Find Intersection Of 2 Sub-spaces $U=span((1,1 ,0),(2 ,0 ,1))$ $W=span((1,1,1),(5,3,1))$
what is $U \cap W$?
can I find the rref of them both and to find the intersections? that mean
$U= \begin{pmatrix} 
1 & 0 & 0.5 \\
0 & 1 & -0.5 
\end{pmatrix}$
$W= \begin{pmatrix} 
1 & 0 & -1 \\
0 & 1 & 2 
\end{pmatrix}$
so is $U \cap W$=$Span \begin{pmatrix} 
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix}$
 A: Well, $U \cap W$ is the space consisting of all vectors that can be formed both as a linear combination of the elements of U and V. So, for any $\vec v \in U \cap W$: $$ \vec v = c_1 (1, 1, 0) + c_2 (2, 0, 1) $$ and
$$ \vec v = d_1(1, 1, 1) + d_2(5, 3, 1)$$
where $c_1, c_2, d_1, d_2 \in \mathbb{R}$. Does that help?
A: From the rrefs you calculated, you know that $\dim U=2$ and $\dim V=2$. (And this can be also seen without calculating rref: We have only two vectors generating a subspace, and neither of these vectors is a multiple of the other one, so they are linearly independent.) 
If you make rref from the matrix containing all four vectors, you will find out, that $\dim(U+V)=3$.
$$\begin{pmatrix}
1 & 0 & \frac12\\
0 & 1 & -\frac12\\
1 & 0 & -1\\
0 & 1 & 2
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}
$$
So from the formula $\dim(U)+\dim(V)=\dim(U+V)-\dim(U\cap V)$ we can calculate 
$$\dim(U\cap V)=1.$$
So the intersection is generated by single vector.
If we notice that
$$(5,3,1)=5(1,0,\frac12)+3(0,1,-\frac12) = 3(1,1,0)+(2,0,1)$$
then we see that the vector $(5,3,1)$ belongs to both $U$ and $V$, i.e., $(5,3,1)\in U\cap V$.
So now, without any other calculations, we can say that
$$U\cap V = \operatorname{span}\{(5,3,1)\}$$

However, in this case we were simply lucky that we only needed one vector and we were able to guess a vector which belongs to both given subspaces. 
If we want to avoid guesswork, a reasonable way to go is to solve a system of equations as suggested in Toxicz's answer. 
