Since \(B-A\ge 0\), positivity gives \(T(B-A)\ge 0\). By linearity, this is \(T(B)-T(A)\ge 0\).

Regard \(T\) as the associated positive linear map (Definition 4.4.1). For positive linear maps between matrix \(C^*\)-algebras, \(T(A^\ast )=T(A)^\ast \). Since the star on each matrix algebra is the conjugate transpose, \(T(A^\dagger )=T(A^\ast )=T(A)^\ast =T(A)^\dagger \).

Write \(X=\sum _{i,j}X_{ij}e_{ij}\) in the standard matrix units and use linearity in \(X\). For \(X=e_{ij}\) one has \(\operatorname{tr}(E^*(\rho )e_{ij})=(E^*(\rho ))_{ji}\), while the definition of \(E^*\) gives \((E^*(\rho ))_{ji}=\operatorname{tr}(\rho E(e_{ij}))\).

Fix \(X\in M_{D}(\mathbb {C})\) and an arbitrary matrix \(N\in M_{D'}(\mathbb {C})\). Applying (??) first to \(E^* : M_{D'}(\mathbb {C}) \to M_{D}(\mathbb {C})\) and then to \(E\) gives

Both \((E^*)^*(X)\) and \(E(X)\) lie in \(M_{D'}(\mathbb {C})\), so, since this holds for every \(N\in M_{D'}(\mathbb {C})\), nondegeneracy of the trace pairing on \(M_{D'}(\mathbb {C})\) gives \((E^*)^*=E\).

Test the hypothesis on rank-one positive semidefinite matrices \(B=xx^\dagger \). The identity \(\operatorname{tr}(Axx^\dagger )=x^\dagger A x\) and the hypothesis give \(x^\dagger A x\geq 0\) for every vector \(x\), so the Hermitian matrix \(A\) is positive semidefinite.

Let \(\rho \geq 0\). First \(E^*(\rho )\) is Hermitian. Indeed, Theorem A.1.2 gives \(E(A^\dagger )=E(A)^\dagger \), and hence, for matrix units \(e_{ij}\),

For every \(B\geq 0\), (??) gives \(\operatorname{tr}(E^*(\rho )B)=\operatorname{tr}(\rho E(B))\). Since \(E(B)\geq 0\), the right-hand side is non-negative by positivity of the trace product of two positive semidefinite matrices. Self-duality of the positive semidefinite cone therefore gives \(E^*(\rho )\geq 0\).

One direction is Theorem A.1.6. The other follows by applying the same theorem to \(E^*\) and using Theorem A.1.4.

Define \(F : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})\) by \(F(X) = f(X)\mathbb {1}\). For \(X \geq 0\) one has \(f(X) \geq 0\), so \(F(X) = f(X)\mathbb {1}\geq 0\) and \(F\) is positive. Set \(\sigma = D^{-1}\mathbb {1}\) and \(\rho = F^*(\sigma )\). Positivity of the trace-pairing adjoint gives \(\rho \geq 0\), while (??) gives

Taking \(X = \mathbb {1}\) yields \(\operatorname{tr}(\rho ) = 1\).

The assertion is immediate when \(\psi =0\). Otherwise set \(s=\langle \psi ,\psi \rangle \), \(P=|\psi \rangle \! \langle \psi |\), and \(Q=s^{-1}P\). Then \(Q\) is the orthogonal projection onto \(\mathbb {C}\psi \). Write \(B=cP-A\geq 0\). Since \(A+B=cP\) vanishes on the kernel of \(Q\), positivity implies \(A(\mathbb {1}-Q)=0\). Taking adjoints gives \((\mathbb {1}-Q)A=0\), and therefore \(A=QAQ\). Finally,

The coefficient is non-negative because \(A\geq 0\) and \(s{\gt}0\).

Write \(P_v=|v\rangle \! \langle v|\). If \(D=0\), then \(M_{D}(\mathbb {C})\) is the zero vector space, so the conclusion holds with \(c=0\). If \(D=1\), the single matrix unit is \(P_{e_0}\). The hypothesis gives \(T(P_{e_0})=c_{e_0}P_{e_0}\), and linearity gives \(T=c_{e_0}\, \operatorname{id}\) on all of \(M_{1}(\mathbb {C})\).

It remains to consider \(D\geq 2\). Set \(c=c_{e_0}\). For \(i\ne j\), the identities

give, from their \((i,i)\), \((j,j)\), and \((i,j)\) entries, \(c_{e_i+e_j}=c_{e_i-e_j}=c_{e_i}=c_{e_j}\). The imaginary parallelogram identity and its image under \(T\) are

Comparing the same three entries gives \(c_{e_i+\mathrm{i}e_j}=c_{e_i-\mathrm{i}e_j}=c_{e_i}=c_{e_j}\). Taking the pair \((0,i)\) shows that \(c_{e_i}=c\) for every \(i\). The hypothesis therefore gives \(T(E_{ii})=cE_{ii}\) for the diagonal matrix units. For \(i\ne j\), the polarization identity

gives \(T(E_{ij})=cE_{ij}\). Thus \(T=c\, \operatorname{id}\) on every matrix unit.