Commuting diagonalizable matrices are simultaneously diagonalizable

Let \[ V=\bigoplus_\lambda E_A(\lambda) \] be the eigenspace decomposition for the diagonalizable operator \(A\).

Because \(AB=BA\), each \(A\)-eigenspace is \(B\)-stable: if \(v\in E_A(\lambda)\), then \[ A(Bv)=B(Av)=\lambda Bv. \] Thus \(B\) restricts to an operator on every \(E_A(\lambda)\).

Since \(B\) is diagonalizable, its minimal polynomial splits into distinct linear factors. The minimal polynomial of each restriction \(B|_{E_A(\lambda)}\) divides the minimal polynomial of \(B\), hence also splits into distinct linear factors. Therefore each restriction is diagonalizable.

Choose a basis of \(B\)-eigenvectors inside each \(E_A(\lambda)\). The union of these bases is a basis of \(V\) consisting of common eigenvectors of \(A\) and \(B\). Hence \(A\) and \(B\) are simultaneously diagonalizable.