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Math Help - Vector Spaces

  1. #1
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    Vector Spaces

    The question:

    Show that the set

    S = \{x \in \mathbb{R}^3 : x_1 \le 0, x_2 \ge 0 \}

    with the usual rules for addition and multiplication by a scalar in \mathbb{R}^3 is not a vector space by showing that at least one of the vector-space axioms is not satisfied. Give a geometric interpretation of this result.

    My attempt:

    There's no solution to this answer in my text. My understanding is that this set is not closed under scalar multiplication, since multiplying by -1 will reverse the conditions on x_1 and x_2. Is this correct?

    As for the second part of the question, it asks for a geometric interpretation, which I'm not sure about. It's a 3D object, but otherwise I don't know how else to interpret it. Any ideas?

    Thanks.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Glitch View Post
    The question:
    My understanding is that this set is not closed under scalar multiplication, since multiplying by -1 will reverse the conditions on x_1 and x_2. Is this correct?
    Yes, it is correct.

    As for the second part of the question, it asks for a geometric interpretation, which I'm not sure about. It's a 3D object, but otherwise I don't know how else to interpret it. Any ideas?
    Bear in mind that S=A \times \mathbb{R} where A is the second quadrant on the x_1x_2 plane.

    Regards.

    Fernando Revilla
    Last edited by FernandoRevilla; November 23rd 2010 at 09:51 PM.
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    Bear in mind that S=A \times \mathbb{R} where A is the second quadrant on the x_1x_2 plane.
    I'm not sure what you mean by this. I'm unfamiliar with 'quadrants' in terms of planes in space (except for the standard Cartesian plane).
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  4. #4
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    Yes, that's what he is talking about. The x and y axes divide the xy-plane into 4 quadrants. The usual convention is that the "first quadrant" is x> 0, y> 0, the "second quadrant" is x< 0, y> 0, the "third quadrant" is x< 0, y< 0, and the "fourth quadrant" is x> 0, y< 0. \{(x, y)| x\le 0, y\ge 0\} is the second quadrant, together with the axes bounding it.

    In fact, that can be extended to three dimensions: the x, y, and z coordinate planes divide the space into 8 "octants". Just take the four quadrants above with z> 0 to get four of them and take the four quadrants above with z< 0 to get the other four.


    S= \{(x_1, x_2, x_3)|x_1\le 0, x_2\ge 0\} is the second and sixth octants, together with the planes bounding them.
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  5. #5
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    Ahh that makes more sense. Thank you!
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