For $x \in \mathbb{R}$, define the fractional linear transformation of $x$ as $f(x)$ where:
$$f(x) = \frac{ax + b}{cx+d}$$
Then $f(x)$ has a matrix representation in $\mathbb{R}^2$ of $F$ where:
$$F = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
Then the function $g(x)= \dfrac{1}{1-x}$ has the matrix representation of:
$$G = \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix}$$
Observe that $g(g(g(x)))=x$. Then why isn't $G^3=I$ ? In fact:
$$G^3 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I$$
How does that make sense where $g^3(x)=x$ and not $-x$ ? Shouldn't the vector
$$ v=\begin{pmatrix} x \\ y \end{pmatrix}$$ get mapped back to itself after 3 iterations of multiplying by $G$ ? What am I missing?
Hint:
For which matrices $M = \pmatrix{a & b\\ c & d}$ is $f_M(x) = \frac{ax + b}{cx+d}$ the identity function? You know that it is when $M = I$, but is that the only matrix? No. For instance, $M = 2I$ also produces a function $f_{2I}$ that's the identify function. The mapping $M \mapsto f_M$ from matrices to fractional linear transformations is not injective.