2013/05/9
eigenvalues under commutation
A and B are two square matrices. The eigenvalues of AB and BA are the same.
Proof: Suppose λ and v are an eigenvalue and the corresponding eigenvector for AB, so that ABv=λv. Let q=Bv. Then Aq=λv, so BAq=λBv=λq. So λ is also an eigenvalue of BA, and its associated eigenvector is q=Bv.