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Math Help - Transponse and Eigenvalues

  1. #1
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    Transponse and Eigenvalues

    Let A be a Matrix with no real eigenvalues. It is possible that the following Matrix B has real eigenvalues?
    B=A-A*AT+AT

    If I multiply a Matrix with it's transponse I get a symmetric matrix, which has two real roots so A*AT has real eigenvalues. If I subtract A and add AT it is not symmetric anymore... but how can I derive if it still has no real eigenvalues?
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  2. #2
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    Re: Transponse and Eigenvalues

    Suppose that \lambda, \underline{x} is an eigenpair of A (and A^{T}),

    then if

    B=A-AA^{T}+A^{T},

    B\underline{x}=(A-AA^{T}+A^{T})\underline{x}=A\underline{x}-AA^{T}\underline{x}+A^{T}\underline{x}=\lambda \underline{x}-A\lambda\underline{x}+\lambda\underline{x}=(2 \lambda-\lambda^{2})\underline{x}.

    \therefore B will only have real eigenvalues if 2 \lambda - \lambda^{2} is real and positive.
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  3. #3
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    Re: Transponse and Eigenvalues

    Ooops ! Sorry about that, 2 \lambda - \lambda^{2} needs to be real but not necessarily positive.

    Also, it would appear that it is possible to come up with a condition for the eigenvalues of B to be real.
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  4. #4
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    Re: Transponse and Eigenvalues

    Thank you, what would happen if A has no real eigenvalue. Is it possible that A+AT has a real eigenvalue?
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  5. #5
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    Re: Transponse and Eigenvalues

    Profuse and sincere apologies for those two responses, I suspect it was those magic mushrooms.

    The idea of post multipling by \underline{x} isn't correct because even though A and A^{T} have the same eigenvalues, they will (usually) have different eigenvectors.

    The basic property that answers both of your questions is that the eigenvalues of a real symmetric matrix are real.

    So, even if the matrix A (with all real elements) has complex eigenvalues, the matrices C=A+A^{T} and D=AA^{T} will both have real eigenvalues since they are both symmetric.

    For your earlier example B=A-AA^{T}+A^{T}, this is also symmetric, (consider it as B=(A+A^{T})-AA^{T}) and will also have real eigenvalues irrespective of whether the eigenvalues of A are real or complex.
    Thanks from infernalmich
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