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Linear Functionals Inner Product

  1. Dec 10, 2006 #1
    Assume that [tex] m<n [/tex] and [tex] l_1,l_2,...,l_m [/tex] are linear functionals on an n-dimensional vector space
    [tex] X [/tex].

    Prove there exists a nonzero vector [tex] x [/tex] [tex] \epsilon [/tex] [tex] X [/tex] such that [tex] < x,l_j >=0 [/tex] for [tex] 1 \leq j \leq m[/tex]. What does this say about the solution of systems of linear equations?


    This implies
    [tex] l_j(x) [/tex] [tex] \epsilon [/tex] [tex] X^\bot [/tex] for [tex] 1 \leq j \leq m [/tex] or [tex] l_j(x)=0 [/tex] for [tex] 1 \leq j \leq m [/tex]. Since it is stated in the problem that [tex] l_1,l_2,...,l_m [/tex] are linear functionals on the vector space X, [tex] l_j(x)=0 [/tex]. Does this reasoning even help me find the proof? I am stuck.

    If you have trouble reading this, it is also at http://nirvana.informatik.uni-halle.de/~thuering/php/latex-online/olatex_33882.pdf
     
    Last edited: Dec 10, 2006
  2. jcsd
  3. Dec 10, 2006 #2

    StatusX

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    I don't know what [itex]X^\perp[/itex] means, nor why you are taking the inner product rather than just applying the functional to the vector. You'll obviously need to involve n and use n>m, and the easiest way to use this is appeal to the existence of a basis with n elements.
     
  4. Dec 10, 2006 #3
    so basically I'm trying to prove that for some nonzero [tex]x[/tex], [tex] l_j(x)=0 [/tex]?
     
  5. Dec 10, 2006 #4

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    Yea, for some x, for all j.
     
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