I am trying to derive equation (7) from this physics paper (sorry for the paywall)
$M_n$ for $n\geq 0$ is a sequence of $2\times 2$ real matrices with determinant $1$. These matrices obey the following recurrence relation for $ i\geq 1$:
$$M_{i+1}=M_{i-1}M_i$$
From this we may obtain
$$M_{i+1}+M_{i-2}^{-1}=M_{i-1}M_i+M_{i-1}M_i^{-1}$$
Let $x_n$ be the half-trace $\frac{1}{2} \mathrm{Tr}(M_n)$. Apparently by taking the trace of the equation above we get the recurrence relation on the half-traces:
$$x_{i+1}=2x_ix_{i-1}-x_{i-2}$$
But I can't figure out how to do the derivation
Best Answer
Note that if \begin{align} A = \begin{pmatrix} a & b\\ c & d \end{pmatrix} \end{align} with $\det A = 1$, then \begin{align} A^{-1}= \begin{pmatrix} d & -b\\ -c & a \end{pmatrix}. \end{align} So the trace are the same.
Edit: Note, by direct calculation, we see that \begin{align} BA + BA^{-1} = B(A+A^{-1}) \end{align} where \begin{align} A+A^{-1} = \operatorname{tr}(A) I. \end{align}