= Solution
Expanding the <characteristic polynomial> gives
$$
\det(A-\lambda I)=(z-\lambda)^2-x^2-y^2=\lambda^2-2z\lambda+z^2-x^2-y^2.
$$
Therefore \b[$a=-2z$ 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 $X$, $b=\det A\ne0$. Its image contains positive values, for example $b(I)=1$, and negative ones, for example $b(\operatorname{diag}(1,-1))=-1$. If $X$ were connected, its continuous real image would be an interval containing both signs and hence zero. That is impossible. Thus \b[$X$ 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.
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