Linear Algebra - Self-adjoint Operators

In summary, we defined an inner-product space P2(R) by the integral <p, q> = \int_0^1p(x)q(x)dx and a linear operator T in L(P2(R)) by T(a_0 + a_1*x + a_2*x2) = a_1*x. We showed that T is not self-adjoint by demonstrating that it does not satisfy the property <Tv, w> = <v, Tw> for all v, w. This was done by finding the inner-product for both sides and showing that they do not match. We also showed that the matrix of T with respect to the basis (1, x, x2) is equal to its
  • #1
steelphantom
159
0

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!
 
Physics news on Phys.org
  • #2
: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
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
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
Dick said:
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
steelphantom said:
I know that a if an operator is self-adjoint, its matrix equals its conjugate transpose,
Why do you think that?
 
  • #7
steelphantom said:
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
I see now. Thanks for the clarification. :smile:
 

1. What is a self-adjoint operator in linear algebra?

A self-adjoint operator in linear algebra is a linear transformation on a vector space that is equal to its own adjoint. This means that the inner product of the operator with any vector in the space is equal to the inner product of the vector with the operator's adjoint. In other words, the operator is symmetric with respect to the inner product on the vector space.

2. How is a self-adjoint operator different from a Hermitian operator?

A self-adjoint operator and a Hermitian operator are essentially the same thing. The only difference is that the latter is used in the context of complex vector spaces, while the former can be used in both real and complex vector spaces. In both cases, the operator is equal to its own adjoint.

3. What is the significance of self-adjoint operators in quantum mechanics?

In quantum mechanics, self-adjoint operators are used to represent physical observables such as position, momentum, energy, and angular momentum. This is because these operators have real eigenvalues, which correspond to measurable quantities in the physical world. Furthermore, the eigenfunctions of self-adjoint operators form a complete orthonormal basis, making them particularly useful in solving equations and making predictions in quantum systems.

4. Can all operators in linear algebra be self-adjoint?

No, not all operators in linear algebra are self-adjoint. In order for an operator to be self-adjoint, it must satisfy certain conditions, such as being a linear transformation on a vector space with an inner product. Additionally, the operator must have a unique adjoint, which is not always the case for all operators.

5. How do self-adjoint operators relate to unitary operators?

Self-adjoint operators are closely related to unitary operators, as they share many similar properties. Both types of operators have real eigenvalues and their eigenfunctions form a complete orthonormal basis. However, unitary operators are also invertible and preserve the norm of vectors, while self-adjoint operators do not necessarily have these properties.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
278
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
964
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
789
  • Calculus and Beyond Homework Help
Replies
8
Views
615
  • Calculus and Beyond Homework Help
Replies
3
Views
489
  • Calculus and Beyond Homework Help
Replies
0
Views
149
Back
Top