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Math Help - Finding Eigenvalues of an Operator

  1. #1
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    Finding Eigenvalues of an Operator

    For a 5-dimensional vector space, I have an operator I'm working with T satisfying
    (T-1)(T^2 - 4) = 0, but T =/=1 and T^2 - 4 =/= 0.

    I'm not sure how to get started on this. I can find them for a matrix, but in this case I have no matrix, only an equation in T. Any push in the right direction would be great, thank you.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The minimal polynomial \mu (\lambda) divides to (\lambda-1)(\lambda-2)(\lambda +2) and \mu(\lambda)\neq \lambda -1,\;\mu(\lambda)\neq (\lambda -2)(\lambda +2) so, ...
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  3. #3
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    Does this imply that the minimal polynomial is (x - 2) or (x + 2) then?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by arcketer View Post
    Does this imply that the minimal polynomial is (x - 2) or (x + 2) then?

    No, if \mu (x)=x-2 i s the minimal polinomial of T then,

    T-2I=0\Rightarrow T^2-4I=(T-2I)\circ (T+2I)=0

    (contradiction) .
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  5. #5
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    Well then this must mean that the minimal polynomial has to be x + 2, as I don't see any other factors dividing the original equation, and a simple check yields

    T + 2I = 0 -> (T+2I)(T+2I) = T^2 + 4T + 4I = 0 is not a contradiction, so the only eigenvalue is -2? That feels right but any input would be great.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    The minimal polynomial \mu (x) divdes to (x-1)(x-2)(x+2).
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  7. #7
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    I don't know where you got that. Fernando Revilla told you that the minimal polynomial must divide (\lambda- 1)(\lambda- 2)(\lambda+ 1) but was NOT \lambda- 1 nor (\lambda- 2)(\lambda+ 2). Since the orginal equation has three distinct first degree factors, it is the minimal polynomial.

    (The "I don't know where you got that" was in response to arcketer's last post.)
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  8. #8
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    Oh, I see. I think I was thrown off by the wording. So having the minimal polynomial, it is obvious the corresponding eigenvalues are 1, 2, -2 each with single multiplicity, and this allows me to make some claims about the possible Jordan Canonical forms. Thank you both very much for your help
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Since the orginal equation has three distinct first degree factors, it is the minimal polynomial.

    Also \mu(\lambda)=(\lambda-1)(\lambda-2) and \mu(\lambda)=(\lambda-1)(\lambda+2) can be minimal polynomials of T . For example

    T\equiv \textrm{diag}(1,1,1,2,2) and T\equiv \textrm{diag}(1,1,1,-2,-2) .

    In both cases, T-I\neq 0 and T^2-4I\neq 0 .
    Last edited by FernandoRevilla; March 15th 2011 at 11:39 AM.
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  10. #10
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    Good point. I had misinterpreted "T satisfying (T-1)(T^2 - 4) = 0, but T =/=1 and T^2 - 4 =/= 0" as meaning that (T-1)^2(T^2- 4) was the characteristic polynomial but clearly that is not necessarily true.
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