Proof of Theorem HMRE

Let $\vect{x}\neq\zerovector$ be one eigenvector of $A$ for the eigenvalue $\lambda$. Then by Theorem PIP we know $\innerproduct{\vect{x}}{\vect{x}}\neq 0$. So \begin{align*} \lambda &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\lambda\innerproduct{\vect{x}}{\vect{x}} &&\knowl{./knowls/property.MICN.knowl}{\text{Property MICN}}\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\innerproduct{\vect{x}}{\lambda\vect{x}} &&\knowl{./knowls/theorem.IPSM.knowl}{\text{Theorem IPSM}}\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\innerproduct{\vect{x}}{A\vect{x}} &&\knowl{./knowls/definition.EEM.knowl}{\text{Definition EEM}}\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\innerproduct{A\vect{x}}{\vect{x}} &&\knowl{./knowls/theorem.HMIP.knowl}{\text{Theorem HMIP}}\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\innerproduct{\lambda\vect{x}}{\vect{x}} &&\knowl{./knowls/definition.EEM.knowl}{\text{Definition EEM}}\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\conjugate{\lambda}\innerproduct{\vect{x}}{\vect{x}} &&\knowl{./knowls/theorem.IPSM.knowl}{\text{Theorem IPSM}}\\ &=\conjugate{\lambda} &&\knowl{./knowls/property.MICN.knowl}{\text{Property MICN}} \end{align*}

If a complex number is equal to its conjugate, then it has a complex part equal to zero, and therefore is a real number.