Is it possible to swap vectors into a basis to get a new basis? Let $V$ be a vector space in $\mathbb{R}^3$. Assume we have a basis, $B = (b_1, b_2, b_3)$, that spans $V$. Now choose some $v \in V$ such that $v \ne 0$. Is is always possible to swap $v$ with a column in $B$ to obtain a new basis for $V$?
My initial thought is that it is not always possible since if $v$ is some linear combination of any 2 vectors in $B$ then if we were to swap $w$ into $B$, $B$ would no longer span $V$. Am I missing a nuance here?

I believe the answer to my  question is that yes, we can swap $w$ with some vector in the basis to obtain a new basis.
$Proof$:
Let $V$ be a vector space in $\mathbb{R}^n$ and $B = (b_1, \ldots, b_n)$ be a basis for $V$. Let $w \in V$. As noted in Ittay's response, swapping $w$ with $b_i$ results in a basis iff $w \notin$ span$\{v_j~|~j\ne i\}$.
Thus, to not have a basis after swapping implies that $w$ is contained in the span of all ${n \choose n-1} = n$ subsets of $B$'s basis vectors. However, the $i^{th}$ subset is missing $b_i$ which implies the $i^{th}$ coordinate of $w$ must be $0$ (otherwise $w \notin $ span$\{v_j~|~i\ne j\}$). It follows that $w = 0_n$ and hence if $w \ne 0_n$ then we can swap $w$ with one of the vectors in $B$ and get a new basis for $V$.
