Let \(A,B\) be endomorphisms of a finite-dimensional vector space \(V\) over a field \(K\). Prove or disprove:
- Every eigenvector of \(AB\) is also an eigenvector of \(BA\).
- Every eigenvalue of \(AB\) is also an eigenvalue of \(BA\).
Solution
1. Statement (1) is false.
Take \(V=K^2\) with basis \(e_1,e_2\), and define endomorphisms by \[ A(e_1)=e_1,\qquad A(e_2)=0, \] and \[ B(e_1)=e_2,\qquad B(e_2)=e_1. \] Then \[ AB(e_1)=A(e_2)=0, \] so \(e_1\) is an eigenvector of \(AB\) with eigenvalue \(0\). On the other hand, \[ BA(e_1)=B(e_1)=e_2, \] which is not a scalar multiple of \(e_1\). Thus \(e_1\) is not an eigenvector of \(BA\).
2. Every nonzero eigenvalue of \(AB\) is an eigenvalue of \(BA\).
Let \(\lambda\neq 0\) be an eigenvalue of \(AB\), and choose \(0\neq v\in V\) such that \[ ABv=\lambda v. \] Then \(Bv\neq 0\), since otherwise \(ABv=0\), contradicting \(\lambda v\neq 0\). Applying \(B\) to the eigenvalue equation gives \[ BA(Bv) = B(ABv) = \lambda Bv. \] Hence \(Bv\) is a nonzero eigenvector of \(BA\) with eigenvalue \(\lambda\).
3. If \(0\) is an eigenvalue of \(AB\), then \(0\) is an eigenvalue of \(BA\).
If \(0\) is an eigenvalue of \(AB\), then \(AB\) is singular, so \[ \det(AB)=0. \] By multiplicativity of the determinant, \[ \det(BA) = \det(B)\det(A) = \det(A)\det(B) = \det(AB) = 0. \] Thus \(BA\) is singular and has a nonzero kernel. Therefore \(0\) is an eigenvalue of \(BA\).
4. Statement (2) is true.
Step 2 proves the claim for every nonzero eigenvalue of \(AB\), and step 3 proves it for the eigenvalue \(0\).
5. Q.E.D.
Step 1 disproves statement (1), and step 4 proves statement (2).