Let \(A\) and \(B\) be endomorphisms of a finite-dimensional vector space \(V\) over a field \(F\). Prove or disprove each statement:
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 \(V=F^2\) and, in the standard basis, let \[ A= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad B= \begin{pmatrix} 1&0\\ 1&0 \end{pmatrix}. \] Then \[ AB= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad BA= \begin{pmatrix} 1&0\\ 1&0 \end{pmatrix}. \] For \[ e_1= \begin{pmatrix} 1\\ 0 \end{pmatrix}, \] one has \[ ABe_1=e_1, \] so \(e_1\) is an eigenvector of \(AB\). But \[ BAe_1= \begin{pmatrix} 1\\ 1 \end{pmatrix}, \] which is not a scalar multiple of \(e_1\). Thus \(e_1\) is not an eigenvector of \(BA\).
2. Let \(\lambda\neq0\) be an eigenvalue of \(AB\). Then \(\lambda\) is an eigenvalue of \(BA\).
Choose a nonzero vector \(v\) such that \[ ABv=\lambda v. \] One must have \(Bv\neq0\), because otherwise \(ABv=0\), contradicting \(\lambda v\neq0\). Applying \(B\) to the eigenvalue equation gives \[ BA(Bv) = B(ABv) = \lambda Bv. \] Thus the nonzero vector \(Bv\) is an 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 not invertible. Suppose, for contradiction, that \(BA\) were invertible.
If \(Ax=0\), then \[ BAx=0, \] so invertibility of \(BA\) gives \(x=0\). Hence \(A\) is injective and therefore invertible because \(V\) is finite-dimensional.
Also, for every \(y\in V\), surjectivity of \(BA\) gives some \(x\in V\) such that \[ BAx=y; \] thus \(y=B(Ax)\) lies in the image of \(B\). Hence \(B\) is surjective and therefore invertible. Thus both \(A\) and \(B\) would be invertible, forcing \(AB\) to be invertible, a contradiction.
Therefore \(BA\) is not invertible, so \(0\) is an eigenvalue of \(BA\).
4. Statement 2 is true.
Step 2 handles every nonzero eigenvalue of \(AB\), and step 3 handles the eigenvalue \(0\).
5. Q.E.D.
Statement 1 is disproved in step 1, while statement 2 is proved in step 4.