If $A\ne V$ is a subspace of $V$ and $B$ is linearly independent subset of $V$, then $B$ can be completed to a basis of $V$ Let $A \ne V$ be a subspace of $V$ and $B$ a linearly independent subset of $V$. Prove that $B$ can be completed to a basis of $V$ with vectors from $V \setminus A$.
OK, I started with: 
$K=\operatorname{lin}(v_1,v_2,v_3,\dotsc)$
and if any of $v_1,v_2,\dotsc$ belongs to $A$ we can replace this vector by a linear combination of vectors which don't belong to $A$ 
So $v_{x_1},v_{x_2},\dotsc \in A$
$v_{x_1}\in\operatorname{lin}(w_{1,1},w_{1,2},\dotsc)$
$v_{x_2}\in\operatorname{lin}(w_{2,1},w_{2,2},\dotsc)$
etc.
Hence $V=\operatorname{lin}(w_{1,1},w_{1,2},\dots,w_{2,1},\dots,w_{i,j})$
and I don't know what to do next, but what I know is that I should use the Steinitz exchange lemma.
Does anybody have an idea how to finish this prove?
 A: Assume $$Dim(V)=m,$$ $$Dim(A)=n<m,$$ $$\text{and } B=\{v_1,...,v_n\}.$$ 
Then $$Span(B)=A\neq V,$$
so $\exists v_{n+1}\in V\backslash A$, which means that there is no $n$-tuple $(c_1,...,c_n)$ with $v_{n+1}=c_1v_1+...+c_nv_n$ , i.e. $B\cup \{v_{n+1}\}$ is linearly independent. If $n+1=m$, we are done because we have a linearly independent set of $m$ vectors, which must be a basis for $V$. Otherwise, repeat the process.
A: this should do it: Let {$B:=v_1,..,v_n$} and consider a basis $B_V$:={$v_1,..,v_m$}  (i.e., let $DimV=m$ )for $V$; $n>m$ . Construct a  matrix $M_{ij}$ with $n+m$ rows ,where the first $n$ rows are the vectors in $B$ and the remaining $m$ rows are the vectors of $B_V$. Now do Gaussian elimination on $M$ . The result will be a collection of $m$ linearly-independent vectors, the first $n$ of which will be the vectors in $B$.
A: I assume that in the hypotheses you have that $V$ is finite dimensional.
Lemma. The only subspace of $V$ containing $V\setminus A$ is $V$.
Proof. Let $v\in V$. If $v\notin A$ we have nothing to prove. If $v\in A$, then $v=(v-w)+w$ where $w\in V\setminus A$ (one such vector exists by hypothesis). Since $v-w\notin A$, we are done. QED
If $B=\{v_1,\dots,v_m\}$ spans $V$ we have nothing to prove, because we extend it to a basis by taking zero vectors from $V\setminus A$.
If $B$ doesn't span $V$, then the lemma shows that $V\setminus A$ is not contained in the span of $B$. Take $v_{m+1}\in V\setminus A$ which doesn't belong to the span of $B$. Then $B_1=B\cup\{v_{m+1}\}$ is linearly independent.
Now we can reapply the argument starting from $B_1$. Finite dimensionality of $V$ ensures this algorithm terminates.
This can be formalized in the following way. Suppose $k$ is the maximum number of vectors from $V\setminus A$ that can be added to $B$ to still give a linearly independent set (it's possible that $k=0$). Choose one such set of vectors $\{v_{m+1},\dots,v_{m+k}\}$ and consider $B'=\{v_1,\dots,v_{m+k}\}$. If $B'$ doesn't span $V$, then the argument above shows that there exists $v\in V\setminus A$ such that $B'\cup\{v\}$ is linearly independent, contrary to the maximality of $k$.

This holds also for infinite dimensional spaces (but we need Zorn's lemma).
Consider the set $\mathcal{F}$ of subsets $C$ of $V\setminus A$ such that $B\cup C$ is linearly independent and order it by inclusion. Since $\emptyset\in\mathcal{F}$, $\mathcal{F}$ is not empty.
Let $\mathcal{G}$ be a totally ordered subset of $\mathcal{F}$ and let $D=\bigcup\mathcal{G}$. Let's prove that $B\cup D$ is linearly independent.
If it isn't, then there are $w_1,w_2,\dots,w_k\in D$ such that $B\cup\{w_1,\dots,w_k\}$ is linearly dependent. Since $\mathcal{G}$ is totally ordered by inclusion, there is $C\in\mathcal{G}$ such that $w_1,\dots,w_k\in C$. But then $B\cup C$ is linearly dependent: contradiction.
Thus $D$ is an upper bound of $\mathcal{G}$ and we can apply Zorn's lemma to find $E$ maximal in $\mathcal{F}$. If $B\cup E$ doesn't span $V$, we find $v\in V\setminus A$ such that $B\cup E\cup\{v\}$ is linearly independent. In particular $v\notin E$ so $E\cup\{v\}\in\mathcal{F}$ contrary to the maximality of $E$.
