Basis of a subspace of a vectorspace

[JamesTheBond]

Is the basis for the subspace W of a vectorspace V spanned by the basis of the vectorspace V? If so how?

 

[radou]

Is the basis for the subspace W of a vectorspace V spanned by the basis of the vectorspace V? If so how?

What exactly do you mean? A basis for V is a span for W.

 

[matt grime]

Arguably it is not, since elements in a basis of V are not in general going to be elements of W. W would certainly be in the span of a basis of V, but that is I would suggest subtley different from them being a spanning set for W.

Anyway, the original question is not very clear. There is no such thing as *the* basis of a vector space. And I don't know what it means for one basis to span a basis of subset.

If you mean: given W<V, and a basis set B for V, does B have a subset that is a basis for W, then the answer is no.

 

[JamesTheBond]

I am sorry, I misworded my question. At any rate, what I wanted to know was a confirmation of basis extension. For W < V, B_W (some basis of W) is linearly independent and spans W, therefore, there exists some B_V (basis of V) s.t. B_W < B_V.

I believe this is the basis extension theorem... not sure.

 

[radou]

For W < V, B_W (some basis of W) is linearly independent and spans W, therefore, there exists some B_V (basis of V) s.t. B_W < B_V.

I believe this is the basis extension theorem... not sure.

I'm not sure what B_W < B_V is supposed to mean, but if you meant that every linearly independent set in a vector space V (and hence a basis of a subspace W, too, since it's a linearly independent set) can be 'extended to the basis' of V, then the answer is yes.

 

[JamesTheBond]

Sorry, I meant to say: for some B_W \subset B_V for some basis of V.

 

[HallsofIvy]

And the answer, as radou said, is "yes".

Let {v_i} be a basis for W. If it is also a basis for V we are done. If not, since it is clearly independent, it must not span V- there exist some v which is not a linear combination of {v<sub>i[/sup]}. Append v to that set of vectors and show that it is still a set of independent vectors. If it spans V, we are done. If not, repeat.

That works for a finite dimensional vector space. If you are working with an infinite dimensional vector space, you will have to use "transfinite" induction- use Zorn's lemma to prove that any collection of independent vectors in V (i.e. a basis for W) is contained in a basis for V.</sub>

 

[mathwonk]

here is an example to think about. (1,0)n and (0,1) give a basis of R^2. now consider the subspace where y=x, i.e. the line at 45degrees through the origin. how woulkd you get a basis of that subspace from the given basis of R^2?

 

[HallsofIvy]

here is an example to think about. (1,0)n and (0,1) give a basis of R^2. now consider the subspace where y=x, i.e. the line at 45degrees through the origin. how woulkd you get a basis of that subspace from the given basis of R^2?

Interesting but simple question. Since that is a one dimensional subspace and (1,0)+ (0,1)= (1,1) is in the space, {(1,0)+ (0,1)} is a basis for the subspace.

However, in case this confuses any one, the original question asked here was "if you had a basis for that subspace, how would you get a basis for R² that contained the original basis?" Any basis for the subspace must be of the form {(a,a)} where a is a non-zero number. It is easy to show that (a, -a) is orthogonal to that and so not a multiple. The set {(a, a), (a, -a)} is a basis for R² containing the original basis for the subspace. Actually, any vector in R² that is NOT a multiple of (a,a) would work as a second vector in the basis.