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

  • Thread starter CrazyIvan
  • Start date
45
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1. Homework Statement
Prove or give a counterexample: the product of any two self-adjoint operators on a finite-dimensional inner-product space is self-adjoint.


2. Homework 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?
 

Answers and Replies

Dick
Science Advisor
Homework Helper
26,258
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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.
 
45
0
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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