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Image of Normal Operator

  1. Mar 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that if T is a normal operator on a finite dimensional vector space than it has the same image as its adjoint.


    2. Relevant equations
    N/A


    3. The attempt at a solution
    I have been able to show that both T and [itex]T^{*}[/itex] have the same kernel. Thus, by using the finite dimension property and the rank nullity theoremit just suffices to show containment one way.

    However, if you suppose that a vector v is in the Im(T) I haven't been able to find some representative w such that T*(w) = v.

    Does anyone have any idea how to proceed?

    Thanks!
     
  2. jcsd
  3. Mar 16, 2014 #2

    Dick

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    You should also be able to show that Im(T) is orthogonal to Ker(T). Use that.
     
  4. Mar 16, 2014 #3
    Is that necessarily true? And if it is (I know that you can show that Im(T) is orthogonal to Ker(T*) and vice versa), but how does that specifically help?
     
  5. Mar 16, 2014 #4

    micromass

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    The idea is to show both ##\textrm{im}(T)## and ##\textrm{im}(T^*)## to be orthogonal to ##\textrm{ker}(T)##. For former you will need normality, the latter should be easy.
     
  6. Mar 16, 2014 #5
    I understand that but does that necessarily imply that they have to be equal? I recognize that they have to have the same dimensions in that case, but that doesn't get me anywhere.
     
  7. Mar 16, 2014 #6

    micromass

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    If you accept the fact in my last post to be true for now, then what would ##\textrm{ker}(T)^\bot## be?
     
  8. Mar 16, 2014 #7
    Ah it would be Im(T) and it would be Im(T*) so thus they have to be equal?
     
  9. Mar 16, 2014 #8

    micromass

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    Right!
     
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