\begin{split}
&\hbox{(a)} \;\; \sum_\alpha \hat{A}_\alpha \hbox{Tr}(\hat{M}\hat{A}_\alpha) = N\hat{M} \\
&\hbox{(b)} \;\; \sum_\alpha \hat{A}_\alpha \hat{M} \hat{A}_\alpha = N(\hbox{Tr}\hat{M})\hat{I}.
\end{split}$$ Starting with Eq. (\[Pdef\]), we multiply both sides by $\hat{A}_\gamma$ and use Eq. (\[properties\]a) to get $$\sum_\gamma P_{\alpha\gamma}\hat{A}_\gamma = \sum_{\beta\delta} {\mathcal B}_{\beta\delta} \hat{A}_\delta \hat{A}_\alpha \hat{A}_\beta.$$ Now multiply on the left by $\hat{A}_\nu$ and on the right by $\hat{A}_\mu \hat{A}_\alpha$, sum over $\alpha$ and use Eq. (\[properties\]b): $$\sum_{\alpha\gamma} P_{\alpha\gamma} \hat{A}_\nu \hat{A}_\gamma \hat{A}_\mu \hat{A}_\alpha = N \sum_{\beta\delta} {\mathcal B}_{\beta\delta} \hat{A}_\nu \hat{A}_\delta \hbox{Tr}(\hat{A}_\beta \hat{A}_\mu).$$ Finally, take the trace of both sides and use the fact that $\hbox{Tr}(\hat{A}_\nu \hat{A}_\delta) = N\delta_{\nu\delta}$ to get $$\label{inverse}
{\mathcal B}_{\mu\nu} = \frac{1}{N^3} \sum_{\alpha\gamma} \hbox{Tr}(\hat{A}_\mu \hat{A}_\alpha \hat{A}_\nu \hat{A}_\gamma) P_{\alpha\gamma} = \frac{1}{N^2} \sum_{\alpha\gamma} \Xi_{\mu\alpha\nu\gamma} P_{\alpha\gamma}.$$ Comparing this equation to Eq. (\[PXi\]), we see that $P$’s and the ${\mathcal B}$’s are related to each other in a symmetric way. One consequence of Eq. (\[inverse\]) is that the values ${\mathcal B}_{\beta\delta}$ are uniquely determined by the quantum transformation. (In this respect they differ from the set of operators $\hat{B}_j$, for which one can choose among many equivalent sets.)

In the preceding paragraph, we began by assuming that the $P$ values we were given could be expressed in the form (\[Pdef\]). But how do we 