Suppose we have 1-dimensional subspaces $U_1, U_2$ of $V = \mathbb{R^{3}}$ spanned by $U_1 = (1,2,0), U_2 = (1,1,1)$. Is the sum $U_1$ + $U_2$ direct?
I first stated that $U_1 \cap U_2$ = (0,0,0) if it is a direct sum, but I am not sure with what to say about the sum $U_1$ + $U_2$ = (2,3,1) and how to tell if its unique or not.
Any help would be appreciated.
It is correct, the sum is direct because the intersection is the trivial space. In fact You must impose
$\lambda \cdot (1,2,0)=\mu\cdot (1,1,1)$
and it is clear this holds only if $\mu=\lambda=0$
At this point, if you want to find generators for the sum space, will be
$U_1+U_2=\langle \begin{pmatrix}1 \\ 2 \\ 0 \\ \end{pmatrix} , \begin{pmatrix}1\\ 1\\ 1\\ \end{pmatrix} \rangle$
In your case it is simple to see what it’s happening:
$\langle v\rangle$ is the line generated by the vector $v$ and so it is clear that your two no-proportional vectors $v, w$ generate two lines for which the intersection is only $0$ and such that their sum is the plane generated by the two vectors.