$\operatorname{Tr}(A^2) \geq-2$, if $A\in SL(2,R)$

Question

I saw as a hint to an exercise that if $A\in SL(2,R)$ then $$\operatorname{Tr}(A^2)\geq-2.$$

I did the exercise with this hint, but I can't prove why this is true. Also, is there a similar inequality to $A\in SL(3,R)$?

Answer 1

If $\lambda$ and $\mu$ are the (complex) eigenvalues of $A$, then $\lambda\mu=\det(A)=1$, hence $$\mathrm{tr}(A^2)=\lambda^2+\mu^2=(\lambda+\mu)^2-2\lambda\mu=\mathrm{tr}(A)^2-2\geq-2$$ Note that $\mathrm{tr}(A)$ is real even if $\lambda$ and $\mu$ are not.