TITLE: Question on linear independence of particular vectors of $\mathbb{R}^8$
QUESTION [11 upvotes]: Given any four points $z_1,z_2,z_3,z_4\in \mathbb{C}$, all different from zero, consider the following seven vectors of $\mathbb{C}^4$:
$$v_1:=(0,z_2-0,0,0)$$
$$v_2:=(0,0,z_3-0,0)$$
$$v_3:=(z_1-z_2,z_2-z_1,0,0)$$
$$v_4:=(0,z_2-z_4,0,z_4-z_2)$$
$$v_5:=(0,0,z_3-z_4,z_4-z_3)$$
$$v_6:=(z_1-z_3,0,z_3-z_1,0)$$
$$v_7:=(z_1-z_4,0,0,z_4-z_1)$$
Under what conditions on $z_1,z_2,z_3,z_4$ are these seven vectors $v_1,\dots,v_7$ linearly independent as vectors of $\mathbb{R}^8$?
Of course I know how to consider the vectors $v_i$ as vectors of $\mathbb{R}^8$. The difficult part (which is the aim of the question) is to deduce all sufficient conditions for their linear independence.
Edit: User MvG posted an answer which involves the use of a computer algebra system, but I am actually interested in a method which could be solved by hand (without a computer).

REPLY [2 votes]: Because $\mathbb{R}^8_\mathbb{R}$ and $\mathbb{C}^4_\mathbb{R}$ are isomoprhic this is equivalent to finding conditions so that $V = \{v_1, ...,v_7\}$ is $\mathbb{R}$-linear independent.
Let $x = (x_1, ..., x_7) \in \mathbb{R}^7$. We want to obtain conditions on $V$ such that
$$ \sum\limits_{n=1}^7 x_n v_n = 0 \implies \forall n \in 1...7, x_n = 0 $$
Since a zero vector would immediately ruin the linear independence of $V$ assume for a while that $z_1 \neq z_2$, $z_2 \neq z_4$, $z_3 \neq z_4$, $z_1 \neq z_3$ and $z_1 \neq z_4$. Later some of these will be replaced by stronger conditions.
Writing the equation $\sum\limits_{n=1}^7 x_n v_n = 0$ in matrix form we get
$$ \underbrace{ \left[
\begin{array}{ccccccc}
 0 & 0 & z_1-z_2 & 0 & 0 & z_1-z_3 & z_1-z_4 \\
 z_2 & 0 & z_2-z_1 & z_2-z_4 & 0 & 0 & 0 \\
 0 & z_3 & 0 & 0 & z_3-z_4 & z_3-z_1 & 0 \\
 0 & 0 & 0 & z_4-z_2 & z_4-z_3 & 0 & z_4-z_1 \\
\end{array}
\right] }_{\displaystyle A} x = 0 $$
Row reducing $A$
$$ \left[
\begin{array}{ccccccc} \displaystyle
 1 & 0 & 0 & 0 & \dfrac{z_4-z_3}{z_2} & \dfrac{z_1-z_3}{z_2} & 0 \\
 0 & 1 & 0 & 0 & \dfrac{z_3-z_4}{z_3} & \dfrac{z_3-z_1}{z_3} & 0 \\
 0 & 0 & 1 & 0 & 0 & \dfrac{z_1-z_3}{z_1-z_2} & \dfrac{z_1-z_4}{z_1-z_2} \\
 0 & 0 & 0 & 1 & \dfrac{z_3-z_4}{z_2-z_4} & 0 & \dfrac{z_4-z_1}{z_4-z_2} \\
\end{array}
\right] x = 0$$
Multiplying $Ax$ and equating elements yields
$$ x_2 = -x_1 \frac{z_2}{z_3} \tag{1} $$
$$ x_3 = x_1 \frac{z_2}{z_1-z_2} - x_7 \frac{z_1-z_4}{z_1-z_2} - x_5 \frac{z_3-z_4}{z_1-z_2} \tag{2} $$
$$ x_4 = - x_5 \frac{z_3-z_4}{z_2-z_4} - x_7 \frac{z_4-z_1}{z_4-z_2} \tag{3} $$
$$ x_6 = -x_1 \frac{z_2}{z_1-z_3} -x_5 \frac{z_4-z_3}{z_1-z_3} \tag{4} $$
Note that because $\dim\ker A \ge 3$ there will be at least $3$ free variables and this means that there are other ways of manipulating these equations. Nonetheless, every way will result in equivalent conclusions.
Starting by the simpler equation: $(1)$ we make $x_1 = x_2 = 0$ by requiring that $\operatorname{Im}{(\dfrac{z_2}{z_3})} \neq 0$.
Now setting $x_1 = 0$ in $(4)$, the same reasoning leads to requiring $\operatorname{Im}{(\dfrac{z_4-z_3}{z_1-z_3})} \neq 0$. Hence $x_5 = x_6 = 0$.
Similarly for $(3)$, we require $\operatorname{Im}{(\dfrac{z_4-z_1}{z_4-z_2})} \neq 0$. Then $x_4 = x_7 = 0$.
Thus $(2)$ yields $x_3 = 0$.
Therefore we conclude that the following six conditions are sufficient for $V$ to be $\mathbb{R}$-L.I.:

$z_1 \neq z_2$
$z_1 \neq z_3$
$z_2 \neq z_4$
$\operatorname{Im}{(\dfrac{z_2}{z_3})} \neq 0$
$\operatorname{Im}{(\dfrac{z_4-z_3}{z_1-z_3})} \neq 0$
$\operatorname{Im}{(\dfrac{z_4-z_1}{z_4-z_2})} \neq 0$