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Figuring out if subset of R2 is a subspace

  1. Dec 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Let E be the subset of R2 defined by E={(x,y)|x≥0,y[itex]\in[/itex]ℝ}. Is E a subspace of R2?


    2. Relevant equations
    E={(x,y)|x≥0,y[itex]\in[/itex]ℝ}


    3. The attempt at a solution
    I honestly have no idea where to start. Please help !!! I'm not asking for the answer per se, just a starting point
     
  2. jcsd
  3. Dec 14, 2011 #2

    radou

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    What are the conditions for a space to be a subspace of some other space? Can you apply these to find a counterexample, perhaps?
     
  4. Dec 14, 2011 #3
    ahh.. So E is not closed under scalar multiplication since -k(x,y)=(-kx, -ky) and this is not an element of E
     
  5. Dec 14, 2011 #4

    radou

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    Exactly. Note only that you have to take an element (x, y) of E where x > 0 to demonstrate that it works, since if you take an element of the form (0, y) in E, then for some k < 0 you have k(0, y) = (0, ky), which is still in E.
     
  6. Dec 14, 2011 #5
    Thanks! I was also struggling a bit with this problem:
    Let E={(x,2x+1|x∈ℝ}, so E is a subset of R2. Is E a subspace of R2?

    I said no because it is not closed under addition:
    (x,2x+1)+(y,2y+1)=(x+y,2x+1+2y+1)=(x+y,2(x+y)+2) which isn't equal to (x+y,2(x+y)+1)
    Is this correct?
     
  7. Dec 14, 2011 #6

    radou

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    Yes. Also, it's even easier to see that (0, 0) is not in E in this case.
     
  8. Dec 14, 2011 #7
    Oh that is much easier! Thank you so much!
     
  9. Dec 14, 2011 #8
    Sorry, no one is replying to my other thread so I figured I would ask you one more thing here.

    Let E={(2a,a)|a∈ℝ}. Let B={(b,b)|b∈ℝ}.
    Is E∪B a subspace of R2?
    What is E+B

    My solution:
    E∪B={(2a,a),(b,b)|a,b∈ℝ}
    I don't know how to show if tis is a subspace of R2 or if that is the correct union

    For the second part, I know that E+B = span (E∪B)
    but I don't know if I have the right union?
     
  10. Dec 14, 2011 #9

    radou

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    A union of subspaces of a given space need not be a subspace of that space. For example, take a non-zero a in R, and let (2a, a) and (a, a) be elements of E U B. Then (2a, a) + (a, a) = (3a, 2a) is neither in E nor in B.

    The sum of two subspaces is again a subspace of that space.
     
  11. Dec 14, 2011 #10
    So, is E+B= span(EUB)=
    a(2,1)+b(1,1)=(2a,a)+(b,b)=(2a+b, a+b)?
     
  12. Dec 14, 2011 #11

    radou

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    You have two subspaces of a space V, E and B. The span of E U B is defined to be the set of all linear combinations of elements from E U B. Let a and b be scalars, and let v1 and v2 be elements of E U B. Then av1 + bv2 is an element of E U B, and hence E U B is a subspace of V.

    Correction: let v1 and v2 be elements of span(E U B) (note that v1 and v2 are linear combinations).
     
    Last edited: Dec 14, 2011
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