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vector spaces

Started by Fooshnucka Apr 19, 2002 at 1:09 PM 14 replies 1k views

Original Post

Apr 19, 2002 at 1:09 PM

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.

Fooshnucka

Author

Fooshnucka

Apr 19, 2002 at 1:11 PM

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"

Dobbs

Apr 19, 2002 at 1:50 PM

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.

Fooshnucka

Author

Fooshnucka

Apr 19, 2002 at 1:57 PM

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?

BloodScourge

Apr 19, 2002 at 3:46 PM

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).

Dobbs

Apr 19, 2002 at 6:06 PM

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]

_SHO_NUFF

Apr 19, 2002 at 6:24 PM

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

I just wrote an exams yesterday on that..

BloodScourge

Apr 19, 2002 at 7:26 PM

Dobbs, the answer to your doubt is here

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

Timkin

149

Timkin

Apr 19, 2002 at 9:39 PM

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]

Dobbs

Apr 20, 2002 at 2:13 PM

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.

Cedric

Apr 20, 2002 at 2:46 PM

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

BloodScourge

Apr 20, 2002 at 3:17 PM

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]

BloodScourge

Apr 20, 2002 at 5:46 PM

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]

Timkin

149

Timkin

Apr 20, 2002 at 7:34 PM

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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