Is $U+V$ equivalent to $U \cup V$? $U$ und $V$ are subspaces of a $R^n$
Is $U+V$ equivalent to $U \cup V$?
Additional question:
If I compare basis' instead of subspaces it would be correct? There is a basis(U+V), that is equivalent to a basis($U \cup V$) 
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 A: Note that in general $U \cup V$ isn't a subspace: If there are $u \in U \setminus V$ and $v \in V \setminus U$ then $u+v \not\in U \cup V$. Hence $U \cup V$ is a subspace iff $U \subseteq V$ or $ V \subseteq U$.
$U+V$ is by definition the smallest subspace containing $U$ and $V$, that is $U+V = {\rm span}\, (U \cup V)$. 
A: No. $U+V$ is usually strictly bigger than the union (which is usually not even a subspace). Take for example $U=\mathbb{R}\oplus \{0\}$, $V=\{0\}\oplus \mathbb{R}$. Then $U+V=\mathbb{R^2}$, but the union does not contain $(1,1)$.
We can prove more: over infinite field (such as $\mathbb{R}$), no finite union of proper subspaces can be a full space. Thus, for any finite collection of subspaces such that none contains all the others, the union won't be a subspace.
On the other hand, it's possible for spaces over finite fields. Take $\mathbb{F}_2^2$ and $U=\langle(0,1)\rangle$, $V=\langle(1,0)\rangle$, $W=\langle(1,1)\rangle$. Then $U\cup V\cup W=U+V+W=\mathbb{F}_2^2$.
A: It need not be.
Example: $U = \{(x,0): x \in \mathbb{R}\}$, and $V = \{(0,y): y \in \mathbb{R}\}$, then $U + V = \mathbb{R}^2$, but $U\cup V \neq \mathbb{R}^2$
