Proof of Theorem EOMP

Let $\vect{x}\neq\zerovector$ be one eigenvector of $A$ for $\lambda$. Suppose $A$ has size $n$. Then we proceed by induction on $s$ (Proof Technique I). First, for $s=0$, \begin{align*} A^s\vect{x}&=A^0\vect{x}\\ &=I_n\vect{x}\\ &=\vect{x}&&\knowl{./knowls/theorem.MMIM.knowl}{\text{Theorem MMIM}}\\ &=1\vect{x}&&\knowl{./knowls/property.OC.knowl}{\text{Property OC}}\\ &=\lambda^0\vect{x}\\ &=\lambda^s\vect{x}\\ \end{align*} so $\lambda^s$ is an eigenvalue of $A^s$ in this special case. If we assume the theorem is true for $s$, then we find \begin{align*} A^{s+1}\vect{x}&=A^sA\vect{x}\\ &=A^s\left(\lambda\vect{x}\right)&&\knowl{./knowls/definition.EEM.knowl}{\text{Definition EEM}}\\ &=\lambda\left(A^s\vect{x}\right)&&\knowl{./knowls/theorem.MMSMM.knowl}{\text{Theorem MMSMM}}\\ &=\lambda\left(\lambda^s\vect{x}\right)&&\text{Induction hypothesis}\\ &=\left(\lambda\lambda^s\right)\vect{x}&&\knowl{./knowls/property.SMAC.knowl}{\text{Property SMAC}}\\ &=\lambda^{s+1}\vect{x} \end{align*}

So $\vect{x}\neq\zerovector$ is an eigenvector of $A^{s+1}$ for $\lambda^{s+1}$, and induction tells us the theorem is true for all $s\geq 0$.