Q: Find $U + W$ and $U \cap W$ 
Find $U + W$ and $U \cap W$ for $U =\{\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} \in \mathbb{R}^5 | \left\{\begin{array}{l}x_2=2x_1-x_3\\x_4=3x_5\end{array}\right.  \}$, $W=\{\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} \in \mathbb{R}^5 | x_3+x_4=0 \}.$

So I found out that:
$U = Sp\{ \begin{pmatrix}1\\2\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\-1\\1\\0\\0\end{pmatrix},\begin{pmatrix}0\\0\\0\\3\\1\end{pmatrix} \}$
and
$W=Sp\{ \begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\\-1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\0\\0\\1\end{pmatrix} \}$
Now... how do I calculate $U + W$ and $U \cap W$?

A: First recall the formula
 $dim(U+W)+dim(U\cap W)= dim W + dim U$.
From the definitions and your computations it it clear that $dim U=3$ and $dim W=4$. Thus if we find a vector in $U$ but not in $W$ it is clear that $U+W= \mathbb{R}^5$: $(0,-1,1,0,0)$ which you presented does the trick. Now the formula gives us $dim (U \cap W)=3+4-5=2$: if we can find two linearly independent vectors in which are in $U$ and $W$ we have found basis for the intersection subspace. Notice $(1,2,0,0,0)$ and $(0,1,-1,1, 1/3)$ satisfy the requirements.

A: Observe that all vectors in U are of the form $\begin{pmatrix}x_1 \\ 2x_1- x_3 \\ x_3 \\ 3x_5 \\ x_5\end{pmatrix}= x_1 \begin{pmatrix}1 \\ 2 \\ 0 \\ 0 \\ 0 \end{pmatrix}+ x_3\begin{pmatrix}0 \\ -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}+ x_5\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix}$ so U is a three dimensional subspace spanned by $\begin{pmatrix}1 \\ 2 \\ 0 \\ 0 \\ 0 \end{pmatrix}$, $\begin{pmatrix}0 \\ -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}$, and $\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix}$.  
And vectors in W are of the form $\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ -x_3 \\ x_5\end{pmatrix}=$$ x_1\begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}+$$ x_2\begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}+$$ x_3\begin{pmatrix}0 \\ 0 \\ 1 \\ -1 \\ 0\end{pmatrix}+$$ x_5\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix}$. So W is a four dimensional subspace spanned by those four vectors.  
The question, really, is 'how may of the vectors spanning U are in W'.  If all of them are U+ W= W.  If there 1 that is not (and there cannot be more than 1) then U+ W= $R^5$.  $U\cap W$ is the subspace spanned by those basis vectors of U that are in W.
