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Linear Algebra - Vector Subspaces

  1. Feb 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Give an example of a nonempty subset U of R2 such that U is closed under scalar multiplication, but U is not a subspace of R2.

    2. Relevant equations

    3. The attempt at a solution
    I think I have it, but I just want to make sure it's right:

    Let U = {(x, x + 2)} | x is in R}. Let u be in U. Then au = a(x, x + 2) = (ax, a(x + 2)), which is still in U (i.e. still on the same line). But this is not a subspace of R2 because 0 is not in this subset.

  2. jcsd
  3. Feb 21, 2008 #2


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    You get [itex]<0,0>[/itex] if you multiply by [itex]a=0[/itex].
    (Hint: What are the properties of subspaces?)
  4. Feb 21, 2008 #3
    1) Additive Identity: 0 is in U.

    2) Closed Under Addition: u, v in U implies u + v is in U.

    3) Closed Under Scalar Multiplication: a is in R (or C) and u is in U implies au is in U.

    So I need to come up with a subset that fails for one (or both) of the first two properties, but not for property 3.

    How about checking for property 2? If u = (x, x + 2) and v = (y, y + 2), where u, v are in U, then u + v = (x + y, (x + y) + 4), which is not in U, so U is not a subspace of R2.
  5. Feb 21, 2008 #4


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    ...better, but:
    [tex]2<3,(3+2)>=<6,10> \notin U[/tex]
    so you don't have closure under scalar multiplication.
  6. Feb 21, 2008 #5
    Wow, I guess I don't. I really don't know where to go from here. Any hints on a a type of subset that would work in this situation?
  7. Feb 21, 2008 #6
    I was searching the forums and found pretty much the exact same question. Here's the thread if anyone is interested: https://www.physicsforums.com/showthread.php?t=41884

    A set that works is the following: U = {(x, y) | xy = 0}. This set is closed under scalar multiplication, but not vector addition.
  8. Jan 31, 2010 #7

    You are absolutely wrong.U={(x,x+2)| x is in R} is not closure under multiplication.
    Ex:put x=0,a=2 (0,2) is in U but (0,4) is not in U.
  9. Jan 31, 2010 #8


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    Yes. he was told that almost two years ago!
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