Is $u,v,w$ a basis of $\text{span}(u,v,w)$, and why? Let $u = (1, 2, 0, 1)$, $v = (1, −1, 1, 0)$, $w = (2, 4, 0, 2)$, and suppose $W = \text{span}(u, v, w)$ is a subspace of $\mathbb{R}^4$.
Is $u,v,w$ a basis of $W$? Why or why not?
I'm not sure if it is or not, I reduced it down to echelon form.
$$ \left[
    \begin{array}{cccc}
      1&2&0&1\\
      0&1&-1/3&1/3\\
      0&0&0&0
    \end{array}
\right] $$
But I'm not sure where to go from here.
 A: Recall that a set $\mathcal{B}$ is a basis for $W$ if $Span\mathcal{B}=W$ and $\mathcal{B}$ is linearly independent. In your problem, $\{u,v,w\}$ spans $W$ by definition of $W$. So you only have to check the linear (in)dependence of $\{u,v,w\}$. How can you relate linear independence of rows of a matrix to the number of rows that have all zero entries in the matrix's reduced echelon form?
A: Hint:
The three vectors are a basis iff they are linearly independent, but
 a simple inspection shows that $w=2u$ (as confirmed by your echelon form reduction), so ......
A: If $u, v, w $  are linearly independent and span $ W $, then they form a basis.
But you can see that they are not linearly independent. As your echelon form suggests,  $w$ could be turned into a zero vector by the linear combination of $u$ and $v$. 
For the definition of linear independence, check this wiki article.
More precisely,
$w$ = 2$u$
This means that $w$ is redundant and you only need two vectors i.e. $u$ and $v$     to describe every vector in the subspace $W$.
