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Math Help - Bijection?

  1. #1
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    Bijection?

    Describe what is (almost almost almost) a bijection between (R3)* [* signifying dual space] and planes in R3 through the origin

    Any help with this, I am completely stumped.
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  2. #2
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    Re: Bijection?

    Isn't the dual space to R^3 isomorphic to R^3? And planes going through the origin have equations Ax + By + Cz = 0 for various (but not all) real A, B and C.
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  3. #3
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    Re: Bijection?

    suppose we have an element of (R3)*.

    this is a linear function f:R3-->R. since f is linear, f(x,y,z) = ax + by + cz.

    now, consider the mapping φ:P-->(R3)*, where P is the set of planes through the origin, given by:

    if p is the plane defined by Ax + By + Cz = 0, φ(p) = f, where f(x,y,z) = Ax + By + Cz.

    the question is now: why is this not a bijection?

    it might be instructive to note that:

    1) the planes x + y + z = 0 and 2x + 2y + 2z = 0 are the same plane, but the linear functionals f(x,y,z) = x + y + z and g(x,y,z) = 2x + 2y + 2z are different.

    2) the 0-functional does not correspond to any plane (why?).

    if you are clever enough, can you think of a way to express φ(p) as an inner product of the vector (A,B,C) and something? how does this relate to the normal vector to a plane p?

    since 3 = 1 + 2, can you see some connection here?
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