= Fibonacci powers of the cat-map matrix
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{title2=$A^n=\begin{pmatrix}F_{2n}&F_{2n-1}\\F_{2n-1}&F_{2n-2}\end{pmatrix}$}
With <Fibonacci number>[Fibonacci] indexing $F_0=F_1=1$, powers of $A=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ satisfy $A^n=\begin{pmatrix}F_{2n}&F_{2n-1}\\F_{2n-1}&F_{2n-2}\end{pmatrix}$ for $n\ge1$. The base case is immediate. Multiplication by $A$ advances each entry by the two-index <Fibonacci number>[Fibonacci recurrence], proving the identity by induction.
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