| Thread Closed |
Basis of a Subspace |
Share Thread | Thread Tools |
| Feb8-10, 01:04 PM | #1 |
|
|
Basis of a Subspace
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 :). |
| Feb8-10, 01:59 PM | #2 |
|
Mentor
|
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. |
| Feb8-10, 02:39 PM | #3 |
|
|
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 :). |
| Feb9-10, 03:51 PM | #4 |
|
Recognitions:
|
Basis of a Subspace
I don't think you fully get it yet.
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. |
| Feb10-10, 07:10 AM | #5 |
|
|
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.
|
| Thread Closed |
| Tags |
| basis, linear algebra, subspace |
| Thread Tools | |
Similar Threads for: Basis of a Subspace
|
||||
| Thread | Forum | Replies | ||
| basis and subspace help | Calculus & Beyond Homework | 6 | ||
| Basis of a subspace? | Calculus & Beyond Homework | 4 | ||
| Basis for a Subspace. | Precalculus Mathematics Homework | 6 | ||
| basis for subspace | Linear & Abstract Algebra | 10 | ||