Show that the subspace $M = \{x~:x_1 + x_2 = 0\} \subset \mathbb{R}^2$ is spanned by one vector 
Show that the subspace $M = \{x~:x_1 + x_2 = 0\} \subset \mathbb{R}^2$ is spanned by one vector

I want to show that the system of equations has at least one solution and thus is spanned by the vector, but I am not sure what the system of equations should look like.
Can I take two vectors $x,y \in M$ and make a system like this?
$$\begin{align}
a(x_1 + x_2) = 0 \\
b(y_1 + y_2) = 0
\end{align}$$
Then matrix looks like...
$$\begin{bmatrix}
1 & 1 & 0 \\
1 & 1 & 0  
\end{bmatrix} 
\rightarrow 
\begin{bmatrix}
1 & 1 & 0 \\
0 & 0 & 0  
\end{bmatrix} \rightarrow ~~\textrm{infinite solutions b\c all 0's on the bottom?}$$
 A: Other answers have given you solutions, but none has yet addressed your attempt to solve the problem and why it doesn’t work.  
Starting at the end, it’s not enough to conclude that the system has an infinite number of solutions. You’re supposed to show that $M$ is one-dimensional, so you need to figure out what the dimension of the solution space to your system is. After all, the system might’ve been equivalent to $0=0$, which has all of $\mathbb R^2$ as a solution space. You have to eliminate that possibility.  
However, it’s not at all clear what this system of equations that you’ve created is meant to represent in the first place. Certainly, the augmented matrix that you present doesn’t correspond to this system: if it’s the $x_i$ and $y_i$ that you’re considering to be the variables, then it should have five columns; if it’s $a$ and $b$ that are the variables, then the left side of the matrix should be diagonal. If they’re all variables, then the two equations aren’t linear and can’t be represented by a matrix in this way.  
Perhaps what you meant to do here is to write down two scalar multiples of the defining equation of $M$. This doesn’t provide any new information, since they’re redundant with the original equation, and only create more work for you to do. They also introduce extraneous solutions: you haven’t eliminated the possibility that $a=b=0$, in which case the solution set is all of $\mathbb R^2$.  
You’re already got a perfectly good system of equations to examine in the definition of $M$, so you should use that directly. This system happens to consist of only one equation, but that doesn’t make any difference to the way you go about determining the dimension of its solution space: it just eliminates a lot of work. The augmented matrix that corresponds to this system of equations is $$\left[\begin{array}{cc|c}1&1&0\end{array}\right].$$ This is already in row-reduced echelon form, so you can immediately conclude that the nullity of the left side is one—there’s one non-pivot column—and as this is a homogeneous system, the solution set is a one-dimensional subspace of $\mathbb R^2$.
A: A vector of $M$ is
$$(x_1,x_2)=(x_1,-x_1)$$
$$=x_1(1,-1)$$
thus $M$ is a vectorial line spanned by the vector $(1,-1)$.
A: Let $L : \mathbb{R}^2 \to  \mathbb{R}$ be $Lx = x_1+x_2$. Note that $\dim {\cal R} L = 1$
and so the rank nullity theorem gives $\dim \ker L = 2-1$. Note that $M = \ker L$.
A: Hint:
Your subspace $M$ is one-dimensional and consists of all vectors orthogonal to $(1,1)$. It is therefore spanned by a vector which is orthogonal to $(1,1)$, such as $(1,-1)$.
