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Need help with an Eigenvalue

  1. Dec 29, 2004 #1
    How do you prove that [itex]det(A) = \lambda_1*\lambda_2*...*\lambda_n[/itex], where [itex]\lambda_i[/itex] is the eigenvalues of A? I'm stuck :cry:
     
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  3. Dec 29, 2004 #2

    matt grime

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    It isn't true, so you can't prove it. You should examine the question carefully.
     
    Last edited: Dec 29, 2004
  4. Dec 29, 2004 #3

    dextercioby

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    For that to happen,u must make certain assumptions on the matrix 'A'.
    The most important is that the matrix 'A' is of square form.If it is symmetrical,then:
    a)if A has real elements,then exists a nonsingular orthogonal matrix M which can bring A to diagonal form:
    [tex] \exists M\in O_{n}(R) [/tex],so that [tex]MAM^{T}=A_{diag} [/tex]
    Then it's easy to show that det A=det A_{diag}=product of eigenvalues.
    b)if A has complex elements,then exists a unitary matrix Z which can bring A to diagonal form
    [tex] \exists Z\in U_{n}(C) [/tex],so that [tex]ZAZ^{\dagger}=A_{diag} [/tex]
    Again,it's easy to show that the eigenvalues are on the diagonal and hence the det.is the product of eigenvalues.

    Daniel.
     
  5. Dec 29, 2004 #4

    dextercioby

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    Well,Matt,if u're right and i'm wrong,then i'm gonna kill my QFT teacher since he graduated both physics and maths. :mad: For a square,symmetrical matrix it has to be true.For other cases (meaning square form and nonsymetry),probably not.

    Daniel.
     
  6. Dec 29, 2004 #5

    matt grime

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    No, we're both correct, I said you should be careful, and you showed something inthe special case the matrix is diagonalizable, which is in some sense the notion I meant when I said that you should be careful. This depends upon how we dsitinguish between algebraic and geometric multiplicity.
     
  7. Dec 30, 2004 #6
    Ok,sorry here's the whole text:
    Let A be an nxn matrix, and suppose A har n real eigenvalues [itex]\lambda_1 ... \lambda_n[/itex] repeated accordingly to multiplicities, so that
    [tex] det(A - \lambda I) = (\lambda_1 - \lambda)*(\lambda_2 - \lambda)*...*(\lambda_n - \lambda)[/tex]
    Explain why det(A) is the product of the n eigenvalues of A.
    (Hint: the equation holds for all [itex]\lambda[/itex])
     
  8. Dec 30, 2004 #7

    matt grime

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    let lambda = 0
     
  9. Dec 30, 2004 #8
    Thanks! I know now that I'm stupied :)
     
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