Endomorphism/Eigenvalue/Invariant hyperplane/Fact/Proof

Because of the condition and fact, the mapping ${\displaystyle {}\varphi -\lambda \operatorname {Id} _{V}}$ has a nontrival kernel. Hence, this mapping is not injective and, due to fact, also not surjective. Therefore,

${\displaystyle {}B:=\operatorname {Im} {\left(\varphi -\lambda \operatorname {Id} _{V}\right)}\subset V\,}$

is a strict linear subspace of ${\displaystyle {}V}$. It follows that there exists also a linear subspace ${\displaystyle {}U\subset V}$ of dimension ${\displaystyle {}n-1}$, which contains ${\displaystyle {}B}$. For ${\displaystyle {}u\in U}$, we have

${\displaystyle {}\varphi (u)=\lambda u+(\varphi -\lambda \operatorname {Id} _{V})u\in U+B\subseteq U\,.}$

Hence, the image of ${\displaystyle {}U}$ belongs to ${\displaystyle {}U}$, that is, ${\displaystyle {}U}$ is ${\displaystyle {}\varphi }$-invariant.