Can the minimal solutions of a system of linear equations be found with help of the solution vector of the whole system of linear equations?
Definition:
"A set $S\subset\mathbb{R}^n$ is called [a minimal solution, or] a linear algebraic simplex, if $S$ is minimal linearly dependent, that is $S$ itself is linearly dependent but all its proper subsets $T\subset S$ are linearly independent."
Szalkai, I.; Dósa, G; Tuza, Z.; Szalkai, B.: On minimal solutions of systems of linear equations with applications. Miskolc Mathematical Notes 13 (2012) (2) 529-541
Take e.g. the following homogenous system of linear equations:
$$\left( \begin{array}{} 1&0&0&0&1&1&0&0&0&0&0\\ 0&2&0&0&0&0&2&3&0&0&0\\ 0&0&2&0&0&0&0&1&1&1&1\\ 0&0&0&2&1&2&1&0&1&2&3 \end{array} \right) \cdot \left( \begin{array}{} \nu_1\\\nu_2\\\nu_3\\\nu_4\\\nu_5\\\nu_6\\\nu_7\\\nu_8\\\nu_9\\\nu_{10}\\\nu_{11} \end{array} \right) =\left(\begin{array}{}0\\0\\0\\0\end{array}\right)$$
The solution vector of this system of equations is:
$$\left( \begin{array}{c} 2\nu_4+\nu_6+\nu_7+\nu_9+2\nu_{10}+3\nu_{11}\\-\nu_7+3\nu_3+\frac{3}{2}\nu_9+\frac{3}{2}\nu_{10}+\frac{3}{2}\nu_{11}\\\nu_3\\\nu_4\\-2\nu_4-2\nu_6-\nu_7-\nu_9-2\nu_{10}-3\nu_{11}\\\nu_6\\\nu_7\\-2\nu_3-\nu_9-\nu_{10}-\nu_{11}\\\nu_9\\\nu_{10}\\\nu_{11} \end{array} \right)$$
There are 663 solutions. Among them are 83 minimal solutions.
We see, $\nu_1$, $\nu_2$, $\nu_5$ and $\nu_8$ depend on other $\nu$, but $\nu_3$, $\nu_4$, $\nu_6$, $\nu_7$, $\nu_9$, $\nu_{10}$ and $\nu_{11}$ don't depend on other $\nu$.
I would imagine that it is sufficient to combine the $\nu$ with each other in a suitable way to get only the minimal solutions.