Linear Algebra - Self-adjoint Operators

  • #1
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Homework Statement


Make P2(R) into an inner-product space by defining <p, q> = [tex]\int_0^1p(x)q(x)dx[/tex]. Define T in L(P2(R)) by T(a_0 + a_1*x + a_2*x2) = a_1*x.

(a) Show that T is not self-adjoint.
(b) The matrix of T with respect to the basis (1, x, x2) is

[tex]\left(
\begin{array}{ccc}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{array}
\right)[/tex]

This matrix equals its conjugate transpose, even though T is not self-adjoint. Explain why this is not a contradiction.

Homework Equations



The Attempt at a Solution


To show that T is not self-adjoint, I need to show that T != T*. This amounts to explicitly finding what T* is. I know that <Tv, w> = <v, T*w>.

So we have <T(a_0 + a_1*x + a_2*x2), b_0 + b_1*x + b_2*x2> = <a_1*x, b_0 + b_1*x + b_2*x2> = a_1*x(b_0 + b_1*x + b_2*x2). Now I can't quite figure out where to go from here. I know this has to equal <a_0 + a_1*x + a_2*x2, T*(b_0 + b_1*x + b_2*x2)>. How can I find T*?

For part (b), I know that a if an operator is self-adjoint, its matrix equals its conjugate transpose, but not necessarily the other way around. This is a crappy way of putting it, though. Maybe I'll have a better explanation once I find T*. Thanks for any help!
 

Answers and Replies

  • #2
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0
:rolleyes: I wasn't paying attention to the problem and was calculating the inner-product wrong. It's obviously defined to be that integral between 0 and 1, not the dot product as I was doing before.
 
  • #3
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Can I just do this:

An operator is self-adjoint iff <Tv, w> = <v, Tw> for all v, w. So using the defined inner product, we have <T(a_0 + a_1*x + a_2*x2), b_0 + b_1*x + b_2*x2> = (1/2)*a_1*b_0 + (1/3)a_1*b_1 + (1/4)*a_1*b_2.

But we also have <a_0 + a_1*x + a_2*x2, T(b_0 + b_1*x + b_2*x2)> = (1/2)*a_0*b_1 + (1/3)a_1*b_1 + (1/4)*a_2*b_1.

Clearly <Tv, w> != <v, Tw>, so T is not self-adjoint. Is this correct?
 
  • #4
Dick
Science Advisor
Homework Helper
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That's correct. You could even pick an easier example. <1,T(x)>=<1,x>!=0 and <T(1),x>=<0,x>=0. Not self adjoint.
 
  • #5
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That's correct. You could even pick an easier example. <1,T(x)>=<1,x>!=0 and <T(1),x>=<0,x>=0. Not self adjoint.
Cool. Thanks for the confirmation. Now I'm trying to figure out why I can say that even though my matrix equals its conjugate transpose, it is not a contradiction. I'm thinking it has to do with how sparse the matrix is, specifically on the diagonal, but I'm not sure. Anyone? Thanks!
 
  • #6
Hurkyl
Staff Emeritus
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I know that a if an operator is self-adjoint, its matrix equals its conjugate transpose,
Why do you think that?
 
  • #7
Dick
Science Advisor
Homework Helper
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Cool. Thanks for the confirmation. Now I'm trying to figure out why I can say that even though my matrix equals its conjugate transpose, it is not a contradiction. I'm thinking it has to do with how sparse the matrix is, specifically on the diagonal, but I'm not sure. Anyone? Thanks!
It has to do with which basis you use to express the matrix. The property of matrices being conjugate transposes is not independent of the choice of basis. This is the 'basis' of Hurkyl's skepticism.
 
  • #8
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I see now. Thanks for the clarification. :smile:
 

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