Fibonacci powers of the cat-map matrix ($A^n=\left(\begin{array}{cc}{F}_{2n}& F_{2n-1}\\ F_{2n-1}& F_{2n-2}\end{array}\right)$)

With Fibonacci indexing $F_0=F_1=1$, powers of $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ satisfy $A^n=\left(\begin{array}{cc}{F}_{2n}& F_{2n-1}\\ F_{2n-1}& F_{2n-2}\end{array}\right)$ for $n\ge 1$. The base case is immediate. Multiplication by $A$ advances each entry by the two-index Fibonacci recurrence, proving the identity by induction.