Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Do you mean $\mathbb R^n$ rather than $\mathbb R^3$? –  Geoff Pointer Feb 1 at 4:51
    
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 1 at 5:55
    
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 1 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 2 at 0:33
add comment

3 Answers

up vote 4 down vote accepted

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.

share|improve this answer
    
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 1 at 5:59
1  
@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 1 at 14:08
add comment

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.

share|improve this answer
    
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 1 at 1:51
add comment

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 $V$.

share|improve this answer
    
Are you claiming your proof is correct or are you asking if it's correct? –  Geoff Pointer Feb 1 at 4:55
1  
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 1 at 6:01
    
@GeoffPointer, it was a little bit of both. Your answer states it much more clearly and concisely, though. –  Bill DeRose Feb 1 at 20:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.