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
 A: 
$S$ 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!
A: The vectors are from $W$ and they are linearly independent, hence they are basis of $W$, as $\dim W=2$.
A: Your basis, $A$ is correct and so is $S$.  As you have shown with $A$, the dimension of $W$ is $2$.  If you were to graph both bases, you will find that they refer to the same plane in $\mathbb{R}^3$.
To check that $S$ is in fact a basis for $W$, notice that $(-1,-1,2)$ interpreted as $(x_1, x_2, x_3)$ satisfies the requirement to be in $W$ since $(-1) + (-1) + (2) = 0$.  Similarly for $(-3, 2, 1)$.  Furthermore, a quick check will show that these two vectors are not multiples of one another and so are linearly independent (if the problem gave multiple vectors, you would need to more thoroughly check for independence).  As mentioned, you saw with $A$ that the dimension is $2$, and so any two vectors in $W$ that are linearly independent will span the whole space.
For example, the vectors $(52, -52 + \pi, -\pi)$ and $(1+e, -e + \sqrt{2}, -\sqrt{2} - 1)$ span $W$ as well.
As you get further along, you will find that simply being a basis is helpful, but could be better.  We often like to choose our basis elements with special properties.  We often want to choose them to be orthogonal (perpendicular) to one another, and we often also want the length of each to be 1.  Such a selection is what we refer to as an Orthonormal Basis.  There is a handy algorithm to convert any basis into an orthonormal one known as the http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process.
