We call a vector $x \in \mathbb{R}^{n+1}$ positive if $x_{n+1} > 0$.
$A \in M_{n,n}(\mathbb{R})$ is called a Lorentz matrix if $J = A J A^t$ where \begin{align} J = \begin{pmatrix} 1 & & & \\ & \ddots & & \\ & & 1 & \\ & & & -1 \end{pmatrix}. \end{align}
A Lorentz matrix $A$ is called positive if for every positive $x \in \mathbb{R}^{n+1}$ the vector $Ax$ is positive.
By $PO(n,1)$ we denote the group of all positive Lorentz matrices.
Let us consider the hyperbolic space
$\mathbb{H}^n = \{x \in \mathbb{R}^{n+1} | x_{n+1}>0, \langle x,x\rangle_{n,1} := x_1^2 + \dots + x_n^2 - x_{n+1}^2 = -1\}$.
Now we want to show that $PO(n,1)$ acts transitively on $\mathbb{H}^n$, i.e. for each $x,y \in \mathbb{H}^n$ we can find a $A \in PO(n,1)$ such that $Ax = y$.
We also got the hint that one should use an "orthonormal basis" $v_1,\dots,v_{n+1}$ of $\mathbb{R}^n+1$ wrt. $\langle\cdot,\cdot\rangle_{n,1}$ satisfying $\langle v_i,v_i\rangle_{n,1} = 1$ for $i=1,\dots,n$ and $\langle v_{n+1},v_{n+1}\rangle_{n,1}=-1$.
I see the pattern that we have $n$ times $1$ and one time $-1$ in everything but I really don't see how I can assemble everything together so the proof makes sense.
Could you please help me with this problem? Thank you very much.
Take $x,y\in \Bbb H^n$. Consider $\{v_1,\ldots,v_n\}$ and $\{w_1,\ldots,w_n\}$ orthonormal bases for $x^\perp$ and $y^\perp$, respectively, with all $v_i$ and $w_i$ positive (consider $-v_i$ and $-w_i$ instead as needed). Since $x$ and $y$ are timelike, all the $v_i$ and $w_i$ are spacelike. This ensures that the linear map $A:\Bbb R^{n+1}_1\to \Bbb R^{n+1}_1$ defined by $Av_i =w_i$ and $Ax=y$ is a Lorentz transformation. By construction, $A$ is orthochronous (positive).