Consider the vector $V\in\mathbb{R}^6$:
$$V=\begin{bmatrix} -1&1&-1&-1 \\ -1&1&1&-1 \\ -1&-1&1&-1 \\ -1&1&1&1 \\ -1&-1&1&1 \\ -1&-1&-1&1 \end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$
I'd like to find values of $a,b,c,d$ such that $V$ has exactly $4$ different coordinates (among its 6 coordinates). I am not sure if it is possible with $a,b,c,d$ all different but I am stuck with this apparently simple problem. Any ideas?
Edit The rank of the matrix is 4 and denoting by $R_i$ the $i$-th row of the matrix, $R_5=-R_2+R_3+R_4$ and $R_6=R_1-2R_2+R_3+R_4$.
The row-reduced form of the matrix is $$\begin{bmatrix} 1&0&0&0 \\0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \\ 0&-1&1&1 \\ 1&-2&1&1 \end{bmatrix}$$
As I did not manage to find an elegant way to solve this problem, I solved it systematically. In each step I solved $V_i=V_j$ and $V_k=V_l$ with $(i,j)$ and $(k,l)$ different subsets of $\{(2,1), (3,1),\dots,(6,5)\}$.
Deleting all the duplicates, what I finally obtained is (the "$\to$" are to be interpreted as "$=$"): $$\left( \begin{array}{c} \{\{b\to 0,c\to 0\}\} \\ \{\{c\to 0,d\to 0\}\} \\ \{\{c\to 0,d\to b\}\} \\ \{\{c\to 0,d\to -b\}\} \\ \{\{c\to 0\}\} \\ \{\{c\to b,d\to -b\}\} \\ \{\{c\to b,d\to 0\}\} \\ \{\{c\to b,d\to b\}\} \\ \{\{c\to b,d\to 2 b\}\} \\ \{\{b\to 0,d\to -c\}\} \\ \{\{c\to -b,d\to b\}\} \\ \left\{\left\{c\to -\frac{b}{2},d\to \frac{b}{2}\right\}\right\} \\ \{\{c\to 2 b,d\to -b\}\} \\ \left\{\left\{c\to \frac{b}{2},d\to \frac{b}{2}\right\}\right\} \\ \{\{c\to -b,d\to 2 b\}\} \\ \{\{b\to 0,d\to 0\}\} \\ \{\{d\to b\}\} \\ \{\{b\to 0,d\to c\}\} \\ \{\{b\to 0\}\} \\ \{\{c\to -b,d\to 0\}\} \\ \{\{d\to 0\}\} \\ \{\{c\to -2 b,d\to -b\}\} \\ \{\{c\to -b,d\to -b\}\} \\ \end{array} \right)$$
If I restrict these values to positive values, and add the condition that all $a,b,c,d$ should be all different, nothing is left. That answers my question but I am still interested, out of curiosity, in a more elegant way to find this.