Show that the span of the $\{x_1, x_2, x_3, x_4 \}$ is $\{(a, b,c, d): 2a - b = 0, 2a - 3c - d = 0\}$. Show that the span of the $\{x_1, x_2, x_3, x_4 \}$ is $\{(x, y,z, w): 2x - y = 0, 2x - 3z - w = 0\}$ where $x_1 = (1,2,1, -1), x_2 = (2,4,1,1), x_3 = (-1,-2,-2,-4) \text{ and } x_4 = (3,6,2,0)$.
My approach:
We have to solve for $c = (c_1, c_2, c_3, c_4)$ in the following system.
$$\vec x = c_1\vec v_1 + c_2\vec v_2 + c_3\vec v_3 + c_4\vec v_4$$
$$\left[\begin{array}{cccc|c}  
 1 & 2 & -1 & 3& x\\  
 2 & 4 & -2 & 6 & y\\  
 1 & 1 & -2 & 2 & z\\
 -1 & 1 & -4 & 0 & w\\
\end{array}\right]$$
$$\left[\begin{array}{cccc|c}  
 1 & 2 & -1 & 3& x\\  
 0 & 0 & 0 & 0 & 2x-y\\  
 0 & 1 & 1 & 1 & x-z\\
 0 & 0 & 1 & 0 & {2x -3z - w \over 8}\\
\end{array}\right]$$
In order to have a solution $2x-y = 0$.
Hence according to me, the span should be $\{(x, y,z, w): 2x - y = 0\}$
I don't get why the other condition ie. $2x - 3z - w = 0$ is part of the span. Could someone please explain to me why it has to be included.
 A: I'm afraid you simply messed up the row operations. Allow me to present a proof which doesn't use row operations and is therefore far more difficult to screw up.
Let $S =\{(x, y, z, w) : 2x - y = 0$ and $2x - 3z - w = 0\}$. Then $S$ is the kernel of the linear function $T$ defined by $T(x, y, z, w) = (2x - y, 2x - 3z - w)$ and is therefore a subspace of $\mathbb{R}^4$.
Let $J$ be the span of $x_1, ..., x_4$.

*

*$J \subseteq S$.

Proof: because $S$ is a subspace, it suffices to show that it includes $x_1, ..., x_4$. This is immediate by routine checking.


*$S \subseteq J$.

Proof: $S$ is the kernel of $T : \mathbb{R}^4 \to \mathbb{R}^2$. Clearly, $im(T) = \mathbb{R}^2$. And by the rank-nullity theorem, $S$ therefore has dimension 2. Since $x_1$ and $x_4$ are two independent elements of $S$, they must form a basis for $S$. Therefore, $S \subseteq J$.
Then $S = J$.
A: I suspect it should be $x_3 = (-1,-2,-2,4)$. Then the correct reduced matrix is
$$\left[\begin{array}{cccc|c}  
 1 & 2 & -1 & 3& x\\  
 2 & 4 & -2 & 6 & y\\  
 1 & 1 & -2 & 2 & z\\
 -1 & 1 & 4 & 0 & w\\
\end{array}\right] \sim \left[\begin{array}{cccc|c}  
 1 & 2 & -1 & 3& x\\  
 0 & 0 & 0 & 0 & 2x-y\\  
 0 & 1 & 1 & 1 & x-z\\
 0 & 0 &0 & 0 & {2x -3z - w}\\
\end{array}\right]$$
so a solution exists if and only if
$$2x-y=0, \quad 3x-3z-w=0.$$
