The eigenvectors of a fixed eigenvalue of a linear transformation form a subspace

Let F be a field, V and F-vector space, and \varphi : V \rightarrow V a linear transformation. An element \lambda \in F is called an eigenvalue of \varphi if there exists a nonzero element v \in V such that \varphi(v) = \lambda v. Such a v is called an eigenvector of \lambda. Prove that for any fixed \lambda \in F, the set V_\lambda = \{ v \in V \ |\ \varphi(v) = \lambda v \} is an F-subspace.

Certainly 0 \in V_\lambda, since \varphi(0) = 0 = \lambda 0. Now suppose x,y \in V_\lambda and \alpha \in F. Then \varphi(x+ \alpha y) = \varphi(x) + \alpha\varphi(y) = \lambda x + \alpha \lambda y = \lambda (x + \alpha y), since F is commutative. Thus x+ \alpha y \in V_\lambda. By the submodule criterion, V_\lambda \subseteq V is a subspace.

(Note that our proof does not require that \lambda be an eigenvalue of \varphi. However, if it isn’t, then V_\lambda isn’t very interesting.)

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