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Fooshnucka

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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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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