Let $\mathbb{K}$ be a field (it may have positive characteristic and/or not be algebraically closed) and let $\mathbb{F} = \overline{\mathbb{K}}$ be it's algebraic closure. Let $0 \neq v \in \mathbb{F}^n$.
I want to know if it's possible to find $P_1, \ldots, P_{n-1} \in GL_n(\mathbb{K})$ such that the set $\{ v, P_1 v, \ldots P_{n-1} v\}$ is linearly independent. In case it is I'd also like to find a way to construct said matrices.
The only semi-close attempt I've encountered only works if $char(\mathbb{K}) \neq 2$ where you find some of the vectors by multiplying by 2 the entries of $v$ that aren't 0 and do some permutations for those that are. I don't know how to start the general case or even the characteristic 2 case.
Let us assume without loss of generality that the $0$th entry of $v$ is nonzero. Now let $P_i$ be the matrix given by $P_i(e_0)=e_0+e_i$ and $P_i(e_j)=e_j$ for $j\neq 0$, where $e_i$ is the $i$th standard basis vector. Note that $P_iv=v+ae_i$, where $a$ is the $0$th entry of $v$. It follows that $v,P_1v,\dots,P_{n-1}v$ are linearly independent, since they have the same span as $v,ae_1,\dots,ae_{n-1}$ and $a\neq 0$.