I'm trying to understand the connection between the notions of linear independence and general position. I have no background in geometry, so first I'll start with what I know and then I'll pose specific questions, please bear with me and correct me at any point.
Let $q$ be a prime power, $d$ be a nonnegative integer, and $V$ be a $(d+1)$-dimensional vector space over the finite field $F_q$ with $q$ elements. For $v \in V$ denote $$[v] = \left\{ cv \mid c \in F_q, c\neq 0 \right\}.$$ Then the collection of symbols $[v]$ can be seen as the points of the $d$-dimensional projective space PG(d,q). Furthermore, for a $(k+1)$-dimensional subspace $S$ of $V$, the set $$\left\{ [s] \mid s \in S \right\}$$ is a $k$-flat of $PG(d,q)$.
From what I read in pg. 19 of these notes I assume that this definition of the notion of "general position" is correct:
We say that $m$ points in $PG(d,q)$ are in general position if they are not contained in any $(m-2)$-flat.
So, to my understanding, the following statement is correct:
The points $[v_1], \ldots, [v_m]$ of $PG(d,q)$ are in general position if and only if the vectors $v_1, \ldots, v_m$ are linearly independent.
My proof. The points $[v_1], \ldots, [v_m]$ are in general position iff they are not contained in any $(m-2)$-flat, which is true iff $v_1, \ldots, v_m$ are not contained in any $(m-1)$-dimensional subspace in $V$. This is the same as linear independence of $v_1, \ldots , v_m$.
My questions are: Is the above statement correct? Are the preceding definitions accurate?
PS. The reason for my confusion is that I've read different definitions for "general position" that I don't understand well, as well as discussions were people explain that general position is not equivalent to linear independence (which I thought my statement above implies). While I'm trying to understand and digest things, it would be very helpful to know if I got the above correctly.