Sum and intersections of vector subspaces $U_1+U_2=(U_1 \cap U_2) \oplus W$ Let $U_1,U_2$ be vector subspaces from $\in \mathbb R^5$. 
$$\begin{align*}U_1 &= [(1,0,1,-2,0),(1,-2,0,0,-2),(0,2,1,2,2)]\\
U_2&=[(0,1,1,1,0),(1,2,1,2,1),(1,0,1,-1,0)]
\end{align*}$$ (where [] = [linear span])
Calculate a basis from $U_1+U_2$ and a vector subspace $W \in \mathbb R^5$ so that $U_1+U_2=(U_1 \cap U_2) \oplus W$. ($\oplus$ is the direct sum).
I have the following so far. I calculated a basis from $U_1 \cap U_2$ in the previous exercise and got the following result: $(1,0,1,-1,0)$. I've also calculated a basis from $U_1+U_2$ and got that the standard basis from $\mathbb R^5$ is a basis.
So I suppose now I should solve the following:
standard basis from $\mathbb R^5$ = $(1,0,1,-1,0)\oplus W$
I thought I should get 4 additional vectors and they should also respect the direct sum criterion, that their intersection $= \{0\}$, however my colleagues have this:
$W = \{(w_1,w_2,0,w_3,w_4) | w_1,w_2,w_3,w_4 \in \mathbb R\}$. Where did I go wrong? 
Many many many thanks in advance!

It seems that you're fine. The $W$ given by your colleagues' has as a basis $\{e_1,e_2, e_4,e_5 \}$ where $e_i$ is the standard $i^{\rm th}$ unit vector in $\Bbb R^5$. 
Moreover, their  $W$ does not contain  $ (1,0,1,-1,0)$ (any vector in $W$ has 0 in its third coordinate); thus, this vector together with
$e_1$, $e_2$, $e_4$, $e_5$ will give a basis for $\Bbb R^5$. So, then, $\mathbb R^5$ = $(1,0,1,-1,0)\oplus W$.
Your approach would be more or less the same. I imagine your colleagues' interpreted the question as "exhibit the subspace $W$", rather than "find a basis of the subspace $W$".
