1. Limited time only! Sign up for a free 30min personal 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!

Normal operators

  1. Jul 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that a normal operator on a complex inner-product space
    is self-adjoint if and only if all its eigenvalues are real.


    2. Relevant equations



    3. The attempt at a solution
    Let c be an eigenvalue. Now since T=T*, we have
    <TT*v, v>=<v, TT*v> if and only if TT*v=cv on both sides and not -cv (-c is the complex conjugate of c made possible
    by c being a complex number) on one side and cv on the other side. Therefore c must be real.
     
    Last edited: Jul 30, 2009
  2. jcsd
  3. Jul 30, 2009 #2
    For the other direction, suppose Tv=av with a being real. Now
    <Tv, v>=<av, v>=a<v, v>=<v, av>. Now since Tv=av, <v, av>=<v, Tv>. This shows that <Tv, v>=<v, Tv> or Tv=T*v=av.
    Now <T*Tv, v>=<T*av, v>=<T*va, v>=<a2v, v>=<v, a2v>
    =<v, T*Tv>. Therefore, T*T is self adjoint.
     
  4. Jul 30, 2009 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are going to have to do a lot better than that before I'll even start reading it. Who cares whether T*T is self-adjoint? That has nothing to do with the problem. Try and make a linear progression between what you are assuming and what you are trying to prove. Just this once, for my sake, ok? Start with Tv=cv. Assume T=T*. Show me c is real. And nothing else. Write the complex conjugate of c as (c*), not -c. And do it using a minimal number of digressions. Please, please?
     
  5. Jul 30, 2009 #4
    Ok Tv=cv. T=T*. Therefore <Tv, v>=<v, T*v>. Now <cv, v>=<v, T*v> if and only if T*v=cv.
    Therefore c must be real. If c were to be complex, then T*v=c*v and c*v=/=cv.
     
    Last edited: Jul 30, 2009
  6. Jul 30, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I appreciate the attempt at brevity, thanks. And I do appreciate it, really. But you are still somehow missing the point. Try using <Tv,v>=<v,(T*)v>=<v,Tv> and that <v,cv>=c<v,v> and <cv,v>=(c*)<v,v>. You know that <ax,x>=(a*)<x,x>, right? If c=(c*) what does that mean about c?
     
  7. Jul 30, 2009 #6
    c* is not a complex conjugate. c* and c are the same. I got mixed up thinking that the little star above the scalar
    was used to show that it is a complex conjugate.
     
  8. Jul 30, 2009 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You have a really odd way of expressing yourself. I know what you mean by that, but probably few other people in the world would. Here's what you meant to say: "if c=(c*) then the imaginary part of c is zero, so c is real". The * does mean complex conjugate. What do YOU think complex conjugate means?
     
    Last edited: Jul 30, 2009
  9. Jul 30, 2009 #8
    :rofl: I know.
    For the other direction, suppose Tv=av with a being real. Now
    <Tv, v>=a<v, v>=allvll2. And <v, T*v>=<v, v>a*=a*llvll2
    Since a*=a, allvll2=a*llvll2 Thus a*llvll2=a*<v, v>
    =<a*v, v>=<av, v>=<Tv, v>.
     
    Last edited: Jul 30, 2009
  10. Jul 30, 2009 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's not that funny. What is that supposed to prove? I'm regretting, as I have before, even responding to this thread.
     
  11. Jul 30, 2009 #10
    its supposed to prove the other direction that if all eigenvalues
    are real, then T=T*.
    For the other direction, suppose Tv=av with a being real. Now
    <Tv, v>=<av, v>=a<v, v>. And <v, T*v>=<v, a*v>=a*<v, v>.
    Since a*=a, a<v, v>=a*<v, v>. Thus <v, T*v>=<v, a*v>=a*<v, v> =<a*v, v>=<av, v>=<Tv, v>.
    is this right?
     
    Last edited: Jul 31, 2009
  12. Jul 31, 2009 #11
    hey could someone tell me whether or not im right or wrong>
     
  13. Jul 31, 2009 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's wrong (in that it abuses '*' - for a general complex number c, <cv,v>=(c*)<v,v>). Since you are assuming a=(a*) you can, I suppose, move the '*'s around. The big problem is that it's just factoring 'a' in and out of a product over and over again. How is that supposed to prove T=T*? Your 'thus' concludes <v,T*v>=<Tv,v>. That doesn't prove T=T*. How can you not see that? What might a proof that T=T* look like???
     
  14. Jul 31, 2009 #13
    ugh god your right, meant <v, Tv>=<Tv, v>. ok ill fix this. cant believe I missed
    that. :yuck:
     
  15. Jul 31, 2009 #14
    Ok assume the eigenvalue a is real. And let Tv=av for v is an eigenvector of T.
    Now we have <Tv, v>=<v, T*v>. Or <av, v>=<v, a*v>. Since a is real, a=a*.
    Thus <v, a*v>=<v, av>=<v, Tv>. And since <Tv, v>=<v, T*v>, <Tv, v>=<v, Tv>.
    no factoring a or a* .
     
    Last edited: Jul 31, 2009
  16. Jul 31, 2009 #15

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's still vacuous. Wouldn't a proof that T=T* involve showing T(x)=T*(x) for any vector x? I can't even see where you clearly stated that T(v)=T*(v) for your eigenvector v. And even if you had, how could T(v)=T*(v) for a single vector v tell you T=T* for all vectors?
     
  17. Jul 31, 2009 #16
    well ok Let v be any vector. Tv=av and T*v=a*v. Thus Tv-T*v=av-a*v
    Or (T-T*)v=av-a*v. Since a=a*, (T-T*)v=av-av=0. Therefore (T-T*)v=0
    and since (T-T*)v=0 , T must=T*.
     
  18. Jul 31, 2009 #17

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's better. So Tv=(T*)v. The problem now is that not ALL vectors are eigenvectors.
     
  19. Jul 31, 2009 #18
    Ok let x be any vector in V. Now <Tx, x>=<x, T*x>. We represent T with a
    a matrix. Since T "has" eigenvectors, T must be diagonizable. Therefore (after reducing the matrix to a diagonal one, note that TT represents T*) TT=T since all eigenvalues along T's diagonal are real and all real diagonal matrices
    equal their transposes.
     
  20. Jul 31, 2009 #19

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok. Except T isn't diagonalizable because it "has" eigenvectors. T is diagonalizable because it's normal. T* isn't the transpose of T, it's the complex conjugate of the transpose. So for any diagonal element of T, T_kk, T_kk=(T_kk)*. That tells you T_kk is real.
     
  21. Jul 31, 2009 #20
    oh my bad...we're dealing with a complex space. So T's diagonal matrix has scalars of the form a+bi along its diagonal and T*'s diagonal matrix has scalars of
    the form a-bi (since T* is the complex conjugate of the transpose) along its diagonal. But in our case, b=0. So T's diagonal matrix=T*'s.
     
    Last edited: Jul 31, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Normal operators
  1. Normal Operator (Replies: 3)

Loading...