Find a non-zero vector that spans the intersection of the following two subspaces of R 4 : V = span{(1,1,0,-1)^T, (0,1,3,1)^T} and W = span{(0,1,2,-1)^T,(1,2,2,-2)^T}
Don't quite know where to start. The coefficient matrix of V and W are inconsistent after gaussian elimination. Also inconsistent with a 4x4 of V and W. We haven't covered basis/dimension yet, or even intersection for that matter, so I assume this problem can be worked out using an algorithm to find constants. Any insight? Thanks.
Span of V:
$$ \left[
\begin{array}{cc|c}
  1&0&a\\
  1&1&b\\
0&3&c\\
-1&1&d
\end{array}
\right] $$
Row reduced to:
$$ \left[
\begin{array}{cc|c}
  1&0&a\\
  0&1&b-a\\
0&0&c-3b+3a\\
0&0&a2+d-b
\end{array}
\right] $$
So row 3 and 4 are inconsistent, and these vectors don't span R4.
Similar results for Span of W. Am I approaching this problem correctly?
 A: Let $v \in V \cap W$. We then have
$$v=a_1 v_1 + a_2 v_2 = b_1 w_1 + b_2 w_2$$
where $a_i, b_i \in \mathbb{R}$. Thus
\begin{align*}
0&=a_1 v_1 + a_2 v_2 - b_1 w_1 - b_2 w_2 \\
&= (a_1, a_1, 0, -a_1)+ (0, a_2, 3a_2, a_2) - (0, b_1, 2b_1, -b_1)-(b_2, 2b_2, 2b_2, -2b_2) \\
&= (a_1-b_2, a_1+a_2-b_1-2b_2, 3a_2 -2b_1-2b_2, -a_1+a_2+b_1+2b_2).
\end{align*}
Thus we now must solve
\begin{cases}
a_1-b_2=0 \\
a_1+a_2-b_1-2b_2=0\\
3a_2 -2b_1-2b_2 =0\\
-a_1+a_2+b_1+2b_2=0.
\end{cases}
According to Wolfram , this system has no solution has infinitely many solutions.
A: You’re misinterpreting the result of your row-reduction. The pivots on the left side of the rref tell you that the two vectors are indeed linearly independent. The last two rows aren’t automatically inconsistent since you don’t yet know what $a$, $b$, $c$ and $d$ are. In fact, these two rows of the rref give you conditions on these variables for the system to be consistent, namely, $3a-3b+c=0$ and $2a-b+d=0$. If you take $a$ and $b$ as free variables, this means that every vector in $V$ is of the form $(a,b,3b-3a,b-2a)^T$.  
You might proceed instead as Andrew Tawfeek suggested in his answer: You want elements of $V$ that are also elements of $W$. Every element of $V$ is a linear combination $a_1\mathbf v_1+a_2\mathbf v_2$ of its two generators and every element of $W$ is likewise some linear combination $b_1\mathbf w_1+b_2\mathbf w_2$. Therefore, the vectors in their intersection can be found by solving the equation $a_1\mathbf v_1+a_2\mathbf v_2=b_1\mathbf w_1+b_2\mathbf w_2$.
