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Eigenvalue proof. (2nd opinion if my proof is right please)

  1. Jun 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that if two linear operators A and B commute and have non-degenerate eigenvalues then the two operators have common eigenfunctions.

    2. Relevant equations

    [tex][A,B]= AB - BA= 0[/tex]

    [tex] Af=af[/tex]

    [tex] Bg=cg,\ let\ g=(f+1) --> B(f+1)=c(f+1)\ where\ a\neq c[/tex]



    3. The attempt at a solution

    [tex]Af[B(f+1)]-B(f+1)[Af]=0[/tex]
    [tex]Af(Bf+B)-(Bf+B)Af=0[/tex]

    I have stopped here because I feel that I am on the wrong track. I have not used the fact that the eigenvalues are non-degenerate in this proof. Although continuation of what I am doing should show that the two operators commute..
     
  2. jcsd
  3. Jun 19, 2009 #2

    Dick

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    You don't want to show they commute. You already know that. You want to show they have common eigenfunctions. Look at [A,B]f where Af=af. That's ABf-BAf=A(Bf)-a(Bf)=0. That tells you Bf is also an eigenfunction of A corresponding to the eigenvalue a. Just like f. Now what does 'nondegenerate' tell you?
     
  4. Jun 19, 2009 #3
    Ah that totally makes sense! Non-degenerate eigenvalues means different ie. (not the same) eigenvalues does it not?
     
  5. Jun 19, 2009 #4

    Dick

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    Sort of. It means there is only one eigenvector (up to a constant multiple) corresponding to every eigenvalue.
     
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