Let \(\phi,\psi\) be commuting endomorphisms of a finite-dimensional vector space \(E\) over a field \(k\).
- If \(k\) is algebraically closed, prove that \(\phi\) and \(\psi\) have a common eigenvector.
- If both operators are diagonalizable, prove that they are simultaneously diagonalizable.
Solution
1. Assume \(k\) is algebraically closed. Then \(\phi\) has an eigenvalue \(\lambda\), and its nonzero eigenspace\[ E_\lambda=\ker(\phi-\lambda I) \]is invariant under \(\psi\), because for \(v\in E_\lambda\),\[ \phi(\psi v)=\psi(\phi v)=\lambda\psi v. \]The restriction \(\psi|_{E_\lambda}\) has an eigenvector \(0\ne v\in E_\lambda\) since \(k\) is algebraically closed. Then \(v\) is an eigenvector of both operators.
2. Now suppose both \(\phi\) and \(\psi\) are diagonalizable. Decompose\[ E=\bigoplus_\lambda E_\lambda \]into eigenspaces of \(\phi\). As above, every \(E_\lambda\) is \(\psi\)-invariant. Since \(\psi\) is diagonalizable, its minimal polynomial splits into distinct linear factors; the minimal polynomial of each restriction \(\psi|_{E_\lambda}\) divides it, so each restriction is diagonalizable. Choose an eigenbasis for \(\psi\) inside each \(E_\lambda\). The union of these bases is a basis of \(E\) consisting of simultaneous eigenvectors of \(\phi\) and \(\psi\).