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Math Help - The Eigenvector of a Transpose Problem

  1. #1
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    Question The Eigenvector of a Transpose Problem

    Let S \in Mnxn(R) and D \in Mnxn(R) be n \times n matrices such that DTSD = S
    where S is non degenerate. Let \mu \neq 0 be an eigenvalue of D with eigenvector
    x.
    (i) Show that Sx is an eigenvector of DT and find its eigenvalue.
    (ii) Show that \frac{1}{\mu} is an eigenvalue of D.
    [Hint: det(A) = det(AT ) for any square matrix A.]

    This question has been giving me some real trouble and all help is welcome
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  2. #2
    GJA
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    Re: The Eigenvector of a Transpose Problem

    Hi zook,

    I'll try to get your going in the right direction.

    i) Remember that for \overline{x} to be an eigenvector of M we must have two things: (1) \overline{x}\neq \overline{0} and there is \lambda (in whatever field you're working with, presumably the real numbers) such that M\overline{x}=\lambda\overline{x}. This means that if we want to show S\overline{x} is an eigenvector of D^{T}, we must establish (1) and (2).

    To prove (1) we note that \overline{x}\neq 0 because it is an eigenvector of D and so cannot be zero (by definition of eigenvector). Now also note that S is assumed nonsingular/nondegenerate and so S\overline{x}\neq 0, because \overline{x}\neq 0.

    To prove (2) we us D^{T}SD=S. This implies D^{T}SD\overline{x}=S\overline{x}. Since \overline{x} is an eigenvector of D with corresponding eigenvalue \mu we have D^{T}S(\mu\overline{x})=S\overline{x}. Pull the constant through to get \mu D^{T}(S\overline{x})=S\overline{x}. Now divide by \mu.

    Does this help? Let me know if anything is unclear. Good luck!
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  3. #3
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    Re: The Eigenvector of a Transpose Problem

    Yes that is brilliant Thank you
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