### owen by the way

composition of linear fractional transformations
compared to two-by-two matrix multiplications.

consider
$\lbrack x \mapsto {{Ax +B}\over{Cx+D}} \rbrack \circ \lbrack x \mapsto {{ax +b}\over{cx+d}} \rbrack\,.$

in other words,
let f(x) = (Ax+B)/(Cx+D) and
let g(x) =(ax+b)/(cx+d) and
consider the function $f\circ g$ (“f\circ g”,
i.e. f-composed-with-g). recall
(or trust me on this) that
[f\circ g](x) = f(g(x)); i.e.,
functions compose right-to-left
(“first do gee to ex; then plug in
the answer and do eff *to* gee-of-ex”…
first g, then f… alas. but there it is).

so we have
$\lbrack x \mapsto {{Ax +B}\over{Cx+D}} \rbrack \circ \lbrack x \mapsto {{ax +b}\over{cx+d}} \rbrack$
$= \lbrack x \mapsto { {A{{ax +b}\over{cx+d}} + B}\over{C{{ax +b}\over{cx+d}} +D} }\rbrack$
$= \lbrack x \mapsto { {A(ax+b) + B(cx+d)}\over{C(ax+b) + D(cx+d)}}\rbrack$
$= \lbrack x\mapsto { {(Aa+Bc)x + (Ab+Bd)}\over{(Ca+Dc)x + (Cb+Dd)} }\rbrack\,.$
thus
$\lbrack x \mapsto {{Ax +B}\over{Cx+D}} \rbrack \circ \lbrack x \mapsto {{ax +b}\over{cx+d}} \rbrack =\lbrack x\mapsto { {(Aa+Bc)x + (Ab+Bd)}\over{(Ca+Dc)x + (Cb+Dd)} }\rbrack\,.$

whereas one also has
$\begin{pmatrix} A & B \\ C & D\end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} Aa +Bc & Ab + Bd \\ Ca+Dc & Cb + Dd \end{pmatrix}\,.$

so the matrix-multiplication equation
can be obtained from the function-composition equation
merely by applying an eraser here and there.

(my lecture-note-blogging of winter 09 include some
remarks on \mapsto notation and much more
about linear fractional (“mobius”) transformations.)