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Let $N \in \mathbb{N}$, and let $A$ be a $(2N+1) \times (2N+1)$ tridiagonal matrix of the following form $$A= \begin{bmatrix} \dots & \dots & \dots & \dots & \dots & \dots & \dots \\ \vdots &(\nu-2)^2 & 1& 0&0&0&\vdots\\ \vdots & 1 &(\nu-1)^2 &1&0&0&\vdots\\ \vdots & 0&1&\nu^2&1&0&\vdots\\ \vdots & 0&0&1&(\nu+1)^2&1&\vdots\\ \vdots & 0&0&0&1&(\nu+2)^2&1\\ \dots & \dots & \dots & \dots & \dots & 1 &\dots & \end{bmatrix} $$ The matrix is "centered" in the sense that $A_{ii}=A_{2N+2-i}=(N+1-i)^2$

I would like to calculate its eigenvalues, I tried to expand the determinant of $A-\lambda \; \mathrm{Id}$ with respect to the first and last column to have a recurrence relation, but it does not seem to bring an easy recurrence relation I could solve...

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    What is your recurrence relation? You could try row operations to make it triangular. – 2017-02-21
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    So the central element is zero and the diagonal is $\{\ldots, 9,4,1,0,1,4,9,\ldots\}$, right? – 2017-02-21
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    @JackD'Aurizio Yes this is correct ! – 2017-02-21
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    @MichaelBurr I haven't managed yet to get this relation, I will try to make it triangular ! thanks :) – 2017-02-21

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