# How to calcute a determinant of this matrix? [closed]

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

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.

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

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• 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. – Minus One-Twelfth Mar 22 at 7:10
• @Andrey Komisarov I don't understand your Fibonacci number comment. – Max Mar 22 at 7:27
• @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)... – Andrey Komisarov Mar 22 at 7:30
• I think you mean "summand". Edit: Google says "addend" is valid (though I still think "summand" is more common). – Max Mar 22 at 7:35
• @Max Sorry, I mixed. – Andrey Komisarov Mar 22 at 7:37

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 $$n-2k$$.