<s>[INST] Matrix satisfying $A-I = A^{-1}$ Recall the (infinitely) continued fraction definition of the golden ratio
\begin{align}
\phi = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}}}
\end{align}
This is equivalent to the expression
\begin{align}
\phi = 1+\frac{1}{\phi}
\end{align}
Saying that the inverse of $\phi$ is equal to $\phi-1$. Inspired by this fact, consider a square matrix $A$ satisfying:
\begin{align}
A-I = A^{-1}
\end{align}
Does this relationship have any significance?
 [/INST] $$\begin{align}A-I = A^{-1}\end{align}$$
$$\begin{align}A = I + A^{-1}\end{align}$$
$$\begin{align}A^2 = A + I\end{align}$$
$$\begin{align}A^3 = A^2 + A\end{align}=2A+I$$
$$\begin{align}A^4 = A^3 + A^2\end{align}=3A+2I$$
The pattern is now suggesting, 
$$A^n = F_n A+F_{n-1}I$$ where $F_n$ is the Fibonacci's sequence.
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