Suppose that the vector $\vect{x}\neq\zerovector$ is an eigenvector of $A$ for the two eigenvalues $\lambda$ and $\rho$, where $\lambda\neq\rho$. Then $\lambda-\rho\neq 0$, and we also have
\begin{align*}
\zerovector
&=A\vect{x}-A\vect{x}&&\knowl{./knowls/property.AIC.knowl}{\text{Property AIC}}\\
&=\lambda\vect{x}-\rho\vect{x}&&\knowl{./knowls/definition.EEM.knowl}{\text{Definition EEM}}\\
&=(\lambda-\rho)\vect{x}&&\knowl{./knowls/property.DSAC.knowl}{\text{Property DSAC}}
\end{align*}
By Theorem SMEZV, either $\lambda-\rho=0$ or $\vect{x}=\zerovector$, which are both contradictions.