Proof of Theorem HMIP

(⇒) This is the “easy half” of the proof, and makes the rationale for a definition of Hermitian matrices most obvious. Assume $A$ is Hermitian, \begin{align*} \innerproduct{A\vect{x}}{\vect{y}} &=\innerproduct{\vect{x}}{\adjoint{A}\vect{y}}&&\knowl{./knowls/theorem.AIP.knowl}{\text{Theorem AIP}}\\ &=\innerproduct{\vect{x}}{A\vect{y}}&&\knowl{./knowls/definition.HM.knowl}{\text{Definition HM}}\\ \end{align*}

(⇐) This “half” will take a bit more work. Assume that $\innerproduct{A\vect{x}}{\vect{y}}=\innerproduct{\vect{x}}{A\vect{y}}$ for all $\vect{x},\,\vect{y}\in\complex{n}$. We show that $A=\adjoint{A}$ by establishing that $A\vect{x}=\adjoint{A}\vect{x}$ for all $\vect{x}$, so we can then apply Theorem EMMVP. With only this much motivation, consider the inner product for any $\vect{x}\in\complex{n}$. \begin{align*} \innerproduct{A\vect{x}-\adjoint{A}\vect{x}}{A\vect{x}-\adjoint{A}\vect{x}} &= \innerproduct{A\vect{x}-\adjoint{A}\vect{x}}{A\vect{x}}- \innerproduct{A\vect{x}-\adjoint{A}\vect{x}}{\adjoint{A}\vect{x}} &&\knowl{./knowls/theorem.IPVA.knowl}{\text{Theorem IPVA}}\\ &= \innerproduct{A\vect{x}-\adjoint{A}\vect{x}}{A\vect{x}}- \innerproduct{A\left(A\vect{x}-\adjoint{A}\vect{x}\right)}{\vect{x}} &&\knowl{./knowls/theorem.AIP.knowl}{\text{Theorem AIP}}\\ &= \innerproduct{A\vect{x}-\adjoint{A}\vect{x}}{A\vect{x}}- \innerproduct{A\vect{x}-\adjoint{A}\vect{x}}{A\vect{x}} &&\text{Hypothesis}\\ & =0 &&\knowl{./knowls/property.AICN.knowl}{\text{Property AICN}} \end{align*}

Because this first inner product equals zero, and has the same vector in each argument ($A\vect{x}-\adjoint{A}\vect{x}$), Theorem PIP gives the conclusion that $A\vect{x}-\adjoint{A}\vect{x}=\zerovector$. With $A\vect{x}=\adjoint{A}\vect{x}$ for all $\vect{x}\in\complex{n}$, Theorem EMMVP says $A=\adjoint{A}$, which is the defining property of a Hermitian matrix (Definition HM).