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Math Help - Eigen values ....again :(

  1. #1
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    Eigen values ....again :(

    Hi,
    Here is the question that I am stuck on
    B = \left[\begin{array}{cc}1&1\\1&2\end{array}\right]
    Construct the matrix X that reduces B to diagonal form through the transformation
    D = X^{-1}BX. Confirm that D is diagonal, that the diagonal elements of D are the
    eigenvalues of B and that det(B) = det(D).

    So
    det(B - \lambda I) = \left[\begin{array}{cc}1 - \lambda &1\\1&2 - \lambda \end{array}\right]

    I then get to the point where \lambda^{2} - 3\lambda + 1 and therefore my eigen values are \lambda = \frac{3 \pm \sqrt{5}}{2} and my eigen vectors as \left[\begin{array}{c}-\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right] and \left[\begin{array}{c}\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right].

    The matrix X is made from the eigen values of B. So det(X) =\left[\begin{array}{cc}-\frac{(1 + \sqrt{5})}{2}&\frac{(1 + \sqrt{5})}{2}\\1&1\end{array}\right]. I get as being  -1 -\sqrt{5}

    Now I have to find the cofactor of X and then transpose of the cofactor in order to eventually get the inverse of X.

     X_{cof} = \left[\begin{array}{cc}1&-1\\ \frac{(-1 -\sqrt{5})}{2}&\frac{-1 -\sqrt{5})}{2} <br />
\end{array}\right]

     X_{cof}T = \left[\begin{array}{cc}1&\frac{-1 -\sqrt{5})}{2}\\ -1&\frac{-1 -\sqrt{5})}{2} <br />
\end{array}\right]

     X^{-1} = \frac{1}{det(X)} \cdot X_{cof}T which I get as \left[\begin{array}{cc}\frac{1 -\sqrt{5}}{4}&\frac{1}{2}\\ \frac{\sqrt{5} - 1}{4}&\frac{1}{2}\end{array}\right]

    I now have all the components required to find D.

    So when I multiply them all together i get \left[\begin{array}{cc}0&\frac{\sqrt{5} - 1}{2}+1\\ 0&1 +\sqrt{5}\end{array}\right]

    Which is evidently wrong. Can anyone see roughly where i could have gone wrong and then i will post a more detailed process of what i was doing at that stage.

    Thanks for reading and any help that come with it
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Beard View Post
    ... my eigen values are \lambda = \frac{3 \pm \sqrt{5}}{2} and my eigen vectors as \left[\begin{array}{c}-\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right] and \left[\begin{array}{c}\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right].
    I agree with the eigenvalues and the first eigenvector. But I think you'll find that the second eigenvector is \begin{bmatrix}\frac{-1+\sqrt5}2\\1\end{bmatrix}, not \begin{bmatrix}\frac{1+\sqrt5}2\\1\end{bmatrix}.
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