Proof that the intersection of basis' is a basis for the intersection of their respective subspaces I'm doing some linear algebra review and I'm trying to prove two things.
Let $U$ and $W$ be subspaces of $V$.
Firstly, suppose $U+W$ has finite dimensions. Let $B_1$ be a basis for $U$ and Let $B_2$ be a basis for $W$. I need to show that I can choose $B_1$ and $B_2$ such that $B_1 \cap B_2$ is a basis for $U \cap W$.
Secondly, I need to show that the previous statement is not true in general. I think a counter example would suffice but I can't think of any.
 A: Select first a basis ${\cal E}=(e_1,...,e_d)$ for $U\cap W$ then complement ${\cal E}$ with $(u_1,...,u_p)$ to get a basis for $U$ and complement ${\cal E}$ with $(w_1,...,w_q)$ to get a basis for $W$. Then all the vectors are linearly independent since e.g. a non-trivial vector in the span of the $w_k$'s can not be in $U$.
In ${\Bbb R}^3$ with canonical basis let $e_1+e_2,e_2$ be a basis for $U$ and $e_1+e_3,e_3$ be a basis for $W$. Then $U\cap W$ is generated by $e_1$ which is not among the given basis vectors.
A: The key is to observe that $U \cap W$ is a subspace of both $U$ and $W$. Thus we can pick a basis $b_1, \ldots, b_k$ of $U \cap W$ and then complete the basis (we know given any subset of linearly independent vectors of a vector space we can add on vectors until we get a basis), in both $U$ and $W$ respectively to get basis $B_1 = b_1, \ldots, b_k, c_1, \ldots, c_l$ and $B_2 = b_1, \ldots, b_k, d_1, \ldots, d_m$. Note that $B_1 \cap B_2 = \{b_1, \ldots, b_k\}$. One containment is clear. On the other hand if some $c_i$ or $d_i$ were in $B_1 \cap B_2$, this would imply it was in $U \cap W$. However $\{b_1, \ldots, b_k, c_i\}$ is linearly independent since it is a subset of the basis for $U$, contradicting the minimality of our basis $b_1, \ldots, b_k$ for $U \cap W$.
Now consider $V = \mathbb{R}^3$ let $e_i$ be the standard basis, $U = span \{ e_1, e_2 \}, W = span \{ e_1, e_3 \}, U \cap W = span\{ e_1\}$. 
Now pick $B_1 = \{ e_1, e_1 + e_2\}, B_2 = \{e_1 + e_3 , e_3\}$, so $B_1 \cap B_2 = \emptyset$.
