Self-Adjoint Operators problem

  • Thread starter Thread starter genjuro911
  • Start date Start date
  • Tags Tags
    Operators
Click For Summary

Homework Help Overview

The discussion revolves around the properties of self-adjoint operators in the context of linear operators on inner product spaces. The original poster is attempting to prove that if a linear operator T is self-adjoint on an inner product space V, then its restriction Tw to a T-invariant subspace W is also self-adjoint.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to extend a basis from the subspace W to the entire space V, suggesting that if T is self-adjoint, then the basis for W can also be orthonormal and consist of eigenvectors of T.
  • Some participants question the validity of the original proof, particularly regarding the selection of an orthonormal basis for W and the implications of self-adjointness.
  • Others seek clarification on the definition of self-adjoint operators and how it applies to the restriction of T to W, raising concerns about the immediate nature of the proof.
  • There is a discussion about the relationship between the bases of the subspace and the containing space, with one participant providing a counterexample to illustrate potential misconceptions.

Discussion Status

The discussion is active, with participants exploring various interpretations of the proof and the properties of self-adjoint operators. Some guidance has been offered regarding the definitions involved, but there is no explicit consensus on the original poster's proof or its implications.

Contextual Notes

Participants are navigating assumptions about the nature of bases in subspaces versus their containing spaces, as well as the implications of self-adjointness in this context. The discussion reflects a need for clarity on these foundational concepts.

genjuro911
Messages
3
Reaction score
0

Homework Statement


T a linear operator on inner product space V and W a T-invariant subspace of V. Then if T is self-adjoint then Tw is self-adjoint.


Homework Equations


Thm: T is self-adjoint iff \exists an orthonormal basis for V consisting of e-vectors of T.


The Attempt at a Solution


Let \beta1 be a basis for Tw and by thm can extend to a basis \beta for V, s.t. \beta1\subseteq\beta. But by above thm, \beta is ON and consists of e-vectors of T, so then \beta1 is also ON and consists of e-vectors of T, and Tw is self-adjoint.

Does my proof make any sense?? Thanks everyone!
 
Physics news on Phys.org
Tw is the restriction of T to W, right? Which makes sense as an operator on W since W is invariant? If so, then the proof is immediate using the definition of 'adjoint'. As for your proof... If you start with a basis for W and extend to V, how can you know the basis you started with can be selected to have orthonormal eigenvectors if you haven't proved Tw is self-adjoint yet?
 
Dick said:
Tw is the restriction of T to W, right? Which makes sense as an operator on W since W is invariant? If so, then the proof is immediate using the definition of 'adjoint'.
Would it be possible to elaborate a bit more? <T(v),w> = <v,T*(w)> for v,w in W is the def from what I remember but how is that immediate?

Dick said:
As for your proof... If you start with a basis for W and extend to V, how can you know the basis you started with can be selected to have orthonormal eigenvectors if you haven't proved Tw is self-adjoint yet?
I was kind of thinking the same thing but don't the basis vectors for Tw have to come from the basis for T?

Thanks
 
genjuro911 said:
Would it be possible to elaborate a bit more? <T(v),w> = <v,T*(w)> for v,w in W is the def from what I remember but how is that immediate?


I was kind of thinking the same thing but don't the basis vectors for Tw have to come from the basis for T?

Thanks

Nooo. Take V=R^2. Take v=(1/sqrt(2),1/sqrt(2)). Take W to be the subspace t*v for real t. One orthonormal basis for V is e1=(0,1) and e2=(1,0). v isn't in that basis. A basis of a subspace isn't automatically a part of the basis of the containing space. You have to arrange it to be so. You are having a hard time seeing the obvious solution because it's, uh, obvious. That does make things hard to see.

T is self adjoint in V. So T=T* in V. Let Tw be the restriction of T to W. Which makes sense because T(W) is contained in W. So Tw:W->W. But W is contained in V. So if <v,T(w)>=<T(v),w> (T is self adjoint) for v,w in V, then <v,T(w)>=<T(v),w> for v,w in W. What does this tell you about the relation between Tw and (Tw)*?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K