Finding the basis of an intersection of subspaces 
We have subspaces in $\mathbb R^4: $ 
$w_1= \operatorname{sp} \left\{ 
\begin{pmatrix} 1\\  1 \\ 0 \\1 \end{pmatrix} ,
\begin{pmatrix} 1\\  0 \\ 2 \\0 \end{pmatrix},
\begin{pmatrix} 0\\  2 \\ 1 \\1 \end{pmatrix} \right\}$, 
  $w_2= \operatorname{sp} \left\{
\begin{pmatrix} 1\\  1 \\ 1 \\1 \end{pmatrix} ,
\begin{pmatrix} 3\\  2 \\ 3 \\2 \end{pmatrix},
\begin{pmatrix} 2\\  -1 \\ 2 \\0 \end{pmatrix} \right\}$
Find the basis of $w_1+w_2$ and the basis of $w_1\cap w_2$.

So in order to find the basis for $w_1+w_2$, I need to make a $4\times 6$ matrix of all the six vectors, bring it to RREF and see which vector is LD and the basis would be the LI vectors. 
But the intersection of these 2 spans seems empty, or are they the LD vectors that I should've found before ?
In general, how is the intersection of subspaces defined ?
 A: If you consider $\text{rref}(B_1|B_2)$ then you can easily ascertain which vectors in $B_1$ fall in $w_2=\text{span}(B_2)$. Likewise, $\text{rref}(B_2|B_1)$ will tell you which vectors in $B_2$ fall in $w_1=\text{span}(B_1)$. Once you know both of these you ought to be able to put the answer together. 
(the thought above is my initial answer which was not based on calculating the particulars of your problem, this is the general strategy as I see it. In fact, this problem is a bit simpler than what we face in general)
In particular, let $B_1 = \{ u_1,u_2,u_3 \}$ and $B_2 = \{v_1,v_2,v_3 \}$ then we can calculate:
$$ \text{rref}[u_1,u_2,u_3|v_1,v_2,v_3] = \left[ 
\begin{array}{ccc|ccc} 
1 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 & -1 \\
0 & 0 & 0 & 1 & 1 & 1 
\end{array}\right]$$
Therefore, by the column correspondence theorem, none of the vectors in $B_2$ can be written as linear combinations of the vectors in $B_1$. It follows that $w_1 \cap w_2 = \{ 0 \}$. Of course, given this, we could stop here. But, I think it may be instructive to consider the other half of my initial suggestion.
$$ \text{rref}[v_1,v_2,v_3|u_1,u_2,u_3] = \left[ 
\begin{array}{ccc|ccc} 
1 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 & -1 \\
0 & 0 & 0 & 1 & -1 & -1 
\end{array}\right]$$
This calculation shows no vector in $B_1$ is in the span of $B_2$. Again, if this happened to be the first thing we calculated then the conclusion to draw is that $w_1 \cap w_2 = \{ 0 \}$
Oh well, the column correspondence theorem is not too exciting here. That said, for the sake of learning something from all this, notice we can read that:
$$ u_1 = -v_1+v_2-u_2 $$
Why? Simply this, whatever linear combination we see between the columns of the reduced row echelon form of a matrix must also be present in the initial unreduced matrix. I invite the reader to verify the linear combination above is easy to see in the second rref calculation. Just to be explicit,
$$ u_1 = \left[ \begin{array}{c} 1 \\1 \\ 0 \\ 1 \end{array}\right] = 
-\left[ \begin{array}{c} 1 \\1 \\ 1 \\ 1 \end{array}\right]
+\left[ \begin{array}{c} 3 \\2 \\ 3 \\ 2 \end{array}\right]
-\left[ \begin{array}{c} 1 \\0 \\ 2 \\ 0 \end{array}\right]= -v_1+v_2-u_2.$$
A: Hint: the intersection of these two spans is NOT empty. What you need to do is find a new spanning set for $w_2$ that contains some of the vectors from the spanning set for $w_1$. The common vectors will span the intersection. Now that you have a basis for $w_1\cap w_2$, you can extend it to a basis of $w_1+w_2$ by adding on the vectors that are in $w_1$ or $w_2$ but not in $w_1\cap w_2$.
