How to determine if set of vectors is a basis for W Consider the subspace $$
W =\left\{ \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3}\end{bmatrix} \in \mathbb{R}^3 \,| x_{1}+x_{2}+x_{3} = 0 \right\} 
$$
Is the set S a basis for W? $$
S= \left\{ \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix} , \begin{bmatrix} -3 \\ 2 \\ 1\end{bmatrix} \right\}
$$
I'm not too sure how to go about this question but this is what I've tried. Let $x_{2} = s$ and $x_{3}=t$. Then I have $$
\begin{bmatrix} -s-t \\ s \\ t\end{bmatrix} =s \begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix} + t \begin{bmatrix} -1 \\ 0 \\ 1\end{bmatrix}
$$
Therefore, the basis for $W$ is $$
A= \begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix} , \begin{bmatrix} -1 \\ 0 \\ 1\end{bmatrix}
$$
Since the vectors in $S$ is not a scalar multiple of the vectors in $A$, $S$ is not a basis for $W$.
Could anyone confirm if I am correct? Thanks


$S$ is a basis for $W$ since: (2) span $W$; (1) is a linearly independent set. 

To prove (1) you just have to solve: $\alpha (-1,-1,2)+\beta(-3,2,1)=0$ for $\alpha$ and $\beta$ to get $\alpha=0=\beta$. 
To prove (2): Let $(x_1,x_2,x_3)\in W$ (i.e. $x_1+x_2+x_3=0$) you have to find $\alpha$ and $\beta$ such that $\alpha (-1,-1,2)+\beta(-3,2,1)=(x_1,x_2,x_3)$ but $\beta=-\frac{x_1+x_2}{5}$ and $\alpha=\frac{x_3-\frac{x_1+x_2}{5}}{2}$ works.  Then we are done!
