Let be a field, and -vector space, and a linear transformation. An element is called an eigenvalue of if there exists a nonzero element such that . Such a is called an eigenvector of . Prove that for any fixed , the set is an -subspace.
Certainly , since . Now suppose and . Then , since is commutative. Thus . By the submodule criterion, is a subspace.
(Note that our proof does not require that be an eigenvalue of . However, if it isn’t, then isn’t very interesting.)