How to turn span into linear equality constraint? Suppose $$\text{Im}(A) = \text{span}\{\begin{bmatrix} 0 \\ 1 \\ 0 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 0 \\ 1\\ 0 \end{bmatrix} \}$$
How would you go about turning this very set into a linear equality constraint of the form: $$\text{Im}(A) = \{x \in \mathbb{R}^4  | Gx = 0\}$$
In other words, how do you find the $G$?
 A: First your system of vectors obviously has rank $2$. In $\mathbf R^4$, their span requires $4-2=2$ linearly independent equations.
Although one may guess the system of equations, due to the very simple of vectors, I'll show the general method; the vector $u=\begin{bmatrix}x\\y\\z\\t\end{bmatrix}$ lies inthe span of $e_1=\begin{bmatrix}0\\1\\0\\1\end{bmatrix}$ and  $e_2=\begin{bmatrix}1\\0\\1\\0\end{bmatrix}$ if and only if the system of equations in $\lambda,\mu$:
$$u=\lambda e_1+\mu e_2 $$
has a solution. Let's solve this system by Gauß's pivot method applied to the bordered matrix:
$$\begin{bmatrix}0&1&x\\1&0&y\\0&1&z\\1&0&t\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&y\\0&1&x\\0&1&z\\1&0&t\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&y\\0&1&x\\0&0&z-x\\0&0&t-y\end{bmatrix}$$
(we first swapped row 1 and row 2, then substracted row1 from row 3, and row 3 from row 4). Hence the consistency conditions to have a solution are
$$\begin{cases}
x-z=0,\\y-t=0.
\end{cases}$$
A: let $x = a\begin{bmatrix} 0 \\ 1 \\ 0 \\ 1 \end{bmatrix} + b\begin{bmatrix} 1 \\ 0 \\ 1\\ 0 \end{bmatrix} = \begin{bmatrix} b \\ a \\ b \\ a \end{bmatrix}$ for any $a,b \in \mathbb{R}$
Since $Im(A)$ equal to the span, $Gx=0$ 
Now we can just consider G as a row vector, let $G=[g_1,g_2,g_3,g_4]$
$Gx=0 \implies g_1b+g_2a+g_3b+g_4a=0 \implies (g_1+g_3)b+ (g_2+g_4)a=0$
now since this must be true for all $a$ and $b$ in $\mathbb{R}$, 
$(g_1+g_3)=0$ and $(g_2+g_4)=0$
$g_1=-g_3$ and $g_2=-g_4$
Therefore $G=[p,q,-p,-q]$ for any $p$ or $q$ in $\mathbb{R}$
