Showing the difference between a union being a subspace or not I'm having a hard time conceptualizing subspaces and how to determine them. I would appreciate any insight on this problem:
If $U$ and $W$ are subspaces of $V$, show that $U \cup W$ need not be a subspace. However, if $U \cup W$ is a subspace, show that either $U \subseteq W$ or $W\subseteq U$
 A: One key difference is that a subspace is closed under vector addition, but a union of subspaces might not be. For example, if $V=\mathbb R^2$ with $U = \operatorname{span}\{(1,0)^T\}$ and $W = \operatorname{span}\{(0,1)^T\}$, $(1,1)^T = (1,0)^T+(0,1)^T \notin U\cup W$, so it isn’t a subspace of $V$.
A: Hint: For the first one, consider $V = \mathbb{R}^{2}$ and the subspaces $$U = \text{span}\left(\begin{bmatrix} 1 \\ 0 \\\end{bmatrix}\right), \quad W = \text{span}\left(\begin{bmatrix} 0 \\ 1 \\\end{bmatrix}\right)$$ now it is enough to find vectors $u \in V$, $w \in W$ such that $u + w \notin U\cup W$.
For the second point, go from the assumption that $U \cup W$ is a vector space and assume that neither $U\subseteq W$ nor $W \subseteq U$, that is that there is a $u \in U$ such that $u \notin W$ and a $w \in W$ such that $w \notin U$. What can be said about $u + w$?
A: Let be $U$ and $W$ subspaces, $U\cap W=\{0\}$, of $V$ vector space, dim$V \geq 2,$ and suppose that $U\cup W$ is a vector subspace. 
If $u\in U$ and $w\in W$, both non-null, then 
$$[u,w]:= \{(1-t)u+tw \ \ : \ \ t\in[0,1]\}\subset U\cup W,$$
because every vector space is convex (exercise). So for $t=1/2$, we have $\frac{u+w}{2}\in U\cup W$. But 
$$\frac{u+w}{2}\in U\cup W \iff \frac{u+w}{2}\in U \text{ or } \frac{u+w}{2}\in W.$$
If $\frac{u+w}{2}\in U$, then 
$$w=2\left(\frac{u+w}{2}-\frac{u}{2}\right)\in U.$$
This is a contradiction, because $U\cap W=\{0\}$ and $w\neq 0$. The same contradiction we find when we suppose that $\frac{u+w}{2}\in W$. So not necessarily $U\cup W$ is a subspace of $V$.
