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Math Help - how to show the eigenvalues of a jacobi matrix.

  1. #1
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    how to show the eigenvalues of a jacobi matrix.

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  2. #2
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    To find the eigenvalues, you need to evaluate \Delta_n := \det(B-\lambda I). Expand along the top row to see that \Delta_n = (a-\lambda)\Delta_{n-1} - bc\Delta_{n-2}. Thus \Delta_n satisfies the difference equation \Delta_n - (a-\lambda)\Delta_{n-1} + bc\Delta_{n-2} = 0, together with the initial conditions \Delta_1 = a-\lambda and \Delta_2 = (a-\lambda)^2 - bc.

    The auxiliary equation x^2 - (a-\lambda)x + bc = 0 has solutions x = \alpha\pm2ri\sqrt{1-(\alpha/r)^2}, where \alpha=\tfrac12(a-\lambda) and  r = \sqrt{bc}. Standard techniques for solving difference equations (as described here, for example) show that \Delta_n = \frac{r^n}{\sqrt{1-(\alpha/r)^2}}\sin\bigl((n+1)\theta\bigr), where \cos\theta = \alpha/r.

    The eigenvalues are given by \Delta_n=0. So they are the solutions of \sin\bigl((n+1)\theta\bigr) = 0, or (n+1)\theta = k\pi (1\leqslant k\leqslant n). Since \cos\theta = \frac\alpha r = \frac{a-\lambda}{2r}, it follows that \lambda = a-2r\cos\theta = a - 2r\cos\bigl(\tfrac{k\pi}{n+1}\bigr). (You can then use the fact that -\cos\phi = \cos(\pi-\phi) to write the eigenvalues as \lambda_k = a + 2r\cos\bigl(\tfrac{k\pi}{n+1}\bigr).)

    Once you know the eigenvalues, it should be straightforward to get the eigenvectors, by solving the equations (B-\lambda_k)u_k =0.
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