Matrix satisfying $A-I = A^{-1}$

Question

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?

Answer 1

$$\begin{align}A-I = A^{-1}\end{align}$$

$$\begin{align}A = I + A^{-1}\end{align}$$

$$\begin{align}A^2 = A + I\end{align}$$

$$\begin{align}A^3 = A^2 + A\end{align}=2A+I$$

$$\begin{align}A^4 = A^3 + A^2\end{align}=3A+2I$$

The pattern is now suggesting, $$A^n = F_n A+F_{n-1}I$$ where $F_n$ is the Fibonacci's sequence.

Answer 2

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!

Answer 3

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

$$A = \begin{bmatrix} 1/2 & \sqrt5/2 \\ \sqrt5/2 & 1/2 \end{bmatrix}$$