Set

Applying the unital completely positive map blockwise preserves positive semidefiniteness; unitality ensures that the \((2,2)\)-block maps to \(\mathbb {1}\). The Schur complement of this block gives (3).

The operator inequality \(X^\dagger X\leq \| X\| _\infty ^2\mathbb {1}\), positivity of \(E\), and unitality give \(E(X^\dagger X)\leq \| X\| _\infty ^2\mathbb {1}\). Combining this with (3) yields

The \(C^*\)-identity now gives \(\| E(X)\| _\infty ^2\leq \| X\| _\infty ^2\), which is (4).

Choose a nonzero matrix \(X\) such that \(E(X)=\mu X\). Then Theorem 6.1.6 gives \(|\mu |\| X\| _\infty =\| E(X)\| _\infty \leq \| X\| _\infty \). Since \(\| X\| _\infty {\gt}0\), division by this norm proves (5).

The adjoint operators \(\{ K_i^\dagger \} \) satisfy \(\sum _i K_i^\dagger (K_i^\dagger )^\dagger = \sum _i K_i^\dagger K_i = \mathbb {1}\), so they form a unital family. Apply Theorem 6.1.5 to \(\{ K_i^\dagger \} \).

By (3), \(E(X^\dagger X) - E(X)^\dagger E(X) \ge 0\), so \(\operatorname{tr}(E(X^\dagger X)) \ge \operatorname{tr}(E(X)^\dagger E(X))\). Trace preservation gives \(\operatorname{tr}(E(X^\dagger X)) = \operatorname{tr}(X^\dagger X)\).

Expand the right-hand side of (7) using (8), distribute the products, and use unitality \(\sum _i K_i K_i^\dagger = \mathbb {1}\) to recover the Kadison–Schwarz gap.

By (7), the vanishing Kadison–Schwarz gap is \(\sum _i R_i^\dagger R_i=0\), with \(R_i\) given by (8). Each summand is positive semidefinite, so each \(R_i\) vanishes. This gives (9).

Set \(G:=E(X^\dagger X)-E(X)^\dagger E(X)\). By (3), \(G\geq 0\). Trace preservation, \(E(X)=\mu X\), and \(|\mu |=1\) give

A positive semidefinite matrix with zero trace vanishes, which proves (10).

By (9), \(X K_i^\dagger =K_i^\dagger E(X)\) for every \(i\). Taking conjugate transposes gives \(K_iX^\dagger =E(X)^\dagger K_i\). Therefore,

By (9), \(X K_i^\dagger =K_i^\dagger E(X)\) for every \(i\). Hence

Let \(R(A)\) be the right tensor factor of \(E(A)\), so that \(E(A)=\mathbb {1}_m\otimes R(A)\). For \(Y\geq 0\) and \(P_\psi =|\psi \rangle \! \langle \psi |\), the inequality \(Y\leq \operatorname{tr}(Y)\mathbb {1}_m\) and positivity give

By Lemma 6.12.2, for each \(v\in \mathbb {C}^d\) there is a scalar \(c_v\) such that \(R(Y\otimes P_v)=c_vP_v\). By Theorem 6.12.3, applied to the map \(X\mapsto R(Y\otimes X)\), there is a scalar \(c_Y\) such that \(R(Y\otimes X)=c_YX\) for every \(X\in M_{d}(\mathbb {C})\).

Fix a coordinate projection \(P_0\) and define \(f(Y)=R(Y\otimes P_0)_{00}\). Rank-one polarization gives \(R(Y\otimes X)=f(Y)X\) for all \(Y\in M_{m}(\mathbb {C})\) and \(X\in M_{d}(\mathbb {C})\). Positivity of \(R\) gives \(f(Y)\geq 0\) when \(Y\geq 0\), and the retraction property gives \(f(\mathbb {1}_m)=1\). Theorem 6.12.1 therefore supplies a density matrix \(\rho \) satisfying \(f(Y)=\operatorname{tr}(\rho Y)\).

Finally, on simple tensors,

Since matrix units are simple tensors, linearity proves (13) for every \(A\in M_{md}(\mathbb {C})\).

The standard finite-dimensional Wedderburn theorem in [ Wol12 , Proposition 1.5 ] supplies \(U\) and the block form (14). Conjugate by \(U\) to identify the ambient space with \(\bigoplus _k\mathbb {C}^{m_k}\otimes \mathbb {C}^{d_k}\). Let \(P_k\) be the central projection onto the \(k\)-th summand and let \(T_k\) be the \(k\)-th diagonal block of the conjugated map. Since \(T_k(\mathbb {1}-P_k)=0\), positivity implies \(T_k((\mathbb {1}-P_k)A)=T_k(A(\mathbb {1}-P_k))=0\). Hence \(T_k(A)=T_k(P_kAP_k)\). The restriction to the \(k\)-th diagonal summand is a positive retraction onto \(\mathbb {1}_{m_k}\otimes M_{d_k}(\mathbb {C})\), so the one-factor theorem gives the density matrix \(\rho _k\). Taking the direct sum and conjugating back proves (15). Entrywise expansion of the partial trace gives

In finite dimensions, \(\operatorname{ran}(A) \subseteq \operatorname{ran}(B)\) implies \(A=B(B^\dagger B)^\dagger B^\dagger A\), where \((B^\dagger B)^\dagger \) denotes the Moore–Penrose pseudoinverse. Set \(C=(B^\dagger B)^\dagger B^\dagger A\).

The pointwise hypothesis is exactly \(\operatorname{ran}(A)\subseteq \operatorname{ran}(B)\). Apply the preceding range-inclusion factorization theorem.

Define the Kadison–Schwarz gap \(G=E(X^\dagger X)-E(X)^\dagger E(X)\). By Theorem 6.1.5, \(G\succeq 0\). The adjointness identity of Lemma 6.13.1, the fixed-point equation \(E^*(\rho )=\rho \), the eigenvalue equation \(E(X)=\mu X\), and \(|\mu |=1\) give

Since \(\rho {\gt}0\) and \(G\succeq 0\), Lemma 6.13.2 yields \(G=0\), which is the asserted equality.

Suppose first that equality holds at \(A^\dagger A\). Apply (16) to \(A+tY\) and \(A+itY\) for real \(t\). Set

Expansion and the equality at \(A^\dagger A\) give, for every real \(t\),

Nonnegativity for both signs of every real parameter forces \(L+R=0\) and \(L-R=0\). Hence \(L=R=0\), which is precisely

Replacing \(Y\) by \(X^\dagger \) yields \(E(XA)=E(X)E(A)\). The second characterization follows by applying the same argument to \(A^\dagger \). The converse implications follow by evaluating the defining product identities at \(A^\dagger \).

A Kraus map preserves adjoints. The Kadison–Schwarz inequality (3) therefore gives, for every \(A\in M_{D}(\mathbb {C})\),

If \(X\) lies in a one-sided multiplicative domain, evaluate the defining identity at \(Y=X^\dagger \). Conversely, the left-domain implication is exactly Theorem 6.2.4, and the right-domain implication follows by applying Theorem 6.2.4 to \(X^\dagger \).

Closure under addition follows from linearity of \(E\). For closure under multiplication, if \(X,Y\in \mathcal{A}_R(E)\), then \(E((XY)Z)=E(X)E(YZ)=E(X)E(Y)E(Z)\) for all \(Z\), using Theorem 6.8.2.4 iteratively; the \(\mathcal{A}_L(E)\) case is analogous. Containment of \(\mathbb {1}\) follows from unitality of \(E\).

If \(X\in \mathcal{A}(E)=\mathcal{A}_R(E)\cap \mathcal{A}_L(E)\), then

The characterization of Theorem 6.8.2.4 shows that \(X^\dagger \in \mathcal{A}(E)\). Combined with Theorem 6.8.2.5, this gives a \(*\)-subalgebra.

Multiplicativity on \(\mathcal{A}(E)\) follows from the definition. For \(X\in \mathcal{A}(E)\), the Kraus commutation relation (9), its adjoint, and unitality give \(E(X^\dagger )=E(X)^\dagger \), so the restriction preserves adjoints.

By Theorem 6.11.1, \(T(a\mathbb {1})\le T(A)\le T(b\mathbb {1})\). For the lower bound, \(T(a\mathbb {1})=aT(\mathbb {1})\). Since \(a\le 0\) and \(T(\mathbb {1})\le \mathbb {1}\), one has \(aT(\mathbb {1})\ge a\mathbb {1}\). For the upper bound, \(T(b\mathbb {1})=bT(\mathbb {1})\le b\mathbb {1}\) since \(b\ge 0\). These bounds give (17).

For Hermitian matrices, the inclusion \(\operatorname{spec}(A)\subseteq [a,b]\) is equivalent to the order bounds \(a\mathbb {1}\le A\le b\mathbb {1}\). Apply Theorem 6.8.3.1 and translate (17) back into spectral bounds.

Let \(S\) and \(T\) be the symmetric endomorphisms associated with \(A\) and \(B\). Fix \(j\), let \(P\) be the span of the first \(j+1\) eigenvectors of \(S\), and let \(Q\) be the span of the last \(d-j\) eigenvectors of \(T\). Since \(\dim P+\dim Q=d+1\), there is a nonzero vector \(x\in P\cap Q\). The two ordered spectral expansions and positivity of \(T-S\) give

Division by \(\lVert x\rVert ^2{\gt}0\) proves (18).

Choose unitary eigenvector matrices \(V_A,V_B\) in decreasing eigenvalue order and set \(U=V_AV_B^\dagger \). By (18) and monotonicity of \(f\),

Hence the corresponding diagonal matrices are ordered. Conjugating this inequality by \(V_A\) and using \(U=V_AV_B^\dagger \) gives (19). The cross-matrix eigenvalue comparison used here is Theorem 6.8.4.1, proved by the preceding variational argument; it is not assumed separately.

It is a non-negative linear combination of the transpose map and the map \(A \mapsto \operatorname{tr}(A) \mathbb {1}\), and both of these maps send positive semidefinite matrices to positive semidefinite matrices.

Write the Schwarz gap as \(F(A)\). One checks that \(F(A + \lambda \mathbb {1}) = F(A)\) for every scalar \(\lambda \), so it is enough to treat the case \(\operatorname{tr}(A) = 0\). In that case a direct \(2 \times 2\) computation shows that \(F(A)\) dominates \(\frac{1}{2} (A^\dagger A)^T\), hence is positive semidefinite.

Compute the Choi matrix of \(T\) and test it on the antisymmetric vector \(|01\rangle - |10\rangle \). The resulting expectation value is negative, so the Choi matrix is not positive semidefinite.

Write \(C\) for the horizontal concatenation \([A\; B]\). The block matrix \(C^\dagger C\) is positive semidefinite. Applying the \(2\)-positive ampliation entrywise yields the PSD block matrix \(\bigl[\begin{smallmatrix} E(A^\dagger A) & E(A^\dagger B) \\ E(B^\dagger A) & E(B^\dagger B) \end{smallmatrix}\bigr]\). Evaluating the PSD quadratic form on the vector \((v,-w)\) and using the side condition \(E(B^\dagger B)w = E(B^\dagger A)v\) to cancel the off-diagonal and diagonal terms yields \(v^\dagger E(A^\dagger A)v - v^\dagger E(A^\dagger B)w \ge 0\).

A \(2\)-positive map is positive, and positive maps preserve adjoints. Applying this to \(A^\dagger B\) and using \((A^\dagger B)^\dagger = B^\dagger A\) gives the identity.

Rewrite \(E(B^\dagger A)\) as \(E(A^\dagger B)^\dagger \) (Lemma 6.9.2.2) in the ampliated PSD block matrix \(\bigl[\begin{smallmatrix} E(A^\dagger A) & E(A^\dagger B) \\ E(B^\dagger A) & E(B^\dagger B) \end{smallmatrix}\bigr]\) from the proof of Lemma 6.9.2.1: if \(E(B^\dagger B)y=0\) then testing the PSD quadratic form on \((0,y)\) forces \(E(A^\dagger B)y=0\).

The kernel inclusion \(\ker (E(B^\dagger B))\subseteq \ker (E(A^\dagger B))\) (Lemma 6.9.2.3) gives, after taking adjoints, that the support projection of \(E(B^\dagger B)\) absorbs \(E(B^\dagger A)\) on the left. Since \(E(B^\dagger B)E(B^\dagger B)^{+}\) is exactly that support projection, the displayed identity follows.

For every \(v\), the vector \(w = E(B^\dagger B)^{+}E(B^\dagger A)\, v\) satisfies the side condition \(E(B^\dagger B)\, w = E(B^\dagger A)\, v\) of the conditional lemma (Lemma 6.9.2.1): this is exactly Lemma 6.9.2.4 evaluated at \(v\). The kernel inclusion \(\ker (E(B^\dagger B))\subseteq \ker (E(A^\dagger B))\), obtained from the ampliated PSD block matrix of the conditional lemma’s proof, gives \(\operatorname {ran}(E(B^\dagger A))\subseteq \operatorname {ran}(E(B^\dagger B))\), so the pseudoinverse is a genuine right-inverse witness for \(E(B^\dagger A)\) on the support of \(E(B^\dagger B)\). Applying the conditional lemma with this \(w\) for every \(v\) gives the displayed matrix inequality.

By trace-adjoint invariance of \(2\)-positivity, Theorem 4.3.2, the map \(E=T^*\) is itself \(2\)-positive. Apply Theorem 6.9.2.5.

Testing the quadratic form on a vector \(x\) reduces to the scalar statement: if \(tb+t^2d\ge 0\) for every real \(t\), where \(b=x^\dagger Bx\) and \(d=x^\dagger Dx\ge 0\), then \(b=0\). Evaluating at \(t=\pm 1\) forces \(d\ge 0\); evaluating at \(t=-\operatorname {Re}(b)/(d+1)\) and completing the square forces \(\operatorname {Re}(b)=0\), and comparing the imaginary parts at \(t=\pm 1\) forces \(\operatorname {Im}(b)=0\). Since this holds for every \(x\), \(B=0\) by polarization.

Write \(Q(U,V) = E(U^\dagger V) - E(U^\dagger B)E(B^\dagger B)^+E(B^\dagger V)\); the per-\(B\) hypothesis is \(Q(U,U)\ge 0\) for every \(U\), and \(A\in \mathcal{A}_B\) means \(Q(A,A)=0\). Applying the hypothesis to \(A + cX\) for \(c\in \mathbb C\) and expanding bilinearly gives

Taking \(c=t\in \mathbb R\) and applying Lemma 6.9.2.9 to the linear coefficient \(Q(A,X)+Q(X,A)\) and the fixed nonnegative quadratic coefficient \(Q(X,X)\) shows \(Q(A,X)+Q(X,A)\) vanishes; taking \(c=tI\) and applying the same lemma to \(I\, Q(A,X) - I\, Q(X,A)\) shows this also vanishes. Combining the two gives \(Q(A,X)=0\), which is the displayed identity.

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 2.4.1). 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, \(T(A^\dagger )=T(A^\ast )=T(A)^\ast =T(A)^\dagger \).

Testing against the matrix unit \(E_{ji}\) gives \(\operatorname{tr}(M E_{ji}) = M_{ij}\), so \(\operatorname{tr}(M N) = 0\) for all \(N\) forces every entry of \(M\) to vanish.

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 (20) 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 6.11.2 gives \(E(A^\dagger )=E(A)^\dagger \), and hence, for matrix units \(e_{ij}\),

For every \(B\geq 0\), (20) 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 6.11.7. The other follows by applying the same theorem to \(E^*\) and using Theorem 6.11.5.

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 (20) 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.