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

In summary, the question asks if the product of any two self-adjoint operators on a finite-dimensional inner-product space is self-adjoint. The attempt at a solution involved using the definition of adjoint and relevant equations to show that the composition is self-adjoint only if the operators commute. A counterexample was then given by considering two symmetric 2x2 matrices in R^2 with the usual basis and inner product. Therefore, it is not necessarily true that self-adjoint operators commute.
  • #1
CrazyIvan
45
0

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.

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]

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?
 
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  • #2
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.
 
  • #3
Thanks for your help!

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

1. What does it mean for an operator to be self-adjoint?

An operator is self-adjoint if it is equal to its own adjoint. In other words, the operator is symmetric with respect to the inner product of the vector space it acts on.

2. Why is it important to prove that the composition of two self-adjoint operators is self-adjoint?

This is important because self-adjoint operators have many useful and desirable properties, such as real eigenvalues and orthogonal eigenvectors. Proving that the composition of two self-adjoint operators is self-adjoint allows us to extend these properties to the composed operator.

3. What is the process for proving that the composition of two self-adjoint operators is self-adjoint?

The process involves using the definition of self-adjointness and the properties of the operators, such as linearity and commutativity. It also involves using the properties of the inner product, such as conjugate symmetry and linearity. By manipulating the expressions and using these properties, we can show that the composed operator is equal to its own adjoint.

4. Can the composition of two non-self-adjoint operators also be self-adjoint?

No, the composition of two non-self-adjoint operators cannot be self-adjoint. This is because the properties of self-adjointness, such as real eigenvalues and orthogonal eigenvectors, only apply to self-adjoint operators.

5. Are there any real-world applications of proving the composition of two self-adjoint operators is self-adjoint?

Yes, there are many applications in various fields of science and engineering. For example, in quantum mechanics, self-adjoint operators represent physical observables, and proving the composition of two self-adjoint operators is self-adjoint allows us to predict the behavior of a physical system. In signal processing and image analysis, self-adjoint operators are used for noise reduction and data compression.

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