1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Self-Adjoint Operators problem

  1. Apr 29, 2008 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    Thm: T is self-adjoint iff [tex]\exists[/tex] an orthonormal basis for V consisting of e-vectors of T.


    3. 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!
     
  2. jcsd
  3. Apr 29, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  4. Apr 29, 2008 #3
    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
     
  5. Apr 29, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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)*?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Self-Adjoint Operators problem
  1. Self Adjoint Operator (Replies: 9)

  2. Self-adjoint operator (Replies: 1)

Loading...