Proof of Theorem IPSM

The proofs of the two parts are very similar, with the second one requiring just a bit more effort due to the conjugation that occurs. We will prove part 1 and you can prove part 2 (Exercise O.T11). \begin{align*} \innerproduct{\alpha\vect{u}}{\vect{v}} &=\sum_{i=1}^{m}\conjugate{\vectorentry{\alpha\vect{u}}{i}}\vectorentry{\vect{v}}{i} &&\knowl{./knowls/definition.IP.knowl}{\text{Definition IP}}\\ &=\sum_{i=1}^{m}\conjugate{\alpha\vectorentry{\vect{u}}{i}}\vectorentry{\vect{v}}{i} &&\knowl{./knowls/definition.CVSM.knowl}{\text{Definition CVSM}}\\ &=\sum_{i=1}^{m}\conjugate{\alpha}\,\conjugate{\vectorentry{\vect{u}}{i}}\vectorentry{\vect{v}}{i} &&\knowl{./knowls/theorem.CCRM.knowl}{\text{Theorem CCRM}}\\ &=\conjugate{\alpha}\sum_{i=1}^{m}\conjugate{\vectorentry{\vect{u}}{i}}\vectorentry{\vect{v}}{i} &&\knowl{./knowls/property.DCN.knowl}{\text{Property DCN}}\\ &=\conjugate{\alpha}\innerproduct{\vect{u}}{\vect{v}} &&\knowl{./knowls/definition.IP.knowl}{\text{Definition IP}} \end{align*}