# Thread: how to show the eigenvalues of a jacobi matrix.

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

see attachment

2. 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 $(B-\lambda_k)u_k =0$.