Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 2 / b / i / Solution

Expanding the characteristic polynomial givesTherefore $a=-2z$ and $b=z^2-x^2-y^2$. Both are polynomial functions of the three real coordinates, so they are continuous functions, including on the open subset $X$ of invertible matrices.

On $X$, $b=\det A\ne 0$. Its image contains positive values, for example $b(I)=1$, and negative ones, for example $b(\mathrm{diag}(1,-1))=-1$. If $X$ were connected, its continuous real image would be an interval containing both signs and hence zero. That is impossible. Thus $X$ is not connected. More precisely its determinant image is $\mathbb{R}\setminus \{0\}$: take $x=y=0$ for a positive value and $z=y=0$ for a negative one.