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Consider the polynomial recurrence

$$p_{k+1} (x) = (x - \alpha_{k+1})p_k(x) - \beta^2_{k+1}p_{k-1}(x), \quad (k=0,1,\ldots)$$

where $p_0 = 1$, $p_{-1}=0$, and $\alpha_k$ and $\beta_k$ are scalars.

Show that the roots of $p_k(x)$ are the eigenvalues of the below tridiagonal matrix

$$J_k = \begin{bmatrix} \alpha_1 & \beta_2 & & & \\ \beta_2 & \alpha_2 & \beta_3 & & & \\ & & \ddots & & \\ & & \beta_{k-1} & \alpha_{k-1} & \beta_k \\ & & & \beta_k & \alpha_k \end{bmatrix}$$

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I think the recurrence relation should be \begin{eqnarray} p_{k+1}=(\alpha_{k+1}-x)p_k(x)-\beta_{k+1}p_{k-1}(x), p_0=\beta_2, p_1(x)=\alpha_1-x \end{eqnarray}

The recurrence relation can be obtained by the cofactor expansion of $J_{k+1}-xI_{k+1}$ along the last row (or column).

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  • $\begingroup$ Actually, the OP was almost right, the only correction is that $p_0=1$, which I've now edited to fix. Your version does not work. (This is an old question & answer, but it came up in search when I was looking just for this, so I'm commenting to prevent confusion for others in the future). $\endgroup$ – Meni Rosenfeld Aug 8 '20 at 20:35

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