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Math Help - Invertable matrix P and a matrix C

  1. #1
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    Invertable matrix P and a matrix C

    I need to find and invertable matrix P
    Then a matrix C with the form:

    |a -b|
    |b a |
    this comes from complex eigenvalues, and the formula
    A= PC(P^-1)

    my problem is

    |5 -2|
    |1 3 |
    is the matrix

    for the eigenvalues i got 4 +- 2i.
    now i have to find eigenvectors and the C matrix
    Can you help?
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  2. #2
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    By subbing the eigenvalues back into A for lambda, we get:

    \begin{bmatrix}5-(4+i)&-2\\1&3-(4+i)\end{bmatrix}=\begin{bmatrix}1-i&-2\\1&-1-i\end{bmatrix}

    This leads to \begin{bmatrix}1-i&-2\\1&-1-i\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat  rix}=\begin{bmatrix}0\\0\end{bmatrix}

    This leads to eigenvector: t\begin{bmatrix}\frac{2}{1-i}\\1\end{bmatrix}

    For the other eigenvalue, the corresponding eigenvector is: t\begin{bmatrix}\frac{2}{1+i}\\1\end{bmatrix}

    Normalizing and using Gram-Schmidt

    ||u||=\sqrt{|\frac{2}{1-i}|^{2}+1^{2}}=\sqrt{2}

    p_{1}=\begin{bmatrix}\frac{2}{\sqrt{2}(1-i)}\\ \frac{1}{\sqrt{2}}\end{bmatrix}

    p_{2}=\begin{bmatrix}\frac{2}{\sqrt{2}(1+i)}\\ \frac{1}{\sqrt{2}}\end{bmatrix}

    This gives us:

    P=\begin{bmatrix}\frac{2}{\sqrt{2}(1-i)}&\frac{2}{\sqrt{2}(1+i)}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}

    P^{-1}=\begin{bmatrix}\frac{1}{\sqrt{2}i}&\frac{1}{\sq  rt{2}}+\frac{1}{\sqrt{2}}i\\ \frac{1}{\sqrt{2}}i&\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\end{bmatrix}

    Now, P^{-1}AP=\begin{bmatrix}4+i&0\\0&4-i\end{bmatrix}

    Therefore, P diagonalizes A
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