Proof: Suppose \(\lambda\) and \(v\) are an eigenvalue and the corresponding eigenvector for \(AB\), so that \(ABv = \lambda v\). Let \(q = Bv\). Then \(Aq = \lambda v\), so \(BAq = \lambda Bv = \lambda q\). So \(\lambda\) is also an eigenvalue of \(BA\), and its associated eigenvector is \(q = Bv\).