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Homework Help: [Algebra] R4/U?

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Well it isn't so much the problem as it is the notation used within the problem. But here is the question:

    Determine whether or not [itex]\overline{w}[/itex] and [itex]\overline{v}[/itex] are linearly independent in R4/U


    2. Relevant equations
    If v [itex]\in[/itex] V then [itex]\overline{v}[/itex] = v + U


    3. The attempt at a solution

    I don't understand the notation R4/U (I understand R4 and the subspace U, but don't understand the slash between them)
     
  2. jcsd
  3. Feb 5, 2012 #2

    Deveno

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    Science Advisor

    it's called a "quotient space" and its elements are called "cosets" and consist of "parallel translates of a subspace by a vector".

    that's why v has the overline: it's the SET:

    {v+u: u in U}.

    which can be thought of as the entire subspace U, moved in the direction/distance of v.

    another way to think of it, is as regarding the entire space (in this case R4) losing dim(U) dimensions, by regarding all points in U as "equivalent (essentially 0)".

    if dim(U) = 1, each coset is a parallel line, and you need a 3-vector to tell you "which line".

    if dim(U) = 2, each coset is a parallel plane, and you need a 2 vector to tell you "which plane".

    higher dimensions are harder to visualize, but the same sort of logic applies.

    simce v+U is a set, v is just a "representative", and the same coset v+U can have different representatives.

    one common way quotient spaces arise is in analyzing linear maps: often, we don't care about the kernel of a linear map (because everything in it just maps to the 0-vector), so we "mod it out". the resulting quotient space is isomorphic to the image space (this is pretty much equivalent to the rank-nullity theorem, but in a more abstract setting).

    you calculate with elements in R4/U pretty much like you do with elements in R4, but with a "+U" along for the ride:

    v+U + w+U = (v+w)+U
    a(v+U) = av+U

    the overline notation is a bit "cleaner" but hides some of what is going on.
     
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