The Determinant of a Tridiagonal, Periodically Continued Matrix

Abstract

Let A be a t × t matrix with the structure

$$ A = \left( \begin{gathered} a_{11} a_{12} 0 0 \ldots 0 a_{1t} \hfill \\ a_{21} a_{22} a_{23} 0 \ldots 0 0 \hfill \\ 0 \ddots \ddots \ddots 0 \hfill \\ \vdots \vdots \hfill \\ a_{t1} 0 \ldots \ldots \ldots a_{t,t - 1} a_{tt} \hfill \\ \end{gathered} \right) $$

((B.1))

Then det A can be expressed in terms of traces of 2 × 2 matrices formed from the original matrix elements according to [163]

$$ detA = tr\prod\limits_{j = t}^1 {\left( \begin{gathered} a_{jj} - a_{j,j - 1} a_{j - 1,j} \hfill \\ 1 0 \hfill \\ \end{gathered} \right)} + ( - 1)^{t + 1} tr\prod\limits_{j = t}^1 {\left( \begin{gathered} a_{j,j - 1} 0 \hfill \\ 0 a_{j - 1,j} \hfill \\ \end{gathered} \right)} . $$

((B.2))

The proof of the formula is quite analogous to the solution of a Schrödinger equation for a one-dimensional tight-binding Hamiltonian with nearest-neighbor hopping by using transfer matrices [163]. The inverse order of the initial and final indices on the product symbol indicates that the matrix with the highest index j is on the left of the product. The formula should only be applied for t ≥ 3.