Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 65 / 2 / b / Solution

The characteristic polynomial of a real $2\times 2$ matrix is $p(z)=z^2-Tz+D$. Thus $p(1)=1-T+D$ vanishes exactly when $T=D+1$, and $p(-1)=1+T+D$ vanishes exactly when $T+D+1=0$. These are the required eigenvalue conditions.

For $D=1$ and $|T|<2$, the discriminant is negative, andTheir moduli equal one. Taking $\alpha \in (0,\pi )$ with $\cos \alpha =T/2$ gives $z_{\pm }=e^{\pm i\alpha }$. The strict trace inequalities exclude $\alpha =0,\pi$ and hence exclude the repeated real multipliers $+1,-1$.