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?
 A: $$\begin{align}A-I = A^{-1}\end{align}$$
$$\begin{align}A = I + A^{-1}\end{align}$$
$$\begin{align}A^2 = A + I\end{align}$$
$$\begin{align}A^3 = A^2 + A\end{align}=2A+I$$
$$\begin{align}A^4 = A^3 + A^2\end{align}=3A+2I$$
The pattern is now suggesting, 
$$A^n = F_n A+F_{n-1}I$$ where $F_n$ is the Fibonacci's sequence.
A: I think it's worth pointing out we can actually prove Mohammad Riazi-Kermani's
equation 
$A^n = F_nA + F_{n - 1}I \tag 1$
by induction, for (1) yields
$A^{n + 1} = AA^n = A( F_nA + F_{n - 1}I) =$
$F_nA^2 + F_{n - 1}A = F_n(I + A) + F_{n - 1}A = (F_n  + F_{n - 1})A + F_nI = F_{n + 1}A + F_nI, \tag 2$ 
since
$F_{n + 1} = F_n + F_{n - 1}.  \tag 3$
Hmmm . . . curious!
A: The equation $\phi^2 = \phi+1$ has another solution in the Perplex Numbers. (They're like Complex Numbers, but with $j^2={^+}1$ instead of ${^-}1$.)
$$\phi = \frac{1\pm j\sqrt5}{2}$$
Perplex Numbers can be modelled by $2\times2$ matrices:
$$a+bj \cong \begin{bmatrix} a & b \\ b & a \end{bmatrix}$$
So one matrix of the type you're considering (other than the obvious $\frac{1+\sqrt5}{2} I$) is
$$A = \begin{bmatrix} 1/2 & \sqrt5/2 \\ \sqrt5/2 & 1/2 \end{bmatrix}$$
