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If a set S of vectors spans a vector space V, then is it possible to remove one or more vectors from S to create a basis for V ?

I think that this is a tricky question as I am not sure whether S contains a dummy vector that is linearly dependent on the one(s) you left out. In which case it will possible to remove those vectors.

But If for example I have a set {x,y,z} spanning a vector space. Taking y away wont help my set span the same vector space. Right ?

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    $\begingroup$ Assuming you are speaking about finite dimensional vector spaces, then you can work inductively. If there is a linear dependence between the vectors, you can use it to eliminate any one of the vectors that appears with a non-zero coefficient. $\endgroup$
    – lulu
    Feb 3, 2020 at 21:18
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    $\begingroup$ Should say: I don't understand your $\{x,y,z\}$ example. If those are linearly independent, then they already form a basis of the vector space they span. If they are linearly dependent, then you can remove at least one without changing the span. $\endgroup$
    – lulu
    Feb 3, 2020 at 21:20
  • $\begingroup$ I am a bit confused as I have a question asking me whether "If a set S of vectors spans a vector space V, then it is possible to remove one or more vectors from S to create a basis for V." is true or false. To which I replied true as spanning can be done with additional vectors that aren't really the basis of the vector space. Right ? $\endgroup$ Feb 3, 2020 at 21:24
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    $\begingroup$ I agree with your conclusion (again, assuming, for simplicity, that we are speaking of finite dimensional vector spaces) but I do not understand your argument. In my first argument, I suggested that you prove it via induction. $\endgroup$
    – lulu
    Feb 3, 2020 at 21:26
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    $\begingroup$ That's not an argument. For the first part, you just give a single example that works out the way you want. And the second part has nothing to do with what you were asked to prove. Seriously, do it by induction. $\endgroup$
    – lulu
    Feb 3, 2020 at 21:37

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I'll assume that you are working with finite dimensional vector spaces. Let $V$ be the vector space in question and let $d=\dim V$.

The claim we want to prove: any finite collection of vectors in $V$ which spans $V$ contains a basis.

Proof by induction on the number of vectors in the collection.

Base case: if the number of vectors is $d$ then the collection is a basis already (by standard results).

Now suppose we have proven the result for all collections with $n$ vectors (for $n≥d$). We want to prove that the result also holds for collections with $n+1$ vectors, so take such a collection, $S=\{v_1, \cdots, v_{n+1}\}$. Since $n+1>d$ there must be a linear dependence between these vectors. Let's say we have $$\sum_{i=1}^{n+1} \lambda_iv_i=0$$ and that not all the $\lambda_i$ are $0$.

Let $j$ be the greatest index such that $\lambda_j\neq 0$. Then we can write $$v_j=\sum_{i\neq j}-\frac {\lambda_i}{\lambda_j}v_j$$

It follows that we can eliminate $v_j$ from the collection without changing the span. But the set $S'=S-\{v_j\}$ has only $n$ elements so the inductive hypothesis applies to it, and we conclude that $S'$ contains a basis for $V$. But since $S'\subset S$ that basis is also a subset of $S$, and we are done.

Note: if your spanning set $S$ is infinite then it also must contain a basis. To see that, take some basis $\{e_1, \cdots, e_d\}$ for $V$. Since $S$ spans we knwo we can write each $e_i$ as a $\textit {finite}$ linear combination of vectors in $S$. Choose one such expression for each $e_i$. Define $S^*\subset S$ to be the (finite) subset of $S$ consisting of all the vectors from $S$ that are used in those expressions and apply the above to $S^*$.

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  • $\begingroup$ Yes, it is assumed to be in a finite dimensional vector spaces as the questions does not mention anything about the infinite dimensional. "If a set S of vectors is linearly independent in a vector space V, then it is possible to add zero or more vectors to S to create a basis for V". I wrote that as a mean to validate my understanding of the concept of independence, basis and spans. As linearly independent vectors in a finite dimensional sets will span the whole vector space, adding another vector to the independent set will not help create a basis I assume. $\endgroup$ Feb 3, 2020 at 21:58
  • $\begingroup$ I completely understood this answer. Thank you for your time ! $\endgroup$ Feb 3, 2020 at 21:58
  • $\begingroup$ Glad to have helped. Note that your second question, about extending the independent collection to the basis, is different than the one I answered. $\endgroup$
    – lulu
    Feb 3, 2020 at 22:01
  • $\begingroup$ Yes, I am still wondering if that would work. I will continue to read about it. In the case it is not clear I will post a question $\endgroup$ Feb 3, 2020 at 22:26
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    $\begingroup$ Try an inductive approach, as in my argument above. This time you have to apply induction to the "co-dimension". That is, if the dimension is $d$ we assume your collection has $d-i$ elements and apply induction to $i$. $\endgroup$
    – lulu
    Feb 3, 2020 at 22:28

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