Vector Addition and Subspaces

  • Thread starter mrroboto
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I don't understand this, can someone help?:

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.

Thanks!
 

Answers and Replies

  • #2
JasonRox
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Can you think of a subset of R^2 such that it is a subspace?
 
  • #3
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Well you can add things but you can't subtract things. This should be a big enough hint.
 
  • #4
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Well you can add things but you can't subtract things.
He mentioned additive inverses and closure under addition (implying 0 being an element), so he has no problem subtracting things. The key lies in the one part of the definition of a vector space that is left out.
 
  • #5
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He mentioned additive inverses and closure under addition (implying 0 being an element), so he has no problem subtracting things. The key lies in the one part of the definition of a vector space that is left out.
I misread it. I thought the question asked to find something that fails to be a subspace because it's not closed under additive inverses.
 
  • #6
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To JasonRox: Yes, I can think of a subset, V, of R^2 that is a subspace. V= {(0,0)}. But I still don't understand how to approach this problem.
 
  • #7
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first try to solve a simpler problem.
are there nontrivial subgroups of the group [tex] (\mathbb{R},+,-,0) [/tex] ? that means is there a set [tex] A [/tex] with [tex] \{0\} \subset A \subset \mathbb{R} [/tex] such that [tex] A [/tex] is closed under addition and substraction?
once you have found such a subgroup [tex] A [/tex], is [tex] A^2 [/tex] a set with the desired property?
 
  • #8
HallsofIvy
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In order that a subset be a subspace, it must be closed under addition, additive inverses, and scalar multiplication. Since your subset is required to be closed under addition and additive inverse, there's only one place left to look!
 

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