Basis of sum subspace Let $W$ be a vector space and $U, V$ two subspaces with bases $ {u_1,...,u_n}$ and ${v_1, ..., v_s}$ respectively. How can I show rigorously that the basis of $U+V$ is ${u_1, ..., u_n, v_1,...,v_s}$?

My work:
First I have to show that $U+V \subset <{u_1, ..., u_n, v_1,...,v_s}>$. That is,  $h=u+v \implies h = (au_1+ ... + au_n)+ (bv_1 + ... +bv_s) = au_1+ ... + au_n+ bv_1 + ... +bv_s$. So $h \in U+V \implies h \in <{u_1, ..., u_n, v_1,...,v_s}>$.
Second I have to show that $<{u_1, ..., u_n, v_1,...,v_s}> \subset U+V$. That is, $h \in <{u_1, ..., u_n, v_1,...,v_s}> \implies h = au_1+ ... + au_n+ bv_1 + ... +bv_s = (au_1+ ... + au_n)+ (bv_1 + ... +bv_s) = u + v$, for some $ u \in U$ and $v \in V$.
Is this good enough or should I prove that <{u_1, ..., u_n, v_1,...,v_s}> are linearly independent first? If so, how should I do it?
 A: This is true if and only if $V + W = V\oplus W$.  Otherwise the union of the bases is a spanning set; it must be "weeded" to get an actual basis.
A: It is not true that $u_1,\ldots ,u_n,v_1,\ldots ,v_s$ in general will be a basis.  It will span $U+V$, but it won't necessarily be a linearly independent set.  For example, in $\mathbb{R}^2$, take $U$ to be the space spanned by $u_1:=(1,0)$ and $V$ to be the space spanned by $v_1:=(0,1),v_2:=(1,1)$ (so that $V=\mathbb{R}^2$).  Then, $U+V=\mathbb{R}^2$, but $(1,0),(0,1),(1,1)$ is not a basis (the sum of the first and second is the third, and hence they are not linearly independent).
We can, however, show that it spans.  As $\{ u_1,\ldots ,u_n,v_1,\ldots ,v_s\} \subseteq U+V$, we immediately have that $\mathrm{Span}\, \{ u_1,\ldots ,u_n,v_1,\ldots ,v_s\} \subseteq \mathrm{Span}\, (U+V)=U+V$.
For the other direction, let $w=u+v\in U+V$ be arbitrary with $u\in U$ and $v\in V$.  You can write $u$ as a linear combination of $u_1,\ldots ,u_n$ and you can write $v$ as a linear combination of $v_1,\ldots ,v_s$, so that $w$ is a linear combination of $u_1,\ldots ,u_n,v_1,\ldots ,v_s$, i.e. $w\in \mathrm{Span}\, \{ u_1,\ldots ,u_n,v_1,\ldots ,v_s\}$.
