Let $a$,$b$, $c$ be members of a field. (?)
I noticed that $$f:\left(-\infty, \frac{a}{c}\right)\cup \left(\frac{a}{c}, \infty \right)\to \left(-\infty, \frac{a}{c}\right)\cup \left(\frac{a}{c}, \infty \right)$$ $$f(x) = \frac{ax+b}{cx-a}$$ is an involution with the only contingency being that it can't be the case that both $a$ and $c$ are zero.
Meanwhile, for matrices... $$A= \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$$ is an involution with the MUCH STRICTER (to me) contingency that $$a^2+bc=1.$$
My question:
Why is it that the matrix case elicits so much of a stronger restriction than the hyperbola case? Is there something about the algebra of matrices vs. the algebra of real valued functions that makes us have to be so much more specific?
Any insight, or at least direction of where to study further, would be much appreciated.
The map $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \quad\longmapsto\quad \Biggl( x \;\overset{f}{\mapsto}\; \frac{ax + b}{cx + d} \Biggr) $$ that sends a $2 \times 2$ matrix to the Möbius transformation (rational function) determined by its coefficients converts matrix multiplication into function composition. In other words, if the rational function determined by matrix $A$ is called $f_A$, then $$ f_A \circ f_B = f_{AB}. $$ This is a remarkable (and surprising!) fact that you should check yourself. It is a many-to-one correspondence, though, and the fibers (the set of all matrices that determine the same rational function) are related by a scalar: $$ f_A = f_B \quad\iff\quad A = tB \; \text{ for some scalar $t$} $$
The condition of being an involution in matrix world is $A^2 = I$, and of course the corresponding rational functions will satisfy the analogous relation: $\bigl(f_A\bigr)^2 = \operatorname{id}$, where $\operatorname{id}(x) = x$ for all $x$. But any scalar multiple of such a matrix will also determine an involutory function. In other words, all we need is for the matrix to satisfy the more relaxed $B^2 = I$, where $B = tA$ for some scalar $t$. This is enough to make the extra relation moot, since as long as $a^2 + bc \neq 0$, then after scaling by $t$, $$ (ta)^2 + (tb)(tc) = t^2(a^2 + bc), $$ can be made equal to $1$ by a judicious choice of $t$.
By the way, all of this was well-developed in the nineteenth century over the field $\mathbb{C}$ by Möbius, Riemann, and others, but a lot of the algebraic ideas apply more generally. Have a look at this article.