TITLE: Matrix satisfying $A-I = A^{-1}$
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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
$$A = \begin{bmatrix} 1/2 & \sqrt5/2 \\ \sqrt5/2 & 1/2 \end{bmatrix}$$