TITLE: matrix representation of $ax+b$ group
QUESTION [5 upvotes]: In "A Short Course on Spectral Theory", page 10, William Arveson asserts that the "$ax+b$ group", ie. the group generated by all dilations and translations of the real line, is isomorpic to the group of all (real) $2\times 2$ matrices of the form 
$$ \begin{bmatrix} a & b\\ 0 & \frac{1}{a}\end{bmatrix}, \,\,\,\,\,\, a>0 \mbox{ and }, b \mbox{ real}$$
It is very easy to check that the $ax+b$ group is isomorphic to the group of all matrices of the form 
$$ \begin{bmatrix} a & b\\ 0 & 1\end{bmatrix}, \,\,\,\,\,\, a>0 \mbox{ and }, b \mbox{ real}$$
So these two matrix groups should be isomorphic. Is this correct? Can someone give me the isomorphism? I've tried for a while and can't seem to get it.

REPLY [2 votes]: The isomorphism arises simply from factoring $\rm\ \ aa\ X + b\ \  =\ \ a\ (ax+b/a)\:,\ $ namely
$\rm\ \ \ (aa\ X + b)\ \: \circ\: \ (AA\ X+B)\quad =\quad a\ a\ (A\ A\ \ X\ \ +\ \ B)\ \ +\ \ b\quad\ \ \  =\quad\ \ \  aa\ \ AA\ X\ +\ aaB + b$
$\rm\displaystyle\ a\bigg(aX+\frac{b}a\bigg)\circ \:A\:\bigg(AX+\frac{B}A\bigg)\ =\ aA\:\bigg(aX+\frac{b}{aA}\bigg)\circ\bigg(AX+\frac{B}A\bigg)\ =\ aA\bigg(aA\ X\ + \frac{aaB + b}{aA}\bigg)\ $