Why A is diagonalizable? This is a part of a question.
Let $A$ be a real $2\times 2$ matrix such that $\,\det A=1,\operatorname{tr}A>2$.
Why is $A$ diagonalizable? 
The original question asks to prove that $A$ and $B$ have a common eigenvector if $\det B=1,\operatorname{tr}B>2,\operatorname{tr}\left(ABA^{-1}B^{-1}\right)=2$.
 A: If $\lambda_1, \lambda)_2$ are the two (complex) eigenvalues of a real matrix $A$ (size $2\times 2$), then we always have $\det(A) = \lambda_1 \lambda_2$ and $\operatorname{tr} A  = \lambda_1 + \lambda_2$. 
If the eigenvalues would be equal the eigenvalues would be $1,1$ or $-1,-1$ and thus give trace $-2$ or $2$ which is not the case. 
So, the eigenvalues are different, and we now have a base of eigenvectors: the two linearly independent (because they have different eigenvalues) eigenvectors of $A$ must form a base of $\mathbb{R}^2$.
A: This is because the characteristic polynomial of $A$ is
$$\chi_A(\lambda)=\lambda^2-(\operatorname{Tr} A) \lambda+\det A,$$
and that it has two distinct real roots since its discriminant is
$$\Delta=(\operatorname{Tr} A)^2-4\det A>4-4=0$$
by the hypotheses.
A: MooS' first & prompt comment brings up the principal reason: $A$ is diagonalisable because the eigenvalues must be different under these givens.
$A\text{ certainly has two}\:\mathbb C\text{omplex eigenvalues}$
$$\lambda_1 = x+iy\quad\text{and}\quad\lambda_2 = \frac 1{\lambda_1}
 = {x-iy\over x^2+y^2}$$
owing to the condition $\,\det A=1$, in particular both eigenvalues are non-zero. If the imaginary part $\,y\,$ were to be non-zero, then the condition $\:\operatorname{tr}A>2\:$ could not be satisfied. Hence both eigenvalues are real and even greater than  $0\,$.
The trace-condition can be read as AM-GM inequality 
$$\operatorname{tr}A>2\quad\iff\quad
\frac 12\left(\lambda_1 + \frac1{\lambda_1}\right)\:>\:
\lambda_1\cdot\frac1{\lambda_1} = 1$$
which is strict, and strictness implies
$\lambda_1\ne\frac1{\lambda_1}=\lambda_2\,$.
