Linear mapping/Eigenspace for different eigenvalues/Zero/Fact/Proof

Proof

Let ${}v\in \operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\cap \operatorname {Eig} _{\lambda _{2}}{\left(\varphi \right)}$. Then

${}\lambda _{1}v=\varphi (v)=\lambda _{2}v\,.$

Therefore,

${}(\lambda _{1}-\lambda _{2})v=0\,,$

and this implies, because of ${}\lambda _{1}\neq \lambda _{2}$, ${}v=0$.

To fact