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If Ker T != 0 then T is not isomorphism

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that if ker T != 0 then T is not an isomorphism.


    2. Relevant equations



    3. The attempt at a solution
    If Ker T != 0 that means that there are multiple solutions for which T=0 meaning it is not injective and hence not isomorphic? Is that correct? I don't think it is, or is there a better way of proving this? Thanks.
     
  2. jcsd
  3. Oct 24, 2012 #2

    Dick

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    Well, it is correct. If x is a nonzero element of ker T, then Tx=0=T0. Not injective. Not isomorphism.
     
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