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Vector Quotient Spaces

  1. Nov 10, 2006 #1
    I'm having a bit of trouble seeing Vector Quotient Spaces.
    Lets say I have a vector space $V$ and I want to quotient out by a linear subspace $N$. Then $V/N$ is the set of all equivalence classes $[N + v]$ where $v \in V$.

    For example, let me try to take $\mathbb{R}^{2} /$ x-axis. This should be the set of all equivalence classes $[x-axis + r]$ where $r \in \mathbb{R}^{2}$.

    Here is where the difficulty arises I believe. I am told that this set is the class of lines parallel to the x-axis, but I can't see how any coset $x-axis + r$ could yield a line parallel to the x-axis - or maybe my conception of vector space cosets are wrong.
     
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  3. Nov 10, 2006 #2

    Office_Shredder

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    If you have a class that is the set of points a distance r away from the x-axis, you can rewrite it as y=r.

    Try graphing it and see for yourself (pick an r, say, r=1)
     
  4. Nov 10, 2006 #3

    Hurkyl

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    What points lie in $x-axis + r$?
     
  5. Nov 10, 2006 #4
    But if you add a vector to the x-axis, how can that be a line parallel to the x-axis. More importantly, what is an element of the x-axis? Is it just a point?
     
  6. Nov 10, 2006 #5

    Hurkyl

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    Since the x-axis is a vector space... a vector!

    (Of course, sometimes we identify the notions of "point" and "vector" with each other)
     
  7. Nov 10, 2006 #6
    Then i think my conception of adding vectors is incorrect. If i think of vectors as directed line segments, i cant think of a way to add vectors on the x-axis to other vectors to get something parallel.
     
  8. Nov 10, 2006 #7

    Office_Shredder

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    You're not. What you're doing, is adding a vector to the x-axis, and the set of ends of those directed line segments forms a line.
     
  9. Nov 10, 2006 #8
    oh oh oh i see now. thanks
     
  10. Nov 10, 2006 #9
    and the line comes from the vector definition of line.
     
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