Question on linear independence of particular vectors of $\mathbb{R}^8$ 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).
 A: This is not a full answer, but an attempt which would hopefully be helpful.  We can consider $\mathbb{C}^n$ to be $\mathbb{R}^{2n}$ by identifying $\begin{bmatrix}x_1+iy_1\\\vdots\\x_n+iy_n\end{bmatrix}\in\mathbb{C}^n$ with $\begin{bmatrix}x_1\\\vdots\\ x_n\\y_1\\\vdots\\ y_n\end{bmatrix}\in\mathbb{R}^{2n}$.  The given problem looks like an instance of the following setting.
Let $\alpha_1,\ldots,\alpha_n\in\mathbb{C}$ with $\alpha_j=\beta_j+i\gamma_j$.  Let $\beta=\begin{bmatrix}\beta_1\\\vdots\\\beta_n\end{bmatrix},\,\gamma=\begin{bmatrix}\gamma_1\\\vdots\\\gamma_n\end{bmatrix}\in\mathbb{R}^n$.  Then by our identification, we have $\alpha=\begin{bmatrix}\alpha_1\\\vdots\\\alpha_n\end{bmatrix}=\begin{bmatrix}\beta\\\gamma\end{bmatrix}$.
Now we have vectors $v_l,\,l=1,\ldots,s$ in $\mathbb{C}^n$, where with respect to the standard basis, each coordinate of $v_l$ is a $\mathbb{R}$-linear combination of $\alpha_i$'s.  So we have matrices $T_l\in M_{n\times n}(\mathbb{R})$ such that $v_l=\begin{bmatrix}T_l&0\\0&T_l\end{bmatrix}\begin{bmatrix}\beta\\\gamma\end{bmatrix},\quad l=1,\ldots,s$.
For the given example, we have $\beta=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix},\,\gamma=\begin{bmatrix}y_1\\y_2\\y_3\\y_4\end{bmatrix}$.  Further,
$$T_1=\begin{bmatrix}0&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix},\quad T_2=\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix},\quad T_3=\begin{bmatrix}1&-1&0&0\\-1&1&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix},\quad T_4=\begin{bmatrix}0&0&0&0\\0&1&0&-1\\0&0&0&0\\0&-1&0&1\end{bmatrix}$$
$$T_5=\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&1&-1\\0&0&-1&1\end{bmatrix},\quad  T_6=\begin{bmatrix}0&0&0&0\\1&0&-1&0\\0&0&0&0\\-1&0&1&0\end{bmatrix},\quad T_7=\begin{bmatrix}1&0&0&-1\\0&0&0&0\\0&0&0&0\\-1&0&0&1\end{bmatrix}$$
It can be checked that $\{T_1,\ldots,T_7\}$ is linearly independent in $M_{n\times n}(\mathbb{R})$.  Define $\phi:M_{n\times n}(\mathbb{R})\to\mathbb{R}^{2n}$ as $\phi(T)=\begin{bmatrix}T&0\\0&T\end{bmatrix}\begin{bmatrix}\beta\\\gamma\end{bmatrix}$.  Then $\phi(T_i)=v_i,\,i=1,\ldots,7$.  Let $W=\text{span}\{T_1,\ldots,T_7\}$.  Then $\{v_1,\ldots,v_7\}$ are linearly independent if and only if $\phi|_W$ is injective.
