A question on basis of vectorspaces and subspaces Let $V$ be a finite dimensional vector space and $W$ be any subspace . It is known that if $A$ is any basis of $W$ then by "extension-theorem" , there is a basis $A'$ of $V$ such that $A \subseteq A'$. Is the reverse true ? that is if $B$ is any basis of $V$ , does there exist a basis $B'$ of $W$ such that $B'\subseteq B$ ?
 A: No. Take $V=\Bbb R^2$, $W=\operatorname{Span}\left\{e_1+e_2\right\}$, and $B=\{e_1,e_2\}$. Then $B$ is a basis for $\Bbb R^2$ and any basis $B^\prime$ for $W$ is of the form $B^\prime=\{\lambda\cdot(e_1+e_2)\}$ where $\lambda\neq0$. But now $B^\prime\not\subset B$.
A: This is not necessarily true. Take the standard basis $e_1$,$e_2$ for $\mathbb{R}^2$
and let $W$ be the subspace spanned by $e_1 + e_2$ for example.
A: No. Suppose that $V$ is a $k$-dimensional vector space over some infinite field, such as $\mathbb{R}$ or $\mathbb{C}$. Then since every basis contains exactly $k$ members, it follows that any given basis $B$ has exactly $2^k$ subsets, and thus bases for $2^k$ different subspaces of $V$.  
But since $V$ has infinitely many subspaces, it follows that $V$ has infinitely many subspaces which do not have subsets of $B$ as bases! 
The other answers, by Brian Fitzpatrick and jswiegel, give the specific example of $k=2$, $V=\mathbb{R}^2$ and $B=\{\mathbf{e}_{1},\mathbf{e}_{2}\}$.  This is probably the simplest example (it was also the example I used in the original version of this answer), but the result is far more general. 
