Let $A$ be an $n \times n$ nonsingular matrix having distinct eigenvalues and $ B$ is a matrix satisfying $AB = BA^{-1}$

Question

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

I need to show that $B^{2}$ is diagonalizable.
If I prove that $AB^{2} = B^{2}A$ then, it implies that $B^{2}$ is diagonalizable, since it commutes with a diagonal matrix.

But I don't how to prove that $B^{2}$ commute with $A$.
Any hint would be helpful.

Answer 1

Note that\begin{align}AB^2&=(AB)\left(BA^{-1}\right)A\\&=\left(BA^{-1}\right)(AB)A\\&=B\operatorname{Id}BA\\&=B^2A.\end{align}