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Self-Adjoint Operators problem

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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 [tex]\exists[/tex] an orthonormal basis for V consisting of e-vectors of T.


The Attempt at a Solution


Let [tex]\beta[/tex]1 be a basis for Tw and by thm can extend to a basis [tex]\beta[/tex] for V, s.t. [tex]\beta[/tex]1[tex]\subseteq[/tex][tex]\beta[/tex]. But by above thm, [tex]\beta[/tex] is ON and consists of e-vectors of T, so then [tex]\beta[/tex]1 is also ON and consists of e-vectors of T, and Tw is self-adjoint.

Does my proof make any sense?? Thanks everyone!
 

Answers and Replies

  • #2
Dick
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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?
 
  • #3
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?

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
 
  • #4
Dick
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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)*?
 

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