Whether set of 3D vectors span space $\{(x, y, z) | x + y + z = 0\}$ Consider the set of (column) vectors defined by 
$$X = \{x \in \mathbb{R}^3 | x_{1} + x_{2} + x_{3} = 0\},$$
 where $X^{T} = [x_{1}, x_{2}, x_{3}]^{T}$, I need to prove whether(or not) given vectors, $[1, -1, 0]^{T}$, $[1, 0, -1]^{T}$ span $X$? 
I find examples to prove the same for $\mathbb{R}^3$ but not for a subspace.
 A: Let $\textbf{y} = (y_1,y_2,y_3)^T$ be any vector of $X$. We know that $y_1+y_2+y_3=0$ since $\textbf{y} \in X$.
We need to know if it is always possible to find $a,b\in \mathbb{R}$ such that :
$$
a\cdot(1,-1,0)^T + b\cdot (1,0,-1)^T = (y_1,y_2,y_3)^T
$$
Developing we get :
$$
(1)\ \ \ a+b = y_1 \\
(2)\ \ \ -a = y_2 \\
(3)\ \ \ -b = y_3
$$
Doing $(1) + (2) + (3)$ tells us that $0 = y_1 + y_2 + y_3$ which is always true.
A: There is one equation and $3$ variables. So clearly there are 2 Free variables. Let $y$ & $z$ be the free variables. Now for first vector, put $y=0$ and $z=1$, you will get $v_1=[-1,0,1]^T$ and after that put $y=1$ and $z=0$, you will get $v_2=[-1,1,0]^T$. So Subspace v can be written as $-:$
$$v=spam([-1,0,1]^T,[-1,1,0]^T)$$ Note that magnitude of these vectors are highly irrelavant. So you can multiply $v_1$ & $v_2$ with any non-zero number.
Hope this will be helpful !
A: Note that 
$$
\begin{array}{rcl}
\mathrm{X}=\{(x_1,x_2,x_3):x_1+x_2+x_3=0\} 
&=& \{(x_1,x_2,x_3):x_1=-x_2-x_3\} \\
&=& \{(-x_2-x_3,x_2,x_3):x_2,x_3\in\mathbb{R}\}\\
&=& \{x_2(-1,1,0)+x_3(-1,0,1):x_2,x_3\in\mathbb{R}\}\\
&=&\mathrm{span}\{(-1,1,0),(-1,0,1)\}\\
&=&\mathrm{span}\{-(-1,1,0),-(-1,0,1)\}\\
&=&\mathrm{span}\{(1,-1,0),(1,0,-1)\},
\end{array}
$$
Aa requared...
A: The vectors $\begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix}$ and  $\begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix}$ are in the subspace of $\mathbb{R}^3$ where $x_1+x_2+x_3=0$ for vectors of the form $\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}$. 
Note that this subspace has dimension $2$ (since $x_3=-x_2-x_1$, it has only two vectors in it's basis). This could also be inferred by noting that in the standard basis for $\mathbb{R}^3$, $f(x)=x_1+x_2+x_3$ is a linear functional and $x_1+x_2+x_3=0$ is the null space annihilated by this functional and it is a hyperspace, that is of dimension $3-1=2$.
Since the vectors $\begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix}$ and  $\begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix}$ are also linearly independent (their linear combination is zero only in the trivial case), they necessarily form a basis and hence span the space.
