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Bilinear mapping between quotient spaces

  1. Nov 3, 2011 #1
    Problem: Let [itex]L[/itex] and [itex]M[/itex] be finite dimensional linear spaces over the field [itex]K[/itex] and let [itex]g: L\times M \rightarrow K[/itex] be a bilinear mapping. Let [itex]L_0[/itex] be the left kernel of [itex]g[/itex] and let [itex]M_0[/itex] be the right kernel of [itex]g[/itex].

    a) Prove that [itex]dim L/L_0 = dim M/M_0[/itex].

    b) Prove that [itex]g[/itex] induces the bilinear mapping [itex]g': L/L_0 \times M/M_0 \rightarrow K, g'(l+L_0, m+M_0) = g(l,m)[/itex], for which the left and right kernels are zero.

    I am trying to prove b) before a) (I think this would make a) relatively trivial). However, I am still getting used to the notion of quotient spaces as sets of equivalence classes. I am unsure as to how to deal with the bilinear mapping of equivalence classes. All that I have seen so far deal with vector spaces. Perhaps someone can give me some insight as to how to show that this mapping is well defined? How can I visualize this better? My professor is of little help, so I appreciate any insights anyone can give me.

    Thanks all!
     
    Last edited: Nov 3, 2011
  2. jcsd
  3. Nov 3, 2011 #2

    micromass

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    You must prove that if [itex]l+L_0=l^\prime +L_0[/itex] and [itex]m+M_0=m^\prime+M_0[/itex], then [itex]g^\prime(l+L_0,m+M_0) = g^\prime(l^\prime+L_0,m^\prime+M_0)[/itex]. So you must actually prove [itex]g(l,m)=g(l^\prime,m^\prime)[/itex]...
     
  4. Nov 3, 2011 #3

    Deveno

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    but...this is obvious because g is linear in l and m, and l-l' is in L0 and m-m' is in M0

    (g(l,m) = g(l',m') means g(l,m) - g(l',m') = 0.

    but g(l,m) - g(l',m') = g(l,m) - g(l',m) + g(l',m) - g(l',m')

    = g(l-l',m) + g(l',m-m') =....?)
     
  5. Nov 3, 2011 #4

    micromass

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    And what do you know about l-l' and m-m' ?
     
  6. Nov 3, 2011 #5

    Deveno

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    we are assuming l + L0 = l' + L0

    so l-l' is in 0 + L0 = L0.

    the same for m-m', with regard to M0.

    we have to have these conditions, for g' to be well-defined as a function

    (we are defining g' over cosets, by using g defined on elements.

    if we do not have the same image for g(l',m') and g(l,m) when l,l' and m,m'

    are in the same cosets of L0 and M0,

    g' isn't even a function, much less bilinear).
     
  7. Nov 3, 2011 #6
    Wow, okay, so that makes it obvious. Thanks for the help guys, I appreciate it. Neat stuff!
     
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