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

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


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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.