The Jacobian of the inversion of the Lobachevsky space is given by $$J_{\mu\nu} = \delta_{\mu\nu} - 2 \frac{z_\mu z_\nu}{z^2}, $$ where $z_\mu$ are real numbers, $z_0 > 0$ and $z^2 = \delta^{\alpha\beta} z_\alpha z_\beta$. (I'm considering $d+1$ dimensions, so $\mu$ goes from $0$ to $d$.)
Seen as a matrix, this is orthogonal, so we certainly know that its eigenvalues are $\pm 1$.
However, can we say something about the number of positive and negative eigenvalues? There should be $d$ eigenvalues $+1$ and $1$ eigenvalue $-1$.
I'm expanding on the comment by @Hyperon. The matrix you have can be written as $$ J = I - 2 n \otimes n , \qquad n_\mu = \frac{z_\mu}{\sqrt{z^2}}. $$ Let $(n_\mu,t_\mu^i)$ be basis vectors on ${\mathbb R}^{d+1}$ satisfying ($i=1,\cdots,d$) $$ n \cdot n = 1, \qquad t^i \cdot t^j = \delta^{ij} , \qquad n \cdot t^i = 0 . $$ Then, $$ J \cdot n = - n , \qquad J \cdot t^i = + t^i $$ $J$ is a $(d+1) \times (d+1)$ matrix so it has $d+1$ eigenvalues. From the above, we find that it has exactly $1$ negative eigenvalue ($-1$) and $d$ positive eigenvalues ($+1$).