I have a system of equations with infinitely many solutions. I would like to find a "nice" way to write down an explicit solution.
Here, $n,k\geq 1$ are integers, we have $x_1,x_2,\dots, x_{2^k n}$ unknowns and $n$ equations which take the following form:
$$ \begin{cases} x_1+x_{n+1}+x_{2n+1}+\cdots + x_{(2^k -1)n+1}=0\\ x_2 + x_{n+2}+x_{2n+2}+\cdots + x_{(2^k -1)n+2}=0\\ \vdots\\ x_{n} + x_{2n}+ x_{3n}+\cdots + x_{2^kn} =0. \end{cases} $$
Observe two important things: the variables are never repeated, i.e. they only appear once, so the equations are "independent" of each other. Also, the unknowns are ordered vertically, that is, starting from the top left $x_1$ then going down to $x_2,x_3,\dots, x_n$ then up again to the second on the top left $x_{n+1}$ then down again... and so on until we arrive at $x_{2^k n}$. As an example: If $k=2$ and $m=3$ then
$$ \begin{cases} x_1+x_{4}+x_{7}+x_{10}=0\\ x_2 + x_{5}+x_{8}+x_{11}=0\\ x_{3} + x_{6}+ x_{9}+ x_{12} =0. \end{cases} $$
By Rouché-Cappelli's theorem since the ranges of the two matrices are equal but less than $2^k m$ we have infinitely many solutions. I aim at writing something like:
$$w_1 := x_1, w_2 := x_2,\dots, w_m := x_m$$ and then deriving the rest in terms of the $w_j$. Is there a "nice" and "clean" well-shaped pattern for doing this?
I'll be really thankful for any ideas you might have!
The standard way to interpret your system of linear equations is that all variables except $x_1,x_2,\ldots, x_n$ are free, and may be chosen arbitrarily. These first variables are bound, and are determined in terms of the others.
For your small example, $$\begin{cases}\text{For all }x_4,x_5,x_6,x_7,x_8,x_9,x_{10},x_{11},x_{12}\in\mathbb{Z}(?),\\x_1=-(x_4+x_7+x_{10})\\x_2=-(x_5+x_8+x_{11})\\x_3=-(x_6+x_9+x_{12})\end{cases}$$