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Proving that the Composition of Two Self-Adjoint Operators is Self-Adjoint

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove or give a counterexample: the product of any two self-adjoint operators on a finite-dimensional inner-product space is self-adjoint.


    2. Relevant equations
    The only two equations I've used so far are:
    [tex]\left\langle T v, w\right\rangle = \left\langle v, T^{*}w\right\rangle[/tex]
    and
    [tex]\left(T \circ S \right)^{*} = S^{*} \circ T^{*} [/tex]


    3. The attempt at a solution
    I started with the definition of adjoint, using [tex]T \circ S [/tex] :
    [tex]\left\langle \left(T \circ S \right) v, w \right\rangle = \left\langle v, \left(T \circ S \right)^{*} w \right\rangle[/tex]
    Using the second relevant equation, I got
    [tex]\left\langle \left(T \circ S \right) v, w \right\rangle = \left\langle v, \left(S \circ T \right) w \right\rangle[/tex]
    So the composition is self-adjoint only if [tex]T[/tex] and [tex]S[/tex] commute. But I don't know where to go from there.
    Will self-adjoint operators necessarily commute?
     
  2. jcsd
  3. Apr 18, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    Take R^2 with the usual basis and inner product. Then self adjoint operators are the symmetric 2x2 matrices. Can you find two of those that don't commute? It's not that hard. When you find them also notice that their product isn't symmetric.
     
  4. Apr 18, 2008 #3
    Thanks for your help!

    It was, in fact, quite easy to find two such matrices. It only took a few minutes :redface: .
     
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