Is it possible to swap vectors into a basis to get a new basis?

Question

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?

Answer 1

I am first assuming that we are talking about a general $n$-dimensional vector space $V$ with $n$ any finite positive integer.

If $\{\vec v_1, \ldots, \vec v_n\}$ is a basis for $V$ and $\vec w \in V$ is non zero then $\vec w = c_1\vec v_1 + \cdots + c_n\vec v_n$ where at least one $c_i$ is non zero for some $1 \le i \le n$. Thus we get $\frac{1}{c_i}\vec w = \frac{c_1}{c_i}\vec v_1 + \cdots + \frac{c_i}{c_i}\vec v_i + \cdots + \frac{c_n}{c_i}\vec v_n$ and rearranged we get $\vec v_i = \frac{1}{c_i}\vec w - \frac{c_1}{c_i}\vec v_1 - \cdots - \frac{c_n}{c_i}\vec v_n$.

So, if we remove $v_i$ from the basis and add $w$, any $v \in V$ that previously required $v_i$ as part of its representation can now use the $RHS$ of the last step in its place.

Answer 2

If $(v_1,\cdots, v_n)$ is a basis for any vector space $V$, and $w\in V$ is an arbitrary vector, then swapping $w$ for $v_i$ will result in a basis iff $w$ is not in the span of $\{v_j\mid j\ne i\}$. The proof is a good exercise.

Answer 3

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 $V$.