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Normal operators and self adjointness

  1. Jul 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose V is a complex inner-product space and T ∈ L(V) is a
    normal operator such that T9 = T8. Prove that T is self-adjoint
    and T2 = T.


    2. Relevant equations



    3. The attempt at a solution
    Consider T9=T8. Now "factor out" T7 on both sides to get T7T2 =TT7. Now we represent T as a matrix. Since T is normal, it is diagonizable. Therefore it is invertible. Now T2=TT7T-7 so T2=(T)(I)7=T.
     
  2. jcsd
  3. Jul 31, 2009 #2

    Dick

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    The zero matrix is diagonalizable. It's not invertible. And it's not the only such matrix.
     
  4. Jul 31, 2009 #3
    oh right. Can't forget those.
    Let T be a "reduced" matrix. Now let T be a zero transformation. It is clear that T2=T when T=0. Consider T=/=0 (still a "reduced" matrix). Consider T9=T8. Now "factor out" T7 on both sides to get T7T2 =TT7. Now we haveT7T2-TT7=0. Or
    T7(T2-T)=0. Since we know that this equation holds true (equals 0), T2 must=T.
     
  5. Aug 1, 2009 #4
    That's far from a valid proof, T^7 can be 0,even if it's not, that doesn't imply T^2-T=0.
     
  6. Aug 1, 2009 #5

    Dick

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    i) What is 'reduced' supposed to mean? ii) If A and B are matrices, AB=0 and A is not zero, that doesn't imply B=0.
     
  7. Aug 1, 2009 #6
    I'm really stuck here. I mean, is it possible to prove T2=T
    without proving T is self adjoint first? I kept coming up with ways to prove T2=T.
    but they all failed because of the fact that Tn could be 0.
    Also there are many cases where T is normal, and T9=T8 leads to
    T2=T. I mean, T could be an identity, T could be a 0 map,
    T could be an orthogonal projection, let me know if I missed something, lol.
    Sorry but I can't think of a proof for this, if there is, a small hint could do the trick.
    Oh and I COULD say take T-7 and multiply it by both sides. But
    what if T7 is 0? So that's out of the question.
     
    Last edited: Aug 1, 2009
  8. Aug 1, 2009 #7

    HallsofIvy

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    The definition of "normal" is that TTT= TTT. I don't see where you have used that anywhere.

    Also, are you allowed to assume that the vector space is finite dimensional? You seem to be doing that.
     
  9. Aug 1, 2009 #8
    I'm not sure whether or not I'm allowed to assume the space is finite dimensional.
     
  10. Aug 1, 2009 #9

    Hurkyl

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    (Note that the identity and the zero map are orthogonal projections) Do you think these are all of the cases? Can you prove it?


    Yes... what if....

    (did you mean "singular" rather than "zero"?)
     
  11. Aug 1, 2009 #10

    Hurkyl

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    I thought he invoked that to claim the operator was diagonalizable. (I confess I don't know the relevant theorem)
     
    Last edited: Aug 1, 2009
  12. Aug 1, 2009 #11
    you're right on that.
     
  13. Aug 1, 2009 #12
    Why don't you try putting T^9 and T^8 into diagonal form and see what it implies?
    P.S.:Normal matrix is not just diagonalizable, also it can be diagonalized by unitary matrix.
     
  14. Aug 1, 2009 #13
    That is what I meant by reduced. Diagonal form (ie if singular, all 0's).
     
  15. Aug 1, 2009 #14

    Hurkyl

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    That's not correct....
     
  16. Aug 1, 2009 #15
    well ok but when I said reduced I meant diagonal form. Like when you reduce a singular matrix,
    you may get all 0's on the matrix.And when you reduce a non-singular matrix, you get a matrix with all 1's across the diagonal
    and 0's everywhere else.
     
    Last edited: Aug 1, 2009
  17. Aug 1, 2009 #16
    Where did you get this idea? It's not correct in most cases, but lucky you it's just the case in your problem.
     
  18. Aug 1, 2009 #17

    Dick

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    All that you know so far is that the transformation has a basis in which it's diagonal, since it's normal. Now what does T^9=T^8 tell you about the diagonal elements?
     
    Last edited: Aug 1, 2009
  19. Aug 1, 2009 #18
    That all of them are 0 along the diagonal, some but not all are 0 (rest are 1) along
    the diagonal, or every element is 1 along the diagonal.
     
  20. Aug 1, 2009 #19

    Dick

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    In other words, every diagonal element is either 0 or 1. That's ok. But how do you know that? We are trying to do a proof here, right? Not an opinion poll.
     
  21. Aug 1, 2009 #20
    Roger. Now we know that T9=T8. Now is it okay
    if I assume that "T" is reduced all the way to diagonal form? Because that
    would leave us with the three types I mentioned earlier where T has all 0's, some
    1's and some 0's, or all 1's along its diagonal, and 0's everywhere else (a characteristic
    shared by all 3 cases). Then I can split the proof in two: one case where T is 0, the other where
    T has at least one "1" on its diagonal. And life would be alot easier.
     
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