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Basis of a Subspace

by Caspian
Tags: basis, linear algebra, subspace
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Caspian
#1
Feb8-10, 01:04 PM
P: 15
My book made the following claim... but I don't understand why it's true:

If [tex]v_1, v_2, v_3, v_4[/tex] is a basis for the vector space [tex]\mathbb{R}^4[/tex], and if [tex]W[/tex] is a subspace, then there exists a [tex]W[/tex] which has a basis which is not some subset of the [tex]v[/tex]'s.

The book provided a proof by counterexample: Let [tex]v_1 = (1, 0, 0, 0) ... v_2 = (0, 0, 0, 1)[/tex]. If [tex]W[/tex] is the line through [tex](1, 2, 3, 4)[/tex], then none of the [tex]v[/tex]'s are in [tex]W[/tex].

Is it just me, or does this not make any sense? First of all, (1,2,3,4) is a linear combination of (1,0,0,0)...(0,0,0,1), isn't it?

I'm very confused...

Any help would be greatly appreciated :).
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Fredrik
#2
Feb8-10, 01:59 PM
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Yes, but that's not what they're saying. They're saying that none of the four [itex]v_i[/itex] vectors are in W. They're also saying that a basis vector of W (which must be a multiple of (1,2,3,4)) can't be equal to one of the [itex]v_i[/itex].

When they talk about the set of "v's" they really mean a set that only has four members, not the subspace they span.
Caspian
#3
Feb8-10, 02:39 PM
P: 15
Ok, ok... I get it. (1,2,3,4) is in the vector space spanned by (1,0,0,0)...(0,0,0,1), but these vectors aren't a basis for the subspace which is the line through (1,2,3,4) because these vectors span too much space.

Ok, I guess this was a dumb question. Thanks for your help :).

Landau
#4
Feb9-10, 03:51 PM
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Basis of a Subspace

I don't think you fully get it yet.
Quote Quote by Caspian View Post
Ok, ok... I get it. (1,2,3,4) is in the vector space spanned by (1,0,0,0)...(0,0,0,1), but these vectors aren't a basis for the subspace which is the line through (1,2,3,4) because these vectors span too much space.
It should be obvious that B={(1,0,0,0),...,(0,0,0,1)} isn't a basis for W, since B spans the whole space R^4! So they can't be a basis for any proper subspace of V (indeed, they span "too much space").

But that's not what your book is asserting. They are only talking about a subset of B={(1,0,0,0),...,(0,0,0,1)}. So they're saying that even some subset of B cannot be a basis of W. Remember that a basis of W first of all consists of elements of W. None of the vectors in B are in W.
HallsofIvy
#5
Feb10-10, 07:10 AM
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The point you need to keep in mind is that there exist an infinite number of different bases for any given vector space. If W is a subspace of V and we are given a basis for W, then we can extend that to a basis for V. That is, the basis for V will consist of all vectors in the basis for W together with some other vectors. But there can also exist bases for V that do not contain any of the vectors in the basis for W.


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