[proofplan] We prove the three defining properties of an [equivalence relation](/page/Equivalence%20Relation) directly from the definition. Reflexivity is witnessed by the projection itself. Symmetry is obtained by taking adjoints of a witnessing element. Transitivity is obtained by multiplying two witnesses and using the middle projection identity to cancel the common projection. [/proofplan] [step:Use each projection as its own Murray-von Neumann witness] Let $p\in \operatorname{Proj}(A)$. Since $p=p^*$ and $p^2=p$, we have \begin{align*} p^*p=p^2=p \end{align*} and \begin{align*} pp^*=p^2=p. \end{align*} Thus the element $v:=p\in A$ witnesses $p\sim_{\mathrm{MvN}}p$. Hence $\sim_{\mathrm{MvN}}$ is reflexive on $\operatorname{Proj}(A)$. [/step] [step:Take the adjoint of a witness to reverse the equivalence] Let $p,q\in \operatorname{Proj}(A)$ and suppose $p\sim_{\mathrm{MvN}}q$. By definition, there exists $v\in A$ such that \begin{align*} v^*v=p \end{align*} and \begin{align*} vv^*=q. \end{align*} Define $u:=v^*\in A$. Then $u^*=v$, so \begin{align*} u^*u=vv^*=q \end{align*} and \begin{align*} uu^*=v^*v=p. \end{align*} Therefore $u$ witnesses $q\sim_{\mathrm{MvN}}p$. Hence $\sim_{\mathrm{MvN}}$ is symmetric. [guided] We want to reverse the direction of the equivalence. The definition of $p\sim_{\mathrm{MvN}}q$ gives a single element $v\in A$ whose initial projection is $p$ and whose final projection is $q$: \begin{align*} v^*v=p \end{align*} and \begin{align*} vv^*=q. \end{align*} The natural candidate for a witness in the reverse direction is the adjoint. Define $u:=v^*$. Since $A$ is closed under the involution, $u\in A$. Also $u^*=v$. Substituting $u=v^*$ into the two expressions required by the definition gives \begin{align*} u^*u=vv^*=q \end{align*} and \begin{align*} uu^*=v^*v=p. \end{align*} Thus $u$ has initial projection $q$ and final projection $p$, so $u$ witnesses $q\sim_{\mathrm{MvN}}p$. This proves symmetry. [/guided] [/step] [step:Multiply compatible witnesses to prove transitivity] Let $p,q,r\in \operatorname{Proj}(A)$ and suppose $p\sim_{\mathrm{MvN}}q$ and $q\sim_{\mathrm{MvN}}r$. Choose $v,w\in A$ such that \begin{align*} v^*v=p \end{align*} \begin{align*} vv^*=q \end{align*} \begin{align*} w^*w=q \end{align*} and \begin{align*} ww^*=r. \end{align*} Define $u:=wv\in A$. Then, using associativity of multiplication in $A$, \begin{align*} u^*u=(wv)^*(wv)=v^*w^*wv=v^*qv. \end{align*} Since $q=vv^*$, this becomes \begin{align*} u^*u=v^*(vv^*)v=(v^*v)(v^*v)=p^2=p. \end{align*} Similarly, \begin{align*} uu^*=(wv)(wv)^*=wvv^*w^*=wqw^*. \end{align*} Since $q=w^*w$, this becomes \begin{align*} uu^*=w(w^*w)w^*=(ww^*)(ww^*)=r^2=r. \end{align*} Thus $u\in A$ witnesses $p\sim_{\mathrm{MvN}}r$. Hence $\sim_{\mathrm{MvN}}$ is transitive. [/step] [step:Combine the three properties] We have shown that $\sim_{\mathrm{MvN}}$ is reflexive, symmetric, and transitive on $\operatorname{Proj}(A)$. Therefore Murray-von Neumann equivalence is an equivalence relation on the set of projections in $A$. [/step]