$$\text{Let } T_n=\begin{Bmatrix} a & b & \boldsymbol{0} \\ b & a & \ddots \\ \boldsymbol{0} & \ddots & \ddots \end{Bmatrix}\text{ a symmetric tridiagonal Toeplitz matrix of size }n\text{.}$$

I need to find its determinant in terms of $n$.
We know $\det T_1=a$ and $\det T_2=a^2-b^2$.
We can prove that, $\forall n\in\mathbb{N}^*$, $$\det T_{n+1}=a\det T_n-b^2\det T_{n-1}$$

But I would like to get an absolute formula (with no recurrence relation).

  • 1
    $\begingroup$ There is a well-known method for solving linear homogeneous recurrence equations with constant coefficients. It works very well for second-order recurrences. The first reference that I came up with on Google is math.berkeley.edu/~arash/55/8_2.pdf but the method is described in many places. $\endgroup$
    – Ira Gessel
    Jun 2, 2021 at 21:06

1 Answer 1


$$T_n=\frac{1}{2^{n+1}} \left(c_-^n+c_+^n+\left(c_+^n-c_-^n\right)\frac{c_++c_- }{c_+-c_-}\right),\;\;\text{with}\;\; c_\pm=a\pm\sqrt{a^2-4 b^2}.$$


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