Proof of Theorem CUMOS

The proof revolves around recognizing that a typical entry of the product $\adjoint{A}A$ is an inner product of columns of $A$. Here are the details to support this claim. \begin{align*} \matrixentry{\adjoint{A}A}{ij} &=\sum_{k=1}^{n}\matrixentry{\adjoint{A}}{ik}\matrixentry{A}{kj} &&\knowl{./knowls/theorem.EMP.knowl}{\text{Theorem EMP}}\\ &=\sum_{k=1}^{n}\matrixentry{\transpose{\conjugate{A}}}{ik}\matrixentry{A}{kj} &&\knowl{./knowls/theorem.EMP.knowl}{\text{Theorem EMP}}\\ &=\sum_{k=1}^{n}\matrixentry{\,\conjugate{A}\,}{ki}\matrixentry{A}{kj} &&\knowl{./knowls/definition.TM.knowl}{\text{Definition TM}}\\ &=\sum_{k=1}^{n}\conjugate{\matrixentry{A}{ki}}\matrixentry{A}{kj} &&\knowl{./knowls/definition.CCM.knowl}{\text{Definition CCM}}\\ &=\sum_{k=1}^{n}\conjugate{\vectorentry{\vect{A}_i}{k}}\vectorentry{\vect{A}_j}{k}\\ &=\innerproduct{\vect{A}_i}{\vect{A}_j} &&\knowl{./knowls/definition.IP.knowl}{\text{Definition IP}} \end{align*}

We now employ this equality in a chain of equivalences, \begin{align*} &\text{$S=\set{\vectorlist{A}{n}}$ is an orthonormal set}\\ &\iff \innerproduct{\vect{A}_i}{\vect{A}_j}= \begin{cases} 0 &\text{if $i\neq j$}\\ 1 & \text{if $i=j$} \end{cases}&&\knowl{./knowls/definition.ONS.knowl}{\text{Definition ONS}}\\ &\iff \matrixentry{\adjoint{A}A}{ij}= \begin{cases} 0 &\text{if $i\neq j$}\\ 1 & \text{if $i=j$} \end{cases}\\ &\iff \matrixentry{\adjoint{A}A}{ij}=\matrixentry{I_n}{ij},\ 1\leq i\leq n,\ 1\leq j\leq n &&\knowl{./knowls/definition.IM.knowl}{\text{Definition IM}}\\ &\iff \adjoint{A}A=I_n &&\knowl{./knowls/definition.ME.knowl}{\text{Definition ME}}\\ &\iff \text{$A$ is a unitary matrix} &&\knowl{./knowls/definition.UM.knowl}{\text{Definition UM}} \end{align*}