I am looking to prove the following statement:
Let $\{a_i\}^n_{i=0} \subseteq L$ be a set of points, then we have
$\{a_i\}^n_{i=0}$ is geometrically independent $\Longleftrightarrow \{a_i - a_0\}^n_{i=1}$ is linearly independent
Before going into the proof, I'll state for being complete geometrical independence
A set of points $\{a_i\}^n_{i=0} \subseteq L$ is called geometrically independent if for any scalar $\gamma_i$ it holds \begin{equation} \sum_{i=0}^n \gamma_i = 0 \;\; \land \;\; \sum_{i=0}^n \gamma_i a_i = 0 \;\;\Longrightarrow \;\; \gamma_0 = \gamma_1 = \dots = \gamma_n = 0 \end{equation}
Proof:
Let $\{(a_i, \gamma_i)\}_{i=0}^n$ be a set of weighted geometrically independent points. We have
\begin{align} \sum_{i=0}^n \gamma_i a_i &= 0 \\ \sum_{i=1}^n \gamma_i a_i + \gamma_0 a_0 &= 0 \end{align} Note, $a_i$ are geometrically independent, hence $\gamma_0 = \ldots = \gamma_n = 0$, we can write \begin{align} \sum_{i=1}^n \gamma_i a_i - \sum_{i=1}^n \gamma_i a_0 = 0\\ \sum_{i=1}^n \gamma_i a_i - \gamma_i a_0 = 0\\ \sum_{i=1}^n \gamma_i (a_i - a_0) = 0\\ \end{align}
Is this direct proof sufficient for the equivalence statement, i.e. does it show both directions sufficiently, since we can go top-down/bottom-up?