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!
 A: 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 $W$".
A: You could also find your $W$ like this: all you have to do is complete the matrix
$$
\begin{pmatrix}
1   & * & * & * & *  \\
0   & * & * & * & *  \\
1   & * & * & * & *  \\
-1  & * & * & * & *  \\
0   & * & * & * & *  \\
\end{pmatrix}
$$
in such a way that it has rank $5$. You can do this almost as you want, but maybe an easy strategy is like this: sure enough these firs two columns are linearly independent, aren't they?
$$
\begin{pmatrix}
1   & 1 & * & * & *  \\
0   & 0 & * & * & *  \\
1   & 0 & * & * & *  \\
-1  & 0 & * & * & *  \\
0   & 0 & * & * & *  \\
\end{pmatrix}
$$
So, you keep going the same way. These first three columns are linearly independent too:
$$
\begin{pmatrix}
1   & 1 & 0 & * & *  \\
0   & 0 & 1 & * & *  \\
1   & 0 & 0 & * & *  \\
-1  & 0 & 0 & * & *  \\
0   & 0 & 0 & * & *  \\
\end{pmatrix}
$$
So, let's try again with
$$
\begin{pmatrix}
1   & 1 & 0 & 0 & *  \\
0   & 0 & 1 & 0 & *  \\
1   & 0 & 0 & 1 & *  \\
-1  & 0 & 0 & 0 & *  \\
0   & 0 & 0 & 0 & *  \\
\end{pmatrix}
$$
Still four linearly independent columns. Now, the only risk is with the last one:
$$
\begin{pmatrix}
1   & 1 & 0 & 0 & 0  \\
0   & 0 & 1 & 0 & 0  \\
1   & 0 & 0 & 1 & 0  \\
-1  & 0 & 0 & 0 & 1  \\
0   & 0 & 0 & 0 & 0  \\
\end{pmatrix}
$$
If you keep on just putting the next vector of the standard basis of $\mathbb{R}^5$, this matrix has rank four. Never mind: just replace your last column with this one:
$$
\begin{pmatrix}
1   & 1 & 0 & 0 & 0  \\
0   & 0 & 1 & 0 & 0  \\
1   & 0 & 0 & 1 & 0  \\
-1  & 0 & 0 & 0 & 0  \\
0   & 0 & 0 & 0 & 1  \\
\end{pmatrix}
$$
This matrix has rank $5$. Hence, your vector $(1,0,1,-1,0)$ and $e_1, e_2, e_3, e_5$ are linearly independent vectors. Hence the sum $[(1,0,1,-1,0)] + [e_1, e_2, e_3, e_5]$ is a direct sum and is equal to $\mathbb{R}^5$. So, you can take $W = [e_1, e_2, e_3, e_5]$. (The solution of this problem is far away from being unique.)
