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Vector Spaces, Subsets, and Subspaces

 
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Nov8-07, 11:33 PM   #1
 

Vector Spaces, Subsets, and Subspaces


1. The problem statement, all variables and given/known data


What is an example of a subset of R^2 which is closed under vector addition and taking additive inverses which is not a subspace of R^2?

R, in this question, is the real numbers.


2. Relevant equations

I know that, for example, V={(0,0)} is a subset for R^2 that is also a subspace, but I can't figure out how something can be a subset and not a subspace.



3. The attempt at a solution

Does this have anything to do with scalar multiplication being closed on the vector space?
 
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Nov9-07, 12:29 AM   #2

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Think about the set of all (x,y) where x and y are both integers.
 
Nov9-07, 01:38 AM   #3
 
So for example, if we let the subset = (a,b) s.t. a,b are elements of Z. Then it is closed under addition but not under scalar multiplication. i.e. Let (a,b) = (1,3) and multiply by 1/2 for example (which is the example we used to figure it out). Then you get (1/2, 3/2), neither of which are in Z.
 
Nov9-07, 08:15 AM   #4

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Vector Spaces, Subsets, and Subspaces


Sure, but it does have additive inverses.
 
Nov9-07, 03:59 PM   #5
 
Quote by Dick View Post
Think about the set of all (x,y) where x and y are both integers.
or both rational numbers
 
Nov9-07, 05:52 PM   #6
 
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Quote by mrroboto View Post
1. The problem statement, all variables and given/known data


What is an example of a subset of R^2 which is closed under vector addition and taking additive inverses which is not a subspace of R^2?

R, in this question, is the real numbers.


2. Relevant equations

I know that, for example, V={(0,0)} is a subset for R^2 that is also a subspace, but I can't figure out how something can be a subset and not a subspace.
It's very easy to be a subset without being a subspace! Just look at any subset that does not satisfy the requirements for a subpace- what about { (1, 0)}?



3. The attempt at a solution

Does this have anything to do with scalar multiplication being closed on the vector space?
In order for a subset to be a subspace, it must be closed under addition, have additive inverses, and be closed under scalar multiplication. Since you are asked about a subset that IS closed under addition and has additive inverses, looks like there is only one thing left!
 
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