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vector spaces

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If anyone has any insight on the following passage I would love to hear it. I have reread it a thousand times, and it is still not making sense to me. "Every vector space can be generated by a linear combination of a subset of vectors, called a basis for the vector space" Ok, so is this basically saying that given a subset of a set of vectors, you can generate all vectors belonging to the set of which the subset belongs? or am I just sounding like a babbling idiot? thx for your time.

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sorry that should have read

"Every vector space can be generated by linear combinations of a subset of vectors, called a basis for the vector space"

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quote:
Original post by Fooshnucka
"Every vector space can be generated by a linear combination of a subset of vectors, called a basis for the vector space"

Ok, so is this basically saying that given a subset of a set of vectors, you can generate all vectors belonging to the set of which the subset belongs?


Yes, but not every subset has this property. Those that do are called spanning subsets (I think, it''s been a while) and form a basis for the vector space. This means that any other vector in the space can be written as a linear combination of the basis vectors.

For instance coordinates in 3 dimensions are often given as (x,y,z), and it is implied that this means x*(1,0,0)+y*(0,1,0)+z*(0,0,1), making (1,0,0) (0,1,0) and (0,0,1) the basis. You could use many other bases, such as (5,0,1) (0,-1,0) (0,0,1), the coordinates (x,y,z) would just be different to represent the same point in space.

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ok, so then the vectors that make up the basis, would be linearly independant of eachother, and all other vectors in the set would be linearly dependent on the basis vectors? Am I on the right track of thinking here?

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Yeah, absolutely :

In an n-dimensionnal vector space, any vector can be written as a linear combination of basis'' vectors (that contains n linearly independant vectors).

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It takes N vectors to form the basis for an N-dimensional space, but you can always have more. If you have N you have an efficient basis. It's said to be an efficient basis because if you have N+n vectors then n of them can be written as linear combinations of the others and thus removed from the basis without changing the space it spans. Although I don't feel like showing a proof of that right now.

[edited by - Dobbs on April 19, 2002 7:09:07 PM]

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Lol.. are you studying for first-year vector geometry?

I just wrote an exams yesterday on that..

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The other way to think of this is to go down to the level of the basis set...

Consider a set {A} of N orthogonal (and hence linearly independent) vectors. The vector space spanned by this basis set of vectors is the set consisting of all linear transformations (scaling, addition, subtraction) of the basis set.

You can think of a minimal or efficient basis like this...

Consider your first basis set {A} and its corresponding vector space... if there exists a set of vectors {B} of size M (M < N) with which you can describe the original basis {A} (as a set of linear combinations of {B}) then {B} forms an efficient or sometimes called a minimal basis for the vector space spanned by {A}.

Cheers,

Timkin

[edited by - Timkin on April 19, 2002 10:42:51 PM]

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quote:
Original post by BloodScourge
Dobbs, the answer to your doubt is here

[edited by - Bloodscourge on April 19, 2002 8:26:51 PM]


Sorry, what was I doubting exactly?

Oh and according to that document R2 is a subspace of R3. That''s just plain wrong, so that doc isn''t going to do much to counter the doubt you think I have.

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Hey Dobbs. I don''t have any qualms with your original posts, but in your last one, you said that R2 is not a subset of R3. I am not so sure about that. My textbook tells me that W is a subspace of V if V contains W and W supports the same operations as V. It then goes on to say that every vector space V has two trivial subspaces, {0} and V. So I am tempted to believe that if R0 is a subspace of R3, then R1 and R2 are too.

I know this is just semantics, and I can''t prove you wrong, but I am curious: why do you think R2 is not a subset of R3?

Cédric

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Thus, according to you Dobbs, the plane that contains the x-axis and y-axis is not a subspace of R3(!?)...

[edited by - Bloodscourge on April 20, 2002 4:53:15 PM]

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Guest Anonymous Poster
@ Bloodscourge, Dobbs :
R2 is the set of all pairs of real numbers, R3 is the set of all triples of real numbers. They have no elements in common. Strictly speaking, Dobbs is right.

However, Bloodscourge wasn''t wrong either: R3 has the canonic basis (1,0,0),(0,1,0),(0,0,1), while R2 has the slightly different canonic basis (1,0),(0,1). (1,0,0) and (0,1,0) generate a subspace E in R3. One can identify R2 with E.

The same thing happened with real and complex numbers, rational and real numbers, integer and rational numbers. One treats the former as if it was a subset of the latter.

Definition 1: A subset A of a vector space V is called spanning set if any element of V can be written as the linear combination of elements in A.
Definition 2: A spanning set is called a base if there is only one, namely the trivial linear combination (all coefficients zero) that yields the Zero-Vector.
Theorem and Definition 3 (Proof not-so-trivial):
a) Each vector space has a base B.
b) Although there are many different Bs for a given vector space, the Cardinality (the number of elements it contains) of such a base is a constant, it is called the dimension of the vector space.
Remark 4: There are other possibilities to define the term "base".

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Thanks AnonPost for clarifying some points. But :
quote:
Remark 4: There are other possibilities to define the term "base".
Could you be more precise?

[edited by - Bloodscourge on April 20, 2002 6:46:21 PM]

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Well, at least R 2 is a sub-manifold of R 3, otherwise we'd all be in trouble!!! ;-)

Cheers,

Timkin

[edited by - Timkin on April 20, 2002 8:34:54 PM]

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Guest Anonymous Poster
A) A basis is a minimal spanning set (it contains as few vectors as possible).
B) A basis is a maximal set of linearly independent vectors (it contains as many linearly independent vectors as possible).
C) A basis is a spanning set, with only the trivial linear combination yielding the zero-vector.
D) A basis is a spanning set, where every vector has exactly one representation.
E) In the case of R^n : A basis can be represented by a n*n Matrix with no-zero-determinant.

Here comes a crispy one:
F) A basis is a set of n vectors such that
can: R^n -> V , which maps a list of coefficients to the coresponding linear combination, is bijective.

I''m sure some mathematican found an even crispier definition.

HTH
AnonPost

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