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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?

Bill DeRose
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  • Do you mean $\mathbb R^n$ rather than $\mathbb R^3$? – Geoff Pointer Feb 01 '14 at 04:51
  • The answer to your question as stated is obviously "no", because you do not exclude the case that the given vector happens to be $v=0$, and the zero vector can never be member of any basis (or linearly independent family). Also your second paragraph makes no sense: there is no such thing as "any $4$ vectors in $B$" since $B$ has only three vectors. Please rewrite the question so that it is clear what you are asking. – Marc van Leeuwen Feb 01 '14 at 05:55
  • If $V$ is a vector space in $\mathbb R^3$ and has a basis with $3$ vectors then $V$ is $\mathbb R^3$. Similarly in your answer below, if $V$ is a vector space in $\mathbb R^n$ and has a basis with $n$ vectors then $V$ is $\mathbb R^n$. You do need to edit your question to clarify matters. – Geoff Pointer Feb 01 '14 at 14:15
  • @GeoffPointer I used $\mathbb{R}^3$ as a concrete example, but it could just as easily be generalize to $\mathbb{R}^n$. – Bill DeRose Feb 02 '14 at 00:33

3 Answers3

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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.

Ittay Weiss
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  • So to be able to swap $w$ for any column in the basis, $w$ must not be in the span of any such sets ${v_j~|~j \ne i}$ for $i = 1, 2, \ldots, n$. – Bill DeRose Feb 01 '14 at 01:51
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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.

Geoff Pointer
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  • This is the most pertinent answer +1. However it would be easier on the reader if you had used the same notation as in the question. – Marc van Leeuwen Feb 01 '14 at 05:59
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    @MarcvanLeeuwen Yes, I could edit my answer, but I wrote this after reading the OP's attempt at a proof and that has different notation to his question. – Geoff Pointer Feb 01 '14 at 14:08
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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 $V$.

Bill DeRose
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  • Are you claiming your proof is correct or are you asking if it's correct? – Geoff Pointer Feb 01 '14 at 04:55
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    Note that this shows that the answer to your question is "no", since you did not require $w\neq0$ (back in the question when $w$ was still called $v$). Also yout formulation is sloppy: the "$i$-th coordinate" must be taken to mean coordinate with respect to the basis $B$. (Since you are working in $\def\R{\Bbb R}\R^3$, or in $\R^n$, the unqualified meaning of "$i$-th coordinate" would be with respec to the standard basis, in other word the $i$-th component of the vector, and this does not work.) Finally $w\neq\rm span...$ should be $w\notin\rm span...$ – Marc van Leeuwen Feb 01 '14 at 06:01
  • @GeoffPointer, it was a little bit of both. Your answer states it much more clearly and concisely, though. – Bill DeRose Feb 01 '14 at 20:40