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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.

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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}

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