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Self-adjoint problem

  1. May 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Let V be a complex inner product space, and let T be a linear operator on V.

    T_1 = 1/2 (T + T*) and T_2 = (1/2i)(T – T*)

    a) Prove that T_1 and T_2 are self-adjoint and that T = T_1 + T_2
    b) Suppose also that T = U_1 + U_2, where U_1 and U_2 are self-adjoint. Prove that U_1 = T_1 and U_2 = U_2.
    c) Prove that T is normal iff (T_1)(T_2) = (T_2)(T_1)

    2. Relevant equations
    Self-adjoint: T = T*
    Normal: TT* = T*T

    3. The attempt at a solution

    (a) was very easy. However, I cannot get started on (b) at all.
    For (c)
    One direction: assume (T_1)(T_2) = (T_2)(T_1), show T is normal


    (1/2 (T + T*))((1/2i)(T – T*)) = ((1/2i)(T – T*))( 1/2 (T + T*))

    (1/4i)(T^2 – T* ^2) = (1/4i)(T^2 – T*^2)

    I’m not sure what to do now, or if it’s even the right direction.

    Thanks for your help!
  2. jcsd
  3. May 7, 2007 #2


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    Homework Helper

    For b, you must mean

    "Suppose also that T = U_1 + i U_2, where U_1 and U_2 are self-adjoint. Prove that U_1 = T_1 and U_2 = T_2."

    This shouldn't be too hard.
  4. May 8, 2007 #3


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    Staff Emeritus
    Science Advisor

    Are you sure of this? It doesn't look to me like T_2 is self adjoint and it is easy to see that T_1+ T_2 is NOT T! Did you mean T= T_1+ i T_2? This is a lot like breaking ex into cosine and sine but, again, T_2 does not satisfy <T_2 u, v>= <u, T_2 v>. It satisfies <T_2u, v>= -<u, T_2 v>.

    As StatusX said, this should be U_2= T_2.

  5. May 8, 2007 #4
    Sorry about that,

    It should be

    T = T_1 + iT_2

    T_2 is self adoint: T_2* = (-1/2i)(T*-T) = (1/2i)(T-T*) = T_2
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