Subspace and basis problem I am not sure if the problem is correctly stated.But if it is, I need the proof.
 If A is subspace of B and $A\neq B$, proof that there exists basis of B such that it does not have any of the basis vectors of A.
 A: This may mean that, given a basis of $A$, there exists a basis for $B$ with none of the vectors of the given $A$-basis as elements. That was my original interpretation, and to prove that, try doubling all the basis vectors in your basis of $A$, then extending to a basis for $B$.
It may instead mean one of the following:

(i) There is a basis for $B$ without any vectors from any $A$-basis.
(ii) There is a basis for $B$ without any $A$-basis as a subset.

In either of these cases, we will need for $A$ to be a proper subspace of $B$. A basis that works for (i) will also work for (ii), and will also work for my original interpretation in the case that $A$ is a proper subspace of $B$. Start with a basis $C$ for $A,$ and take any vector $x\in B$ that isn't in $A$. Let $D=\{x\}\cup\{v+x:v\in C\}$. Show that this is linearly independent, contains no vectors of $A$, and that $A$ is contained in the span of $D$. Then extend $D$ to a basis of $B$ if necessary. This basis will will then have no vectors of $A$ in it, at all.
I leave it to you to verify the details (that is, make it an actual proof, rather than just a proof idea).
A: The claim is wrong for general vector spaces. Look at the $\mathbb{F}_2$ vector space $B = \mathbb{F}_2$ and also take it as the subspace $A$. There is only a single basis of $A$ and $B$, namely $\{1\}$.
