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Another vector problem

  1. Nov 28, 2007 #1
    Determine if following subsets of R^2 are subspaces of R^2. If the subset is a subspace show that it is closed under vector addition and scalar multiplication. If the subset is not a subspace show why, indicating property that fails.

    [tex] 1) W=\{ \left (x_1,0)\left| x_1\in\Re\} \newline [/tex]
    [tex] 2) W=\{ \left (x_1,0)\left| x_1 > 0\} \newline[/tex]
    [tex]3) W=\{ \left (2c,-3c)\left| c \in\Re\} \newline[/tex]
    [tex]4) W=\{ \left (x_1,x_2)\left| x_1 > 0, x_2>0\} \newline [/tex]

    Answers:
    [tex] 1) (x,0) + (y,0)= (x+y,0) \in W \} \newline [/tex]
    [tex] c(x,0) = (c(x),0) \in W \} \newline [/tex]

    2) Not subspace since x1 can't be 0

    [tex] 3) (2c,-3c) + (x_1,x_2)= (2c+x_1,-3c+x_2) \in W \} \newline [/tex]
    [tex] x(2c,-3c) = (2c(x),-3c(x)) \in W \} \newline [/tex]

    4) Not a subspace since x1 and x2 can't be 0.

    Am I on the right track with this? Thanks
     
  2. jcsd
  3. Nov 28, 2007 #2

    HallsofIvy

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    Yes, you have shown that the set is closed under addition and scalar multiplication.

    And what property of a subspace does that violate? It would be sufficient to point out that the 0 vector (0, 0) is not in that set. Since your problem specifically referred to "closed under addition and scalaar multiplication, I think it would be better to show that 0(x_1, 0)= (0, 0) is not in the set so it is not closed under scalar multiplication.

    This is the only one where I think you have lost track of what you are doing- what are x_1 and x_2? To show "closed under addition", you need to look at (2c, -3c)+ (2d, -3d) where c and d are real numbers.

    Again, since closure under addition and scalar multiplication are specifically mentioned in the problem it would be better to show that 0(x1, x2)= (0, 0) is not in the set so it is not closed under scalar multiplication.

     
  4. Nov 29, 2007 #3
    Okay for the scalar multiplication:


    3)Since [tex]c_2 \ in \ R^2[/tex]
    [tex] c_2(2c_1,-3c_1) = (2c_1c_2,-3c_1c_2) \in W \} \newline [/tex]

    Is this better?
     
  5. Nov 29, 2007 #4

    HallsofIvy

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    Yes, now what about addition?
     
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