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Adjoint of a linear operator

  1. Dec 11, 2007 #1
    Q) Let V be an inner product space and T:V->V a linear operator. Prove that if T is normal, then T and T* have the same image. (i.e. imT=imT*)

    My Attempt:
    <T(v),T(v)>
    =<T*T(v),v>
    =<TT*(v),v>
    =<T*(v),T*(v)>

    =>||T(v)|| = || T*(v)||

    But this doesn't seem to help...

    Thanks!
     
    Last edited: Dec 11, 2007
  2. jcsd
  3. Dec 11, 2007 #2

    HallsofIvy

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    You are trying to prove two sets (Im(T) and Im(T*)) are the same. Start by assuming v is in Im(T). Then there exist u such that v= T(u). You want to show that v= T*(w) for some w in V.
     
  4. Dec 12, 2007 #3
    I understand the definitions, but I have no idea how to prove this.
    Can you give me more hints, please?:smile:
     
  5. Dec 12, 2007 #4
    My attempt:
    v=T(u) for some u E V
    => T*(v)=T*T(u)
    => T*(v)=TT*(u) since T is normal

    And now I am stuck, how can I prove that v=T*(w) for some w E V?
     
  6. Dec 12, 2007 #5

    morphism

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    This seems false unless you're given some other condition on V. Is it finite-dimensional or complete? If so, then from your observation that ||Tv||=||T*v|| we can conclude immediately that kerT=kerT*; on the other hand, it's easy to see that the orthogonal complement of imT* is kerT (i.e. [itex](\text{im} T^*)^{\perp} = \ker T[/itex]). Now use finite-dimensionality to place things together (in case of completeness, use direct sum decompositions).
     
  7. Dec 12, 2007 #6
    We are assuming finite-dimensional vector space.

    v=T(u) for some u E V
    => T*(v)=T*T(u)
    => T*(v)=TT*(u) since T is normal
    I have no idea how to proceed from here...

    How is it possible to prove without using [itex](\text{im} T^*)^{\perp} = \ker T[/itex]?
     
  8. Dec 12, 2007 #7

    morphism

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    Why don't you want to use it?
     
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