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Given S, T, prove that ST and TS have the same eigenvalues.

  1. Mar 4, 2008 #1
    I need help starting/doing this proof.
    Suppose S,T are Linear Operators on a Finite Dimensional Vector Space V. Prove that ST and TS have the same eigenvalues.
    A linear operator is a linear map from a vector space to itself.
    thanks.
     
  2. jcsd
  3. Mar 4, 2008 #2

    mathwonk

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    do you know what a characteristic polynomial is?
     
  4. Mar 4, 2008 #3
    how can u prove that????
    S T must sutysfy more conditions... that u didint typed???

    wich ones??
     
  5. Mar 5, 2008 #4
    thanks but I think I have the answer. If you pick an arbitrary vector in V and define T(u) = au and S(u) = bu, where a,b are eigenvalues for u, then applying S to T(u) and T to S(u) will give the wanted results.
     
  6. Mar 5, 2008 #5
    yes and what is the answer???

    [S,T]=0

    think abou it... very usefull in QM
    ciao
     
  7. Mar 5, 2008 #6

    HallsofIvy

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    But you can't "define" that because you want to show it is true for any T and S. What if T or S don't have eigenvalues?
     
  8. Mar 5, 2008 #7

    HallsofIvy

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    And completely meaningless since you haven't bothered to define "[S, T]"!
     
  9. Mar 5, 2008 #8
    isn't it true that all linear operators on a finite dimensional vector space have eigenvalues? I don't really know what [S, T] is though.
     
  10. Mar 6, 2008 #9

    HallsofIvy

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    No, it is not. Linear operators on a vector space over the complex numbers always have eigenvalues. Linear operators on a vector space over the real numbers may not have (real) eigenvalues.

    I took Marco84 to task for not defining it [S, T]. The standard definition is [S, T]= ST- TS but I really don't see how it will help here. ST and TS always have the same eigenvalues but not the same eigenvectors!

    Suppose [itex]\lambda[/itex] is an eigenvalue for ST. That is, there is some non-zero vector v such that STv= [itex]\lambda[/itex]v. Let u= Tv. Then Su= STv= [itex]\lambda[/itex]v. What happens if you apply T to Su? That shows that [itex]\lambda[/itex] is an eigenvalue for TS as long as u is not 0. Now suppose u= 0.
     
    Last edited: Mar 8, 2008
  11. Mar 7, 2008 #10

    mathwonk

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    i seem to have constructed counter examples from the simplest non trivial 3x3 jordan forms.
     
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