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Thread: stuck on diagonalizabilty problem

  1. #1
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    stuck on diagonalizabilty problem

    is the matrix diagonalizable, if so find an inverted matrix $\displaystyle Q$ and matrix $\displaystyle D$ such that $\displaystyle Q^{-1}AQ = D $

    $\displaystyle A= \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right)$

    so far I have:

    $\displaystyle \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) - x
    \left(\begin{array}{cc} 1 & 0 \\ \\0 & 1 \end{array}\right) = \left(\begin{array}{cc} 1-x & 3 \\ \\3 & 1-x \end{array}\right)$

    $\displaystyle det \left(\begin{array}{cc} 1-x & 3 \\ \\3 & 1-x \end{array}\right) = (1-x)^2-(3)(3) = 1-2x+x^2 -9 = x^2-2x- 8 = (x+2)(x-4)$

    the eigenvalues are $\displaystyle \lambda_1 = -2 , \ \ \lambda_2 = 4$ with algebraic multiplicity 1 for both.

    $\displaystyle \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) -(-2) \left(\begin{array}{cc} 1 & 0 \\ \\0 & 1 \end{array}\right) = \left(\begin{array}{cc} 3 & 3 \\ \\3 & 3 \end{array}\right) = \left(\begin{array}{cc} 1 & 1 \\ \\0 & 0 \end{array}\right)$

    at which point I get stuck.
    Last edited by lllll; Mar 30th 2008 at 06:26 PM.
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  2. #2
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    $\displaystyle \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) -(-2) \left(\begin{array}{cc} 1 & 0 \\ \\0 & 1 \end{array}\right) = \left(\begin{array}{cc} 3 & 3 \\ \\3 & 3 \end{array}\right) = \left(\begin{array}{cc} 1 & 1 \\ \\0 & 0 \end{array}\right)$

    $\displaystyle \left(\begin{array}{cc} 1 \\ \\ -1 \end{array}\right) = \gamma_1$

    $\displaystyle \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) -(4) \left(\begin{array}{cc} 1 & 0 \\ \\0 & 1 \end{array}\right) = \left(\begin{array}{cc} -3 & 3 \\ \\3 & -3 \end{array}\right) = \left(\begin{array}{cc} 1 & 1 \\ \\0 & 0 \end{array}\right)$

    $\displaystyle \left(\begin{array}{cc} 1 \\ \\ 1 \end{array}\right) = \gamma_2$

    $\displaystyle Q= \gamma_1 \bigcup \gamma_2 = \left(\begin{array}{cc} 1 \\ \\ -1 \end{array}\right) \bigcup \left(\begin{array}{cc} 1 \\ \\ 1 \end{array}\right) = \left(\begin{array}{cc} 1 & 1 \\ \\ -1 & 1 \end{array}\right)$

    $\displaystyle Q^{-1} = \left(\begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ -1 & 1 & 0 & 1 \end{array}\right) = \left(\begin{array}{cc|cc} 1 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} \end{array}\right)$

    $\displaystyle D = \left(\begin{array}{cc} -\frac{1}{2} & 0 \\ \frac{1}{2} & \frac{1}{2} \end{array}\right) \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 1 \\ \\ -1 & 1 \end{array}\right)$

    is this right?
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  3. #3
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    Quote Originally Posted by lllll View Post
    $\displaystyle Q^{-1} = \left(\begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ -1 & 1 & 0 & 1 \end{array}\right) = \left(\begin{array}{cc|cc} 1 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} \end{array}\right)$

    $\displaystyle D = \left(\begin{array}{cc} -\frac{1}{2} & 0 \\ \frac{1}{2} & \frac{1}{2} \end{array}\right) \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 1 \\ \\ -1 & 1 \end{array}\right)$

    is this right?
    $\displaystyle Q^{-1}$ is wrong. It should be $\displaystyle \begin{pmatrix}\frac12&-\frac12\\\frac12&\frac12\end{pmatrix}$.

    You can check whether your answer is right by multiplying out the matrix product for D, which ought to be a diagonal matrix with the eigenvalues as its diagonal entries.
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