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Vector subspaces

  1. Oct 9, 2011 #1
    Hello :)

    I've been doing a lot of work on subspaces but have come across this question and need a bit of help!

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

    [itex]W = {(x, y) \in R^{2} | x^{2} + y^{2} = 0}[/itex]

    2. Relevant equations

    1. 0 ∈ W
    2. ∀ u,v ∈ W; u+v ∈ W
    3. ∀ c ∈ R and u ∈ W; cu ∈ W

    3. The attempt at a solution

    Check for 0 vector

    [itex]x^{2} + y^{2} = 0[/itex]

    [itex]0^{2} + 0^{2} = 0[/itex]

    [itex]0 = 0[/itex]

    Check closure under scalar addition

    Let [itex]u = x^{2} + y^{2} = 0[/itex]; let [itex]v = a^{2} + b^{2} = 0[/itex]

    [itex]u + v = (x^{2} + a^{2}) + (y^{2} + b^{2}) = 0 + 0 = 0[/itex]

    Check for closure under scalar multiplication

    [itex]ku = (kx)^{2} + (ky)^{2} = 0[/itex]

    [itex]= k^{2}(x^{2} + y^{2}) = 0[/itex]

    [itex]x^{2} + y^{2} = \frac{0}{k^{2}}[/itex]

    [itex]x^{2} + y^{2} = 0[/itex]

    -----------------------------------

    I have shown that the zero vector is in the set, and that it is closed under scalar multiplication, however; I'm not sure whether or not it is closed under scalar multiplication.

    I have shown that u + v = 0, but, u + v does not have the same form as u and v individually, so I don't think u + v is part of the set?
     
    Last edited: Oct 9, 2011
  2. jcsd
  3. Oct 9, 2011 #2
    Just took another look at this question and might have solved it.

    [itex]u = (a, b), v = (c, d), u + v = (a + c, b + d)[/itex]

    [itex](a + c)^{2} + (b + d)^{2} = 0[/itex]

    [itex]a^{2} + 2ac + c^{2} + b^{2} + 2bd + d^{2} = 0[/itex]

    But, we know from the constraints of the subspace that;

    [itex]a^{2} + b^{2} = 0; c^{2} + b^{2} = 0[/itex]

    So if we cancel those out, we get;

    [itex]2ac + 2bd = 0[/itex]

    So in general, scalar addition breaks the constraint.

    Does this even make sense or do I need more sleep? :grumpy:
     
  4. Oct 9, 2011 #3

    Mark44

    Staff: Mentor

    Yes, you need more sleep.

    Think about what your set looks like - W = {(x, y) [itex]\in[/itex] R2| x2 + y2 = 0}.

    This set consists of a single point at (0, 0). If u [itex]\in[/itex] W and v [itex]\in[/itex] W, what must u and v be? It should be easy to show that u + v [itex]\in[/itex] W.
     
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