## 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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