Prove that $A,B$ have a common eigenvector Let $A,B$ be $2\times2$ real matrices satisfying $\det(A)=\det(B)=1$ and $$\text{tr}(A)>2 , \text{tr}(B)>2, \text{tr}(ABA^{-1}B^{-1})=2$$
Prove that A,B have a common eigenvector.

One possible approach:


*

*Show that each of $A$ and $B$ has distinct real positive eigenvalues.

*Hence $A$ is diagonalisable over $\mathbb R$. By a change of basis, we may assume that $A=\operatorname{diag}(p,\frac1p)$ for some $p>0$. Let $B=\pmatrix{a&b\\ c&d}$ in this basis.

*Use the given conditions and the assumption in (2) to prove that $bc=0$, i.e. $B$ is upper or lower triangular. Now the rest is straightforward.

