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Quick Question

  1. Sep 27, 2007 #1
    Hey guys,

    Does dimension remain unchanged under a linear map? Ie if i have a map f:U->V does dim(U) = dim(img(f))?

    Cheers
    -Z
     
  2. jcsd
  3. Sep 27, 2007 #2
    Not necessarily.
     
  4. Sep 27, 2007 #3

    morphism

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    Homework Helper

    For a trivial example, take the zero map on a space of positive dimension.

    Something that might be of interest to you is the rank-nullity theorem.
     
  5. Sep 27, 2007 #4
    Yeh i kinda figured it didnt hold. Actually i was gonna use it to prove the RL theorem if it was true

    cheers
    Z
     
  6. Sep 27, 2007 #5

    HallsofIvy

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    Staff Emeritus
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    If L is linear, L(U) cannot have dimension higher than U. It can have dimension lower than U. Of course, to do that, L must map many vectors to the 0 vector: its kernel is not empty. The "nullity-rank" theorem morphism mentioned says that the dimension of L(U) plus the dimension of the kernel of U is equal to the dimension of U.
     
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