Let \(A\) and \(B\) be \(n\times n\) matrices with complex entries, such that \(AB-BA\) is a linear combination of \(A\) and \(B\). Prove that there exists a nonzero vector \(v\) that is an eigenvector of both \(A\) and \(B\).
Solution
Let \(AB-BA=C=\alpha A+\beta B\). If \(\alpha=\beta=0\), then \(A\) and \(B\) commute. By a theorem of linear algebra, commuting complex matrices have a common eigenvector. Otherwise, assume without loss of generality that \(\beta\neq0\). Then \(B\) is a linear combination of \(A\) and \(C\), so it suffices to prove that \(A\) and \(C\) have a common eigenvector. Note that \(AC-CA=\beta C\). Since \(A\) has finitely many eigenvalues, it must have one, call it \(\lambda\), such that \(\lambda+\beta\) is not an eigenvalue of \(A\). Let \(v\) be a nonzero vector with \(Av=\lambda v\). Then \(ACv=CAv+\beta Cv=(\lambda+\beta)Cv\), so \(Cv=0\). Hence \(v\) is a common eigenvector of \(A\) and \(C\).