## 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.)