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Math Help - An Exercise

  1. #1
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    Question An Exercise

    Consider An Exercise-1.png
    If An Exercise-2.png compute An Exercise-3.png
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  2. #2
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    Quote Originally Posted by sshh View Post
    Consider Click image for larger version. 

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    If Click image for larger version. 

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ID:	20440 compute Click image for larger version. 

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    Do you just want the transpose of the matrix?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Using well known properties of transpose matrices, it is easy to prove p(A^T)=(p(A))^T


    Fernando Revilla
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  4. #4
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    This is simply "arithmetic".

    With A= \begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix},
    A^*= \begin{bmatrix}1 & -1 \\ 3 & 0\end{bmatrix} so that

    \rho(A^*)= \begin{bmatrix}1 & -1 \\ 3 & 0\end{bmatrix}^3- 5\begin{bmatrix}1 & -1 \\ 3 & 0\end{bmatrix}^2+ 11\begin{bmatrix}1 & -1 \\ 3 & 0\end{bmatrix}+ \begin{bmatrix} 4 & 0 \\ 0 & 4\end{bmatrix}
    Have you done that? With a 2 by 2 matrix, you should find the calculation fairly easy.

    Or, using fernandorevilla's suggestion, calculate
    \rho(A)= \begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^3- 5\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^2+ 11\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}+ \begin{bmatrix} 4 & 0 \\ 0 & 4\end{bmatrix}
    and then take the transpose.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Or, using fernandorevilla's suggestion, calculate
    \rho(A)= \begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^3- 5\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^2+ 11\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}+ \begin{bmatrix} 4 & 0 \\ 0 & 4\end{bmatrix}
    and then take the transpose.
    We needn't calculate anything. By hypothesis:

    p(A)=\begin{bmatrix}{\;\;\;1}&{3}\\{-1}&{0}\end{bmatrix}

    so,

    p(A^T)=(p(A))^T=\begin{bmatrix}{1}&{-1}\\{3}&{\;\;\;0}\end{bmatrix}

    That is all.

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