# Is it possible to remove one or more vectors from S to create a basis for V

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 ?

• 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.
– lulu
Feb 3, 2020 at 21:18
• 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.
– lulu
Feb 3, 2020 at 21:20
• 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 ? Feb 3, 2020 at 21:24
• 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.
– lulu
Feb 3, 2020 at 21:26
• 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.
– lulu
Feb 3, 2020 at 21:37

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 $$S^*$$.

• 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. Feb 3, 2020 at 21:58
• I completely understood this answer. Thank you for your time ! Feb 3, 2020 at 21:58
• Glad to have helped. Note that your second question, about extending the independent collection to the basis, is different than the one I answered.
– lulu
Feb 3, 2020 at 22:01
• 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 Feb 3, 2020 at 22:26
• 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 $i$.
– lulu
Feb 3, 2020 at 22:28