Proof of Theorem ESMM

Let $\vect{x}\neq\zerovector$ be one eigenvector of $A$ for $\lambda$. Then \begin{align*} \left(\alpha A\right)\vect{x}&=\alpha\left(A\vect{x}\right)&&\knowl{./knowls/theorem.MMSMM.knowl}{\text{Theorem MMSMM}}\\ &=\alpha\left(\lambda\vect{x}\right)&&\knowl{./knowls/definition.EEM.knowl}{\text{Definition EEM}}\\ &=\left(\alpha\lambda\right)\vect{x}&&\knowl{./knowls/property.SMAC.knowl}{\text{Property SMAC}} \end{align*}

So $\vect{x}\neq\zerovector$ is an eigenvector of $\alpha A$ for the eigenvalue $\alpha\lambda$.