Internal direct sum of vector spaces Let $U$ be the subspace of $\mathbb{R}^3$ spanned by $\{(1,1,0), (0,1,1)\}$. Find a subspace $W$ of $\Bbb R^3$ such that $\mathbb{R}^3 = U \oplus W$.
As I am having an examination tomorrow, it would be really helpful if one could explain the methodology for doing this problem. I am mostly interested in the methodology, rather than the result. 
Thank you very much in advance.

In general, if you have a subspace $U \subseteq V$ and you want a complement $W$ so that $U \oplus W = V$.  Then first find a basis $\{u_1, \ldots, u_n\}$ of $U$.  Then extend this to a basis $\{u_1, \ldots, u_n, w_1, \ldots, w_m\}$ of $V$.  The complement is the subspace spanned by those additional vectors,
$$W = \mathrm{span}\{w_1, \ldots, w_m\}.$$
So for the specific problem above you need to extend $\{(1, 1, 0), (0, 1, 1)\}$ to a basis of $\mathbb R^3$, i.e., you just need to find a single vector not contained in $U$.  Then $W$ will be the line spanned by that vector.
