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Math Help - jordan in power n

  1. #1
    MHF Contributor
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    jordan in power n

    A is similar to G=diag(J_{2}(-1),J_{2}(1))
    find the jordan form of A^{n}
    i have the formula that A^{n} similar to G^{n}=diag(J_{2}^{n}(-1),J_{2}^{n}(1))
    so J_{2}^{n}(\lambda)=\left(\begin{array}{cc}\lambda^  {n} & n\lambda^{n-1}\\<br />
0 & \lambda^{n}\end{array}\right)
    but in the book its \left(\begin{array}{cc}\lambda^{n} & 1\\0 & \lambda^{n}\end{array}\right)
    why?
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  2. #2
    MHF Contributor

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    Re: jordan in power n

    You are correct. If your text says that \begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}^n= \begin{pmatrix}\lambda^n & 1 \\ 0 & \lambda^n\end{pmatrix}, it is wrong.

    It's easy to check with n= 2:
    \begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}^2= \begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}\begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}= \begin{pmatrix}\lambda^2 & 2\lambda \\ 0 & \lambda^2\end{pmatrix}
    as you say.
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