I am trying to show that any element in $SU(1,1)$ can be expressed as, $$\begin{bmatrix} a & b\\ \overline b & \overline a \end{bmatrix}$$ from the definition. That is, if $A \in U(1,1)$ with $det(A)=1$ then $A \in SU(1,1)$.
I derived a matrix equation for matrices in $U(m,n)$, $$U(m,n)=\{A \in GL(m+n, \mathbb{R})\:;\: \overline A^\intercal I_{m,n}A=I_{m,n}\}$$
I tried to solve this equation, but I cannot do it.
This has already been partly answered here: Description of SU(1, 1) but it seems incomplete (no justification).
They wrote: You precisely get that the general form of such matrix is
$$ U = \left( \begin{array}{cc} \alpha & \beta \\ \gamma & \delta \end{array} \right) $$ where $ \vert \alpha \vert^2 - \vert \beta \vert^2 = 1$, $ \vert \delta \vert^2 - \vert \gamma \vert^2 = 1$ and $(\alpha \, \beta^\ast)^\ast = \gamma \, \delta^\ast$. This implies $\delta = \alpha^\ast$ and $\gamma = \beta^\ast$ as you surely know.
How is that last step justified?
