Let A and B be unkown diagonalizable matrices. If A and B have exactly the same eigenvectors as each other, then show that AB = BA.
$\displaystyle A=P_1^{-1}D_1P_1, \ \ B=P_2^{-1}D_2P_2,$ where $\displaystyle D_1,D_2$ are diagonal and $\displaystyle P_1,P_2$ are matrices with the eigenvectors of $\displaystyle A,B$ as their columns repectively. so $\displaystyle P_1=P_2$ and since diagonal matrices commute
with each other, we'll have: $\displaystyle AB=P_1^{-1}D_1D_2P_1^{-1}=P_1^{-1}D_2D_1P_1=BA. \ \Box$