum_{i=0}^n s_i+\sum_{i=0}^n t_i -t\sum_{i=0}^n t_i\\    &=t(1)+(1)-t(1)\\    &=1    \end{aligned}

Thus the line lies inside the simplex, and thus the simplex is convex.


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