I have a solution set given in the following way and I have to find a system of linear equations with 4 variables.
$\Bigg\{ \begin{align} \begin{bmatrix} m_{1} \\m_{2} \\m_{3} \\m_{4} \end{bmatrix} \end{align}$ + $ \begin{align} m_1\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{bmatrix} \end{align}$ + $\begin{align} m_2\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{bmatrix} \end{align}$;$\:$ $m_1,m_2 \in \mathbb{R} \Bigg\}$ $\subseteq \mathbb{R^4}$
I have tried to construct the equations in like this:
$\begin{align} \begin{bmatrix} m_{1} \\m_{2} \\m_{3} \\m_{4} \end{bmatrix} \end{align} $ = $ \begin{align} \begin{bmatrix} x_{1} \\x_{2} \\x_{3} \\x_{4} \end{bmatrix} \end{align}$ + $ \begin{align} m_1\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{bmatrix} \end{align}$ + $\begin{align} m_2\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{bmatrix} \end{align}$
then getting equation for each $m$ in this way : $m_1 = x_1 +m_1(x_1) + m_2(x_2)$
Now I have set the equations in this manner because there is a solution set for 4 variables namely $m_1, m_2, m_3, m_4$. Now I am not sure if $m_1$ and $m_2$ in my construction $[m_1,m_2,m_3,m_4]^T$ should same as given in the solution set.
I would like to know how to solve such set systems to get linear equations.
P.S: I have not given any example because I want to see how a certain generic solution would look like. If its insufficient mention in comments then I can provide some example.
Your solution vector is of the form
$ x = x_0 + t V_1 + s V_2 $
where $x_0, V_1, V_2 $ are known vectors.
Therefore,
$ (x - x_0) = t V_1 + s V_2 \hspace{48pt} (*)$
We can obtain a linear system in $x$, if we find two vectors $U_1, U_2$ that are orthogonal to both $V_1$ and $V_2$, i.e. we need to solve
$ \begin{bmatrix} V_1^T \\ V_2^T \end{bmatrix} u = 0 $
This will give a solution $u = c_1 U_1 + c_2 U_2 \hspace{48pt} (**)$
From $(*) , (**) $ , it follows that,
$ U_1^T (x - x_0) = 0 $
$ U_2^T (x - x_0) = 0 $
Therefore, the linear system you're looking for is
$ A x = b $
where
$A = \begin{bmatrix} U_1^T \\ U_2^T \end{bmatrix}, \hspace{12pt} b = \begin{bmatrix} U_1^T x_0 \\ U_2^T x_0 \end{bmatrix} $
As a specific example, if
$x_0 = [1, 0, 1, 0]^T , V_1 = [-2, 3, 0, 0]^T, V_2 = [-3, 0, 4, 3]^T $
Then we will solve the system
$\begin{bmatrix} -2 && 3 && 0 && 0 \\ -3 && 0 && 4 && 3 \end{bmatrix} u = 0 $
Using Gauss-Jordan reduction, the matrix is reduced to
$\begin{bmatrix} 1 && 0 && -\dfrac{4}{3} && -1 \\ 0 && 1 && -\dfrac{8}{9} && - \dfrac{2}{3} \end{bmatrix} $
From which,
$U_1 = \begin{bmatrix} 12 \\ 8 \\ 9 \\ 0 \end{bmatrix} $
$U_2 = \begin{bmatrix} 3 \\ 2 \\ 0 \\ 3 \end{bmatrix} $
Now $U_1^T x_0 = 21 $ and $U_2^T x_0 = 3 $, therefore the linear system is:
$\begin{bmatrix} 12 && 8 && 9 && 0 \\ 3 && 2 && 0 && 3 \end{bmatrix} x = \begin{bmatrix} 21 \\ 3 \end{bmatrix}$