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Linear transformations

  1. Jun 28, 2006 #1
    If i am given a linear transformation D:A->A,that is followed by
    A=ImD(+)kerD
    and i am asked to prove that kerD^2=kerD and imD=imD^2.

    instead of trying to work it out the hard way by showing that every element of KerD is an element of kerD^2 , both directions.

    would it not be easier to just say that dimA=dimA and hence the two structures are isomorphic which means that KerD={0} and ImD=A.

    same goes for D^2:A->A
    KerD^2={0}
    ImD^2=A

    => therefore KerD^2=KerD and ImD^2=ImD ?
     
  2. jcsd
  3. Jun 28, 2006 #2

    Hurkyl

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    Why does it mean that?
     
  4. Jun 28, 2006 #3

    AKG

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    What two structures?
    Huh? I'm not sure what in the world you're doing, but just because A is isomorphic to A doesn't mean that every linear map D:A->A is an isomorphism. Is that what you were thinking?
     
  5. Jun 28, 2006 #4
    According to my book two vector spaces of the same dimension are isomorphic to eachother.
    and the proof also apparently seems to be pretty simple.
    If B is a basis for A, then we can easily show that ImD=A
    and that T is injective and if T is injective then KerD={0}
     
  6. Jun 28, 2006 #5

    Hurkyl

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    You're forgetting one of the hypotheses for the theorem -- you need D to be an isomorphism.
     
  7. Jun 28, 2006 #6
    yes akg unfortunately that is what i was thinking, that if the two were isomorphic to eachother, then the map would necessairly be an isormophism.

    just a few minutes before hurkyl put his post up, i was about to say that i went over the theorems in my book again, and that my line of thought was incorrect.

    anyway, thank you all.
     
    Last edited: Jun 29, 2006
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