matrix representation of ax+b group

Question

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) 2x2 matrices of the form

a b

0 1/a

a>0, b real.

It is very easy to check that the ax+b group is isomorphic to the group of all matrices of the form

a b

0 1

a>0, b 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.

Answer 1

The isomorphism is given by mapping $\begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix}$ to $\begin{pmatrix} a^2 & a b \\ 0 & 1 \end{pmatrix}$. The inverse is given by $\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \mapsto \begin{pmatrix} a^{1/2} & a^{-1/2} b \\ 0 & a^{-1/2}\end{pmatrix}.$

Answer 2

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)\ $