It is easy to see that if $B$ is a basis for an arbitrary $n$-dimensional vector space, then for every non zero vector $v$ not in $B$, there exists $u\in B$ such that $(B\setminus \{u\})\cup \{v\}$ is still a basis for $V$.
Is the following generalization also true? :
If $B$ is a basis for an arbitrary $n$-dimensional vector space, then every bunch of $k$ linearly independent vectors given, one can swap them with $k$ vectors in $B$, obtaining a basis again? ($k\leq n$)
It seems to me true, no counter example is known to me. But even $k=2$ does not look easy. Is it a known statement? Is there any “combinatorial proof” (avoiding high linear algebra stuff)?
After choosing an onrdering of your basis $B={b_1,...,b_n}$ you can procede inductively. i.e. consider your set $I_0=I={v_1,...,v_k}$, then you define $I_{i+1}$ by checking if $b_{i+1}\in \langle I_i \rangle$ if so, define $I_{i+1}:=I_i$, otherwise $I_{i+1}=I_{i}\cup\{b_{i+1}\}$. Now you know by construction that all those $I_i$ are linearly independent and that $B\subset \langle I_{n}\rangle$ hence $V \subset \langle I_{n}\rangle$. So we constructed a linearly independent generating set, i.e. a basis! just as you wanted by picking $b_i$. However, be aware that this choices are non canonical (in particular since we chose an order on the basis).