Sudarshan's dynamical maps This is a question about an equation in a paper by E.C.G. Sudarshan, P.M.Matthews and J. Rau. The authors introduce the concept of dynamical maps - objects that determine the time evolution of density matrices. When the system under study is hamiltonian the $B$-map is unitary and it satisfies the equation
\begin{equation}\tag{1}
\sum_{r=1}^n B_{r,r^\prime;r, s^\prime} =  \delta_{r^\prime, s^\prime}.
\end{equation}
The paper uses summation convention but I choose to use the summation explicitly.
In section 3 they mention that using equation (1) one can prove that
\begin{equation}\tag{2}
B^2 - nB = 0.
\end{equation}
I am unable to see how equation (2) follows from equation (1) even after using the fact that $B$ is hermitian. Can someone please help me obtain (2)?
A brief introduction to the $B$ matrix
Sudarshan's paper considers the transformation between density matrices at times $t_0$ and $t_1 > t_0$. One can write
\begin{equation}\tag{3}
\rho_{r,s}(t_1) = A_{r,s;r^\prime, s^\prime}\rho_{r^\prime,s^\prime}(t_0).
\end{equation}
Summation convention is used here and the other equations following this one. The 'object' $A_{r,s;r^\prime, s^\prime}$ can be considered as an $n^2 \times n^2$ matrix labelled by 'double indices' $(rs)$ and $(r^\prime, s^\prime)$. $n$ is the (finite) dimension of the Hilbert space on which the density matrix is defined. The $A$-matrices do not have properties convenient for further analysis. Therefore, the authors introduce the $B$-matrices, defining them as
\begin{equation}\tag{4}
B_{r,r^\prime;s, s^\prime} = A_{r,s;r^\prime, s^\prime}.
\end{equation}
From the hermiticity of $\rho$ one can conclude that
\begin{equation}\tag{5}
B_{r,r^\prime;s, s^\prime} = B_{s, s^\prime; r,r^\prime}^\ast.
\end{equation}
The positive definiteness of $\rho$ gives,
\begin{equation}\tag{6}
z_{rr^\prime}^\ast B_{r,r^\prime;s, s^\prime}z_{ss^\prime} \ge 0.
\end{equation}
Since the density matrices have a unit trace, we require
\begin{equation}\tag{7}
B_{r,r^\prime;r, s^\prime} =  \delta_{r^\prime, s^\prime}.
\end{equation}
This is same as equation (1) except that in the case of (1) I have used the summation sign.
An update 04-Jan-2021
The general form of the $B$-map that satisfies hermiticity condition of equation (5) and the trace condition of equation (7) is
\begin{equation}\tag{8}
B = \begin{pmatrix}
B_{11;11} & B_{11;12} & B_{11;21} & B_{11;22} \\
B_{11;12}^\ast & B_{12;12} & B_{12;21} & B_{12;22} \\
B_{11;21}^\ast & B_{12;21}^\ast & 1 - B_{11;11} & -B_{11;12} \\
B_{11;22}^\ast & B_{12;22}^\ast & B_{11;12}^\ast & 1 - B_{12;12}
\end{pmatrix}
\end{equation}
Unless we impose some more conditions we are unlikely to get the relation $B^2 = 2B$.
 A: I am unable to prove $B^2 = nB$ using equations (5) and (7) in the question. However, the authors mention another way to get this result. When the density matrix evolves unitarily, the $B$-matrix can be written as
\begin{equation}\tag{9}
B_{rr^\prime,ss^\prime} = U_{rr}U^\ast_{ss^\prime}
\end{equation}
so that
\begin{equation}\tag{10}
(B^2)_{rr^\prime, ss^\prime} = \sum_{uv}B_{rr^\prime, uv}B_{uv, ss^\prime}
= \sum_{uv}U_{rr}U^\ast_{uv}U_{uv}U^\ast_{ss^\prime}.
\end{equation}
We can simplify it as
\begin{equation}\tag{11}
(B^2)_{rr^\prime, ss^\prime} = U_{rr}U^\ast_{ss^\prime}\sum_{uv}U^\ast_{uv}
U_{uv} = U_{rr}U^\ast_{ss^\prime}\sum_{uv}U_{uv}(U^\dagger)_{vu}.
\end{equation}
We can write the sum as
\begin{equation}\tag{12}
\sum_{uv}U_{uv}(U^\dagger)_{vu} = \sum_u\left(\sum_v U_{uv}(U^\dagger)_{vu}
\right) = \sum_u \delta_{uu} = n,
\end{equation}
where we have used the fact that $U$ is unitary. From equation (9), (11) and (12) we get
\begin{equation}\tag{13}
(B^2)_{rr^\prime, ss^\prime} = nU_{rr}U^\ast_{ss^\prime} = nB_{rr,ss^\prime},
\end{equation}
where we used equation (9) to get the last term. We can write this
equation as
\begin{equation}\tag{14}
B^2 = nB.
\end{equation}
I think equation (14) is valid only in the case of a unitary evolution and it cannot be proved using the trace condition alone.
