If $B$ satisfies $A B = B A^{-1}$, show that $B^2$ is diagonalisable 
Let $A$ be a $n \times n$ non-singular matrix having distinct eigenvalues. If $B$ is a matrix satisfying  $A B = B A^{-1}$, show that $B^2$ is diagonalisable. 

Answer:
Let $\lambda_i, \ i=1,2,3, \cdots, n$ be the $n$ distinct eigenvalues. 
Now, $AB=BA^{-1} \Rightarrow B=ABA $
But then how to proceed?

We can show that $AB^2=ABABA=B^2A$. So $B^2$ and $A$ commute. Then the claim follows from these duplicates (replacing $B$ by $B^2$):
$AB=BA$. Prove $B$ is diagonalizable.
If $AB=BA$, show that $B$ is diagonalizable.
Indeed, we have $AB^2=(AB)B=ABABA$ and $B^2A=B(BA)=ABABA$.
