# Show that there is a subspace $W^\prime$ of $V$ such that $W \cap W^\prime = U$ and $W + W^\prime = V$.

Let $U$ and $W$ be subspaces of $V$ with $U \subseteq W$. Show that there is a subspace $W^\prime$ of $V$ such that $W \cap W^\prime = U$ and $W + W^\prime = V$.

I'm not quite sure where to begin with this. I've considered using the dimension formula that $\dim(W+W^\prime) = \dim(W) + \dim(W^\prime) - \dim(W \cap W^\prime)$, but am not entirely sure. Any help would be appreciated

Choose a basis $\,A\,$ of $\,U\,$ . Complete it to a base $\,B:=A\cup A'\,$ of $\;W\;$ , and this last one complete it to a basis $\,C:=B\cup B'\;$ of $\;V$ .
Now take a peek at $\,\text{Span}(B')\;\;\text{and to Span of}\ldots$