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\begin{vmatrix} -t & 1 & 0 & 0& \ldots & 0 & 0 &0\\ n & -t & 2 & 0& \ldots & 0 & 0 &0\\ 0 & n-1& -t & 3&\ldots & 0 & 0 &0\\ 0 & 0 & n-2& -t& \ldots & 0& 0 &0\\ \vdots&\vdots&\vdots&\vdots& \ddots&\vdots&\vdots&\vdots\\ 0 & 0 & 0 & 0&\ldots & -t& n-1& 0 \\ 0 & 0 & 0 & 0&\ldots & 2& -t & n \\ 0 & 0 & 0 & 0&\ldots & 0&1 & -t \end{vmatrix}

It is needed to find an eigenvalue. I just know that for $n=m$ there are Fibonacci number $F_{m+2}$ summands in the determinant formula.

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closed as off-topic by Eevee Trainer, Cesareo, GNUSupporter 8964民主女神 地下教會, Rhys Steele, Alex Provost Mar 23 at 1:12

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  • $\begingroup$ You appear to have a tridiagonal matrix. One thing you can do is see the determinant section of that page for a recursive formula for the determinant. $\endgroup$ – Minus One-Twelfth Mar 22 at 7:10
  • $\begingroup$ @Andrey Komisarov I don't understand your Fibonacci number comment. $\endgroup$ – Max Mar 22 at 7:27
  • $\begingroup$ @Max for n=1 there are t^2-1 (2 addend), for n=2 there are (3 addend), for n=3 there are (5 addend)... $\endgroup$ – Andrey Komisarov Mar 22 at 7:30
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    $\begingroup$ I think you mean "summand". Edit: Google says "addend" is valid (though I still think "summand" is more common). $\endgroup$ – Max Mar 22 at 7:35
  • $\begingroup$ @Max Sorry, I mixed. $\endgroup$ – Andrey Komisarov Mar 22 at 7:37
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Your matrix is $-tId$ plus the matrix that appears in Proposition 2 on page 16 of https://arxiv.org/pdf/1210.8062.pdf

The eigenvalues are thus $n-t, n-2-t, \ldots, 2-n-t, -n-t$.

Same solution in more concrete language: Consider the operator $w\frac{\partial}{\partial z}+z\frac{\partial}{\partial w}$ acting on the space of homogeneous degree $n$ polynomials in $z$ and $w$. A natural basis for that space is $z^n, z^{n-1}w, \ldots, w^n$; in that basis the matrix of this operator is like the one in the problem with $t=0$. However, in the basis $(z+w)^{n-k} (z-w)^{k}$ the matrix is diagonal with eigenvalues $n-2k$.

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