Is it always possible to find a vector, result from the sum of 2 elements of 2 different subspaces, which is not in the union of them? $S_{1}$ and $S_{2}$ are subspaces of $V$, which don't contain each other. Is it always possible to find a vector $v=s_{1}+s_{2}$ with $s_{1} \in S_{1}$ and $s_{2} \in S_{2}$ that is not in $(S_{1} \cup S_{2})$? For example if $S_{1}$, $S_{2}$ are subspaces of $\mathbb{R}^{2}$ i.e. two different lines contaning the origin, is it always possible to find a vector (or maybe an entire subspace) that is not in the union? 
 A: Indeed, for any vectors $s_1\in S_1-S_2$, $s_2\in S_2-S_1$ we have that $s_1+s_2$ is not contained in the union $S_1\cup S_2$. To see this we use some group theory. 
Note that $S_1$ and $S_2$ are subgroups of the additive group of the vector space, hence we can speak of their cosets. Two cosets of a subgroup are either disjoint or equal. $s_1+s_2$ is in the coset $s_1+S_2$, and since this coset contains an element not in $S_2$ (namely $s_1$) it in fact contains no elements of $S_2$. In particular $s_1+s_2\notin S_2$. This argument is completely symmetric, so $s_1+s_2\notin S_1$, and hence $s_1+s_2\notin S_1\cup S_2$.
A: Consider the case when $S_1 \cup S_2 \neq V$, since it would be silly otherwise. Let $B_1$ be a basis set for $S_1$ and similarly for $B_2$ (To be entirely clear, $B_1$ and $B_2$ are restrictions of a total basis for $V$). Pick $b_1,b_2 \in B_1 \cup B_2$; we can assume $B_1 \cap B_2 = \emptyset$. Now suppose that $v=b_1 + b_2 \in S_1 \cup S_2$. From here how would you draw a contradiction to prove that $v$ is not in the union?
Note that this works for all vector spaces, assuming the two subspaces don't contain the other. You can extend the result to say that $\text{Span}(b_1+b_2) \cap (S_1 \cup S_2) = \{0\}$
