Write \(x=(f-\mathbb {1}_V)y\). Telescoping gives \(A_N(x)=(f^Ny-y)/N\). Boundedness of the orbit of \(y\) makes the numerator bounded, so the right-hand side tends to zero.

If \(z\) belongs to both summands, then \(A_N(z)=z\) because \(z\in \ker (f-\mathbb {1}_V)\), while Lemma 25.1.2 gives \(A_N(z)\to 0\). Hence \(z=0\). Rank–nullity then gives (??).

By Lemma 25.1.3, decompose

The fixed component has constant Cesàro averages, whereas the averages of the second component tend to zero by Lemma 25.1.2. This proves (??). Since \(P_fx\in \ker (f-\mathbb {1}_V)\), one has \(f(P_fx)=P_fx\), and hence \(fP_f=P_f\). Also, \((f-\mathbb {1}_V)x\in \operatorname{range}(f-\mathbb {1}_V)\) and \(P_f((f-\mathbb {1}_V)x)=0\), so \(P_f(fx)=P_fx\) and hence \(P_ff=P_f\). The range and idempotence identities in (??) follow directly from the projection construction.

For \(X\succeq 0\), positivity and induction on \(n\) give

Thus the orbit lies in a bounded trace section of the positive cone. Apply this to the positive and negative parts of a Hermitian matrix, and then write an arbitrary matrix as a complex linear combination of two Hermitian matrices.

Trace preservation implies trace nonincrease, so Theorem 25.1.6 gives bounded orbits. Theorem 25.1.5 then gives the projection and the fixed-point characterization (??). Its positivity follows from Theorem J.1.1, trace preservation from Theorem J.1.2, and its behavior on the identity from Theorem J.1.3.

Theorem 25.1.7 makes \(P_T\) positive and trace-preserving. Positivity passes to its trace adjoint, while Theorem J.2.1 gives \(P_T^*(\mathbb {1})=\mathbb {1}\). The remaining assertions in (??) are Theorem J.2.2.

The fixed-point equation gives the common difference \(Y=E(P_+)-P_+=E(P_-)-P_-\). Hence \(P_++Y\) and \(P_-+Y\) are positive semidefinite, while trace preservation gives \(\operatorname{tr}(Y)=0\). Diagonalize \(P_+\). Since \(P_+P_-=P_-P_+=0\), each eigenvector belongs to the kernel of \(P_+\) or to the kernel of \(P_-\). Positivity of the corresponding matrix \(P_++Y\) or \(P_-+Y\) shows that every diagonal coefficient of \(Y\) in this basis is non-negative. Their sum is \(\operatorname{tr}(Y)=0\), so every such coefficient vanishes. A positive semidefinite matrix with a zero diagonal coefficient annihilates the corresponding basis vector. Applying this to \(P_++Y\) or \(P_-+Y\) shows that \(Y\) annihilates the entire eigenbasis. Thus \(Y=0\), and consequently \(E(P_\pm )=P_\pm \).

The Kadison–Schwarz equality for fixed points forces both one-sided multiplicative identities, so the fixed point lies in the full multiplicative domain.

Positivity gives \(E(X^\dagger )=E(X)^\dagger \), so fixed points are closed under adjoints. For a fixed point \(X\), set \(\Delta _X=E(X^\dagger X)-E(X^\dagger )E(X)\succeq 0\). The trace-pairing identity, \(E^*(\rho )=\rho \), and \(E(X)=X\) give

Since \(\rho {\gt}0\), Lemma J.3.1 forces \(\Delta _X=0\). Applying the same argument to \(X^\dagger \) gives equality in the other one-sided Schwarz inequality. Hence Theorem 5.6.1.3 places \(X\) in the full multiplicative domain. Consequently, for fixed points \(X,Y\), \(E(XY)=E(X)E(Y)=XY\). Linearity and unitality supply the remaining subalgebra axioms.

Idempotence gives \(E(E(X))=E(X)\), so \(E(X)\) is a fixed point. By the argument in Theorem 25.2.4, the Kadison–Schwarz gap for the fixed point \(E(X)\) vanishes, hence \(E(E(X)^\dagger E(X))=E(X)^\dagger E(X)\). Fixed points are closed under adjoints, so the same argument applied to \(E(X)^\dagger \) gives \(E(E(X)E(X)^\dagger )=E(X)E(X)^\dagger \). The vanishing of both Kadison–Schwarz gaps, again by the argument of Theorem 25.2.4, places \(E(X)\) in the multiplicative domain.

Theorem 25.2.7 places \(E(X)\) and \(E(Y)\) in the multiplicative domain. Therefore \(E(E(X)E(Y))=E(E(X))E(E(Y))=E(X)E(Y)\) by idempotence.

Apply Theorem 25.2.4 to the adjoint family \(\{ K_i^\dagger \} \).

For the inclusion “\(\supseteq \)”, if \(X\) commutes with all \(K_i\) and \(K_i^\dagger \), then \(E^*(X)=\sum _iK_i^\dagger XK_i =X\sum _iK_i^\dagger K_i=X\) by trace preservation. For the inclusion “\(\subseteq \)”, any \(*\)-subalgebra inside \(\operatorname{Fix}(E^*)\) lies in the Kraus commutant, because for \(X\in \operatorname{Fix}(E^*)\) with \(X^\dagger X\in \operatorname{Fix}(E^*)\) the Kadison–Schwarz gap vanishes. This forces \(XK_i=K_iX\) and \(XK_i^\dagger =K_i^\dagger X\).

By Theorem 25.2.12, every \(*\)-subalgebra inside the adjoint fixed points is contained in the Kraus commutant, while the Kraus commutant itself is such a \(*\)-subalgebra.

Write \(S=\sqrt{\sigma }\) and define the Kraus family \(K_{j,i}=|j\rangle \! \langle i|S\), indexed by \((j,i)\in \{ 1,\ldots ,D\} ^2\). Using \(|j\rangle \! \langle i|M|i\rangle \! \langle j|=M_{ii}|j\rangle \! \langle j|\) and summing first over \(i\) and then over \(j\) gives

Here cyclicity of the trace and \(S^2=\sigma \) were used. Thus \(E_\sigma \) is the Kraus map in (??) scaled by the non-negative real number \(\operatorname{tr}(\sigma )^{-1}\), and is therefore completely positive.

Direct computation gives

Expand the transfer map using (??) and use \((\mathbb {1}-P)A^iP=0\) to reduce each Kraus term to the \(P\)-corner.

Write \(E\) in Kraus form \(E(X)=\sum _iK_iXK_i^\dagger \). The relations \(E(\rho )=\rho \) and \(\rho \geq 0\) imply \((\mathbb {1}-P)K_iP=0\) for every \(i\). Apply Lemma 25.4.1.

The support projection \(P\) of \(\rho _\infty \) satisfies \(PE(PXP)P=E(PXP)\) for every \(X\) by Lemma 25.4.2. Irreducibility gives \(P=0\) or \(P=\mathbb {1}\). Since \(\rho _\infty \neq 0\), one has \(P\neq 0\), and therefore \(P=\mathbb {1}\).

Diagonalize \(\rho =U\operatorname{diag}(\lambda )U^\dagger \) with \(\lambda _j\geq 0\) and set \(w:=U^\dagger v\). The hypothesis \(\rho v=0\) gives \(\lambda _jw_j=0\) for every \(j\), so \(w_j=0\) whenever \(\lambda _j{\gt}0\). In the eigenbasis the support projection acts as \(\operatorname{diag}(\chi _{\lambda _j{\gt}0})\), and hence \(Pv=U\operatorname{diag}(\chi _{\lambda _j{\gt}0})w=0\).

Hermiticity follows from Hermiticity of \(\rho \) and the product-star identity. For strict positivity, let \(w\neq 0\) and set \(u:=V^\dagger w\). The identity \(VV^\dagger =\mathbb {1}_k\) gives \(u\neq 0\), while \(V^\dagger V=P\) gives \(Pu=u\). The quadratic form satisfies \(w^\dagger (V\rho V^\dagger )w=u^\dagger \rho u\geq 0\). If it vanishes, then \(\rho u=0\), so Lemma 25.4.1.1 gives \(Pu=0\), contradicting \(Pu=u\neq 0\).

The set in (??) is the fixed-point set of the compressed map \(\widetilde{T}^*\) on the support sector. Since \(\rho \) is fixed by \(T\), Theorem 25.4.2 gives \((\mathbb {1}-Q)K_iQ=0\), so the compressed family \(QK_iQ\) is supported on the corner and is trace-preserving there: \(\sum _i(QK_iQ)^\dagger (QK_iQ)=Q\). Compressing this family to the support sector gives a unital map, while the compression of \(\rho \) is a positive-definite fixed point by Theorem 25.4.1.4. The structure theorem for fixed points of unital Schwarz maps with a positive-definite fixed point (Theorem 25.2.10) makes the compressed fixed points a \(*\)-algebra. The compression isomorphism is multiplicative and star-preserving, intertwines the compressed adjoint map with the corner-restricted map, and transports this structure back to the corner.

The subalgebra is finite-dimensional, hence Artinian, so its Jacobson radical \(J\) is nilpotent. If \(x\in J\), then \(x^*x\in J\) because the radical is a two-sided ideal. Thus \(x^*x\) is a positive semidefinite nilpotent matrix, so all its eigenvalues vanish and hence \(x^*x=0\). Therefore \(x=0\), the radical vanishes, and the subalgebra is semisimple.

Combine semisimplicity (Lemma 25.5.1) with the Wedderburn–Artin structure theorem for finite-dimensional semisimple algebras over the algebraically closed field \(\mathbb {C}\).

Fix \(A\in S\). Since \(S\) is closed under adjoint, \(A^\dagger \in S\), and therefore \(A^\dagger W\subseteq W\). For \(v\in W^\perp \) and \(w\in W\),

Hence \(Av\in W^\perp \).

In a finite-dimensional inner-product space, the orthogonal complement of a subspace is a vector-space complement. Its invariance is Lemma 25.5.3.

Induct on the dimension of \(W\). If \(W=0\), the empty family works; if \(W\) is irreducible, the family \(\{ W\} \) works. Otherwise, \(W\) has a proper nonzero invariant subspace \(U\). By Lemma 25.5.4, the orthogonal complement of \(U\) inside \(W\) is invariant. Both subspaces are strictly smaller than \(W\) and span it. Their inductive decompositions lie in orthogonal subspaces, so their union is the required family.

Apply Lemma 25.5.6 to \(W=\mathbb {C}^D\), which is invariant under every member of \(S\).

The underlying set of an \(S\)-submodule of \(\mathbb {C}^D\) is closed under addition and under the action of every member of \(S\). Since the complex scalars lie in \(S\), it is an invariant complex subspace. Conversely, an invariant complex subspace is closed under addition and the action of every member of \(S\), so it underlies an \(S\)-submodule with the same elements. Under this correspondence, submodules are exactly invariant subspaces. Thus the absence of nonzero proper submodules is equivalent to irreducibility.

The ring \(S\) is semisimple by Lemma 25.5.1, and every module over a semisimple ring is semisimple.

The kernel of \(f\) in \(W\) is invariant under \(S\), so it is either \(W\) or \(\{ 0\} \). In the first case \(f\) vanishes on \(W\); in the second, \(f\) is injective on \(W\). Likewise, \(f(W)\) is an invariant subspace of \(W'\), so it is either \(\{ 0\} \) or \(W'\). Thus \(f\) is zero on \(W\), or it is injective there with image \(W'\).

Restricting \(f\) to \(W\) gives an operator on a nonzero finite-dimensional complex vector space, so it has an eigenvalue \(c\) and an eigenvector in \(W\). The \(c\)-eigenspace in \(W\) is nonzero and invariant under \(S\), because \(f\) commutes with every member of \(S\) on \(W\). By irreducibility it is all of \(W\), so \(f\) acts as multiplication by \(c\) throughout \(W\).

Let \(P_W\) be the orthogonal projection onto \(W\). The operator \(g=P_Wf^*f\) maps \(W\) into itself and commutes there with every member of \(S\). Indeed, for \(A\in S\), the adjoint \(A^*\) also belongs to \(S\), and the intertwining identity for \(f\), paired against vectors in \(W\), gives \(g(Ax)=A(gx)\). By Lemma 25.5.11, \(gx=\lambda x\) on \(W\). Pairing with \(y\in W\) and using that \(P_W\) is invisible against \(W\) gives \(\langle f(x),f(y)\rangle =\overline{\lambda }\langle x,y\rangle \). Taking \(x=y\neq 0\) shows that \(c=\overline{\lambda }=\lVert f(x)\rVert ^2/\lVert x\rVert ^2\) is non-negative, proving (??).

Choose \(x\in W\) with \(f(x)\neq 0\). Then \(\lVert f(x)\rVert ^2=c\lVert x\rVert ^2\), and both squared norms are positive, so \(c{\gt}0\).

If \(W\) and \(W'\) are of the same type, then a nonzero intertwiner is injective on \(W\) with image \(W'\) by Lemma 25.5.10. Conversely, an intertwiner injective on \(W\) with image \(W'\) cannot vanish on \(W\), because \(W'\) is nonzero.

By Lemma 25.5.16, same type gives an intertwiner \(f\) that is injective on \(W\) with image \(W'\). Since \(W'\) is nonzero, \(f\) does not vanish on \(W\), so Lemma 25.5.14 gives \(c{\gt}0\) such that \(\langle f(x),f(y)\rangle =c\langle x,y\rangle \). Set \(u=f/\sqrt{c}\). Scaling by the nonzero factor \(1/\sqrt{c}\) preserves the intertwining identity and the image \(W'\), and gives \(\langle u(x),u(y)\rangle =\langle x,y\rangle \).

The identity operator carries \(W\) into itself, commutes with every member of \(S\), and is nonzero on \(W\) because \(W\) is nonzero.

An isomorphism \(f\) from \(W\) onto \(W'\) has an inverse \(g\), obtained by projecting onto \(W'\), inverting on that subspace, and including into \(W\). For every \(A\in S\) and \(w\in W'\),

Thus \(g\) is an intertwiner from \(W'\) into \(W\), injective on \(W'\) with image \(W\), and the two pieces are of the same type in the reverse order.

Compose an isomorphism \(f:W\to W'\) with an isomorphism \(g:W'\to W''\). For every \(A\in S\) and \(x\in W\),

Thus \(g\circ f\) is an intertwiner from \(W\) into \(W''\), injective on \(W\) with image \(W''\), and is nonzero on \(W\) because \(W''\) is nonzero.

Decompose \(x=Px+P^\perp x\). Each member of \(S\) sends \(Px\) into \(W\) and \(P^\perp x\) into \(W^\perp \) by Lemma 25.5.3. Therefore

The orthogonal projection \(P_{W'}\) onto \(W'\) commutes with every member of \(S\) by Lemma 25.5.21 and carries \(W\) into \(W'\), so it is an intertwiner. By Lemma 25.5.10, it is either an isomorphism, which would make the pieces of the same type, or zero on \(W\). Thus \(P_{W'}x=0\) for \(x\in W\). For \(x\in W\) and \(y\in W'\),

Each pair \(W\in \mathcal D_1\), \(W'\in \mathcal D_2\) is of different type, hence orthogonal by Lemma 25.5.22. Orthogonality to a fixed subspace is preserved by suprema on each side, so the two suprema are orthogonal.

Suppose \(W\) is of a different type from every piece of \(\mathcal D\). By Lemma 25.5.22, \(W\) is orthogonal to every piece of \(\mathcal D\), hence to their supremum and therefore to itself. This contradicts the nonvanishing condition in irreducibility.

Theorem 25.5.7 supplies a finite, pairwise orthogonal family \(\mathcal{D}\) of irreducible pieces with \(\bigvee _{W\in \mathcal{D}}W=\mathbb {C}^D\). For each piece \(W\), let \(\mathcal{D}_W\) be the class of pieces of \(\mathcal{D}\) of the same type as \(W\), and let \(C_W\) be its supremum. Each piece lies in its own class by Lemma 25.5.18, so each class is nonempty, the classes cover \(\mathcal{D}\), and the components span \(\mathbb {C}^D\). Each class is contained in \(\mathcal{D}\), so its pieces are pairwise orthogonal. Two pieces in one class are of the same type through the representative, by Lemmas 25.5.19 and 25.5.20. If two representatives are of the same type, transitivity makes their classes, and hence their components, equal.

Thus distinct components arise from representatives \(W,W'\) that are not of the same type. Write \(\sim \) for the same-type relation. For \(a\in \mathcal{D}_W\) and \(b\in \mathcal{D}_{W'}\), one has \(W\sim a\) and \(W'\sim b\). If \(a\sim b\), symmetry and transitivity would give \(W\sim a\sim b\sim W'\), a contradiction. Hence no piece of \(\mathcal{D}_W\) is of the same type as a piece of \(\mathcal{D}_{W'}\), and Lemma 25.5.23 makes the components orthogonal.

Finally, let \(V\leq C_W\) and \(V'\leq C_{W'}\) be irreducible under \(S\), with \(C_W\neq C_{W'}\). By Lemma 25.5.24, there are pieces \(a\in \mathcal{D}_W\) and \(b\in \mathcal{D}_{W'}\) with \(V\sim a\) and \(V'\sim b\). If \(V\sim V'\), symmetry and transitivity would give \(W\sim a\sim V\sim V'\sim b\sim W'\), again contradicting that \(W\) and \(W'\) are not of the same type.

Fix a piece \(W\in \mathcal{D}\), of dimension \(d\), with an orthonormal basis \(e_0,\ldots ,e_{d-1}\), and enumerate \(\mathcal{D}=\{ W'_0,\ldots ,W'_{m-1}\} \). Every piece of \(\mathcal{D}\) is of the same type as \(W\), by hypothesis for the other pieces and by Lemma 25.5.18 for \(W\) itself. Lemma 25.5.17 therefore provides intertwiners \(u_i\) mapping \(W\) onto \(W'_i\) and preserving the inner product on \(W\). Set \(f_{i,j}=u_i(e_j)\).

Within one copy, the vectors \(f_{i,0},\ldots ,f_{i,d-1}\) are orthonormal because \(u_i\) preserves inner products; across copies they are orthogonal because distinct pieces of \(\mathcal{D}\) are. Since \(u_i\) maps \(W\) onto \(W'_i\), the vectors of the \(i\)-th copy span \(W'_i\). Thus each \(W'_i\) has dimension \(d\), and all the vectors together span \(C\).

For \(A\in S\), invariance of \(W\) gives \(Ae_j=\sum _{j'}B_{j'j}e_{j'}\), where \(B_{j'j}=\langle e_{j'},Ae_j\rangle \). The intertwining identity gives

This is (??).

Theorem 25.6.1 writes \(\mathbb {C}^D\) as the supremum of finitely many pairwise orthogonal components, each the supremum of a finite, nonempty class of pairwise orthogonal irreducible pieces of one type. Applying Theorem 25.6.2 to each class equips each component with its adapted orthonormal basis.

Theorem 25.6.1 writes \(\mathbb {C}^D\) as the supremum of finitely many pairwise orthogonal isotypic components, each the supremum of a finite, nonempty same-type class of pairwise orthogonal irreducible pieces. Theorem 25.6.2 equips the \(k\)-th component with an adapted orthonormal family \((f_{k,i,j})_{i{\lt}m_k,\, j{\lt}d_k}\) spanning it. Here \(m_k\) is the number of pieces in the class, which is positive because the class is nonempty, and \(d_k\) is their common dimension, which is positive because irreducible pieces are nonzero.

Concatenating these families over \(k\) yields an orthonormal family: vectors in one component are orthonormal by construction, and vectors in distinct components are orthogonal. The concatenated family spans \(\mathbb {C}^D\) because each component is spanned by its family and the components have supremum \(\mathbb {C}^D\); hence it is an orthonormal basis. The action property (??) holds componentwise. Each row spans one of the irreducible pieces in its component’s class, and Theorem 25.6.1 shows that irreducible subspaces in distinct components are never of the same type.

Let \(U\) be the change-of-basis matrix from the standard orthonormal basis of \(\mathbb {C}^D\) to \((f_{k,i,j})\), so the columns of \(U\) are the basis vectors. A change of basis between orthonormal bases is unitary. Both bases have \(D\) elements, which identifies the index set of triples with \(\{ 0,\ldots ,D-1\} \) and gives \(\sum _kd_km_k=D\). The matrix \(U^\dagger AU=U^{-1}AU\) represents \(x\mapsto Ax\) in the basis \((f_{k,i,j})\), with entries

This is the block-diagonal matrix in (??).

Let \((f_{k,i,j})\) be the adapted orthonormal basis of Theorem 25.6.4, and let \(U\) be the unitary supplied by Lemma 25.6.5. Every \(A\in S\) acts on the basis by matrices \(B_k\) on the irreducible index, identically across the multiplicity index, so (??) gives (??).

The space \(\mathbb {C}^D\) is a semisimple module over \(S\) by Lemma 25.5.9, and the linear operators commuting with all members of \(S\) are exactly the endomorphisms of this module. The hypothesis therefore makes \(T\) linear over the endomorphism ring of the module. By the Jacobson density theorem, for every finite set of vectors there is a member of \(S\) acting on them as \(T\) does. Applied to a basis of \(\mathbb {C}^D\), this produces a member of \(S\) whose action agrees with \(T\) on a basis, hence equals \(T\).

Evaluate (??) between the basis vectors \(|a,b,c\rangle \) and \(|a',b',c'\rangle \). The identity matrices collapse the sums over the first and third intermediate indices, leaving

This is the \((b,b')\) entry of \(X_{aa'}Y_{cc'}=Y_{cc'}X_{aa'}\).

Choose the adapted orthonormal basis \((f_{k,i,j})\) for \(S\). Formula (??) is its defining action property. If \(T\) commutes with \(S\), the row transport

also commutes with \(S\), and so does \(\tau _{k,i',i}T\). Schur’s lemma makes this composite scalar on each irreducible row and makes its matrix coefficients vanish between distinct isotypic components. Expanding \(Tf_{k,i,j}\) in the orthonormal basis therefore gives (??), with coefficients \((C_k)_{i'i}\) independent of \(j\).

Let \(S\) be the commutant of the family \((Y_{cc'})_{c,c'}\) together with its adjoints. Hermiticity gives \(Y_{cc'}^*=Y_{c'c}\). Theorem 25.6.8 therefore places every \(X_{aa'}\) in \(S\). Conversely, every \(Y_{cc'}\) commutes with all members of \(S\). Apply Theorem 25.6.9 to \(S\): its members act on one tensor factor of each summand, and matrices commuting with \(S\) act on the complementary factor. Specializing (??) and (??) to \(X_{aa'}\) and \(Y_{cc'}\) gives (??).

Use the orthonormal decomposition from Theorem 25.6.10. Let \(U\) be the unitary whose columns are the adapted basis vectors. Assemble the coefficient matrices \((B_q^{aa'})_{a,a'}\) into a matrix \(R_q\) on \(H_A\otimes H_{q,l}\), and similarly assemble \((C_q^{cc'})_{c,c'}\) into a matrix \(S_q\) on \(H_{q,r}\otimes H_C\). Taking matrix entries in the adapted basis turns (??) into (??). Unitary conjugation preserves Hermiticity. Restricting the two conjugated Hermitian operators to a summand, and fixing one basis vector in the nonempty complementary factor, shows that every \(R_q\) and every \(S_q\) is Hermitian.

Let \(V_A=\mathbb {1}_A\otimes U\) and \(V_C=U\otimes \mathbb {1}_C\). Theorem 25.6.11 gives

Since \(V_AV_A^\dagger =\mathbb {1}\) and \(V_CV_C^\dagger =\mathbb {1}\), conjugating these identities by \(V_A\) and \(V_C\), respectively, gives (??).

Let \((f_{k,i,j})\) be the adapted orthonormal basis of Theorem 25.6.4, write \(W_{k,i}\) for the span of the row \((f_{k,i,j})_{j{\lt}d_k}\), and let \(U\) be the unitary supplied by Lemma 25.6.5. Every member of \(S\) satisfies (??), which is the containment direction.

For the reverse inclusion, fix blocks \((B_k)_k\) and let \(T\) act on the adapted basis by \(Tf_{k,i,j}=\sum _{j'}(B_k)_{j'j}f_{k,i,j'}\). By Lemma 25.6.7, it suffices to show that \(T\) commutes with every operator \(g\) that commutes with all members of \(S\). For a component index \(k\) and multiplicity indices \(i,i'\), consider

Because every member of \(S\) acts on the rows \((k,i')\) and \((k,i)\) by the same matrix, \(\tau \) commutes with all members of \(S\), and so does \(\tau \circ g\). The composite maps \(\mathbb {C}^D\) into \(W_{k,i}\), and \(W_{k,i}\) is irreducible. Hence Lemma 25.5.11 gives a scalar \(\gamma ^{(k)}_{i'i}\in \mathbb {C}\) such that \(\tau (gx)=\gamma ^{(k)}_{i'i}x\) for every \(x\in W_{k,i}\). Evaluating at \(x=f_{k,i,j}\) gives

Matrix coefficients of \(g\) between distinct components vanish. Indeed, for \(k'\neq k\), define \(\tau \) using the row \((k',i')\). Then \(\tau \circ g\) is an intertwiner from \(W_{k,i}\) into \(W_{k',i'}\). A nonzero coefficient would make this intertwiner nonzero on \(W_{k,i}\), so the two rows would be of the same type, contrary to Theorem 25.6.4. Expanding \(gf_{k,i,j}\) and using (??) therefore gives

Thus \(g\) has blocks \(C_k\otimes \mathbb {1}_{d_k}\) in the adapted basis, complementary to the blocks of \(T\). The two actions commute on every basis vector, so \(Tg=gT\). By Lemma 25.6.7, \(T\in S\).

Finally, if \(A\) satisfies (??), then \(U^\dagger AU=U^\dagger TU\) for the member \(T\in S\) constructed from the blocks \((B_k)_k\). Hence \(A=T\in S\).

The subalgebra is finite-dimensional as a subalgebra of \(M_{D}(\mathbb {C})\), hence Artinian. Following  [ Wol12 , Theorem 6.14 ] , it suffices to show that the Jacobson radical \(J\) vanishes. If \(x\in J\), then \(x^*x\in J\) by left-ideal closure. The radical of an Artinian ring is nilpotent, so \(x^*x\) is nilpotent. A self-adjoint nilpotent matrix satisfies \(\lVert x^*x\rVert ^{2^k}=\lVert (x^*x)^{2^k}\rVert =0\), hence \(x^*x=0\), and the \(C^*\)-identity gives \(x=0\).

Combine semisimplicity (Theorem 25.6.14) with the Wedderburn–Artin structure theorem for finite-dimensional semisimple algebras over an algebraically closed field.

By Theorem 25.2.4, the fixed-point set \(\operatorname{Fix}(E)\) is a \(*\)-subalgebra of \(M_{D}(\mathbb {C})\). Applying Theorem 25.5.2 gives the asserted product decomposition with each block size \(d_k\geq 1\).

Apply the abstract Wedderburn decomposition (Theorem 25.6.15). Since the adjoint-fixed-point algebra is realized as a subalgebra of \(M_{D}(\mathbb {C})\), this realization also yields multiplicities \(m_k\) and the bound \(\sum _kd_km_k\leq D\).

The fixed points of \(E^*\) form a \(*\)-subalgebra of \(M_{D}(\mathbb {C})\) because \(E\) is trace-preserving with a positive definite fixed point (Theorem 25.2.10). Theorem 25.6.13 supplies the unitary \(U\), the dimensions and multiplicities, and the equivalence between membership and the block form (??).

The fixed points of the unital map \(E\) form a \(*\)-subalgebra of \(M_{D}(\mathbb {C})\) by Theorem 25.2.4, and Theorem 25.6.13 supplies the unitary and the equivalence.

The defining trace identity (??) and nondegeneracy of the trace pairing give the identity and composition formulas in (??). Total-trace preservation gives, for every \(X\in \mathcal A\),

so \(T^*(\mathbb {1}_{\mathcal B})=\mathbb {1}_{\mathcal A}\). Finally, if \(\widehat T(X)=\sum _aK_aXK_a^*\), cyclicity of the trace gives \((\widehat T)^*(Y)=\sum _aK_a^*YK_a\). Thus the adjoint again has a Kraus representation, and compression of positive block-diagonal matrices gives the stated positivity.

For \(A\in \mathcal B\) and \(B\in \mathcal A\), multiplicativity and trace preservation give

Nondegeneracy of the trace pairing gives \(\Phi ^*=\Phi ^{-1}\).

For the matrix \(R_k\) formed from the column blocks of \(A\) outside the \(k\)-th diagonal block, diagonal compression satisfies

Block-diagonal embedding is a positive algebra homomorphism and converts the sum of the block traces into the full trace. It follows that positivity, both stated Schwarz implications, and trace preservation pass to \(\widehat T\). The embedding and compression satisfy

Substituting (??) and using (??) gives \((\widehat T)^*=\widehat{T^*}\).

Since \(\pi \circ \iota =\operatorname{id}\), one fixed-point implication follows from

whenever \(\widehat T(X)=X\). Conversely, if \(T(A)=A\), then

These implications prove (??).

Since \(\pi _{\mathcal B}\iota _{\mathcal B}=\operatorname{id}_{\mathcal B}\),

If \(\widehat T(X)=\sum _iK_iXK_i^*\) and \(\widehat S(Y)=\sum _jL_jYL_j^*\), then

which gives the asserted Kraus representation. For \(Y=\iota _{\mathcal A}(X)\), the \(i\)-th diagonal block of (??) is

which proves the descent assertion. If \(F\) preserves the total trace, then \(\widehat F\) preserves the ordinary trace, so \((\widehat F)^*\) is unital. The adjoint \((\widehat F)^*\) is again a Kraus map, so the Kraus-form Kadison–Schwarz inequality gives

For \(Y=\iota _{\mathcal A}(X)\), use \((\widehat F)^*=\widehat{F^*}\) and project this inequality onto the \(i\)-th diagonal summand. The resulting block identity is

which is (??).

Complete positivity and trace preservation make both trace adjoints unital Schwarz maps. For \(F=T^*\) and \(G=S^*\), set \(\Delta _F(X)=F(X^*X)-F(X)^*F(X)\succeq 0\). Positivity of \(G\) gives \(G(\Delta _F(X))\succeq 0\). The Schwarz inequality for \(G\), together with \(GF=\operatorname{id}\), gives \(-G(\Delta _F(X))\succeq 0\). Hence \(G(\Delta _F(X))=0\), and injectivity of \(G\) implies \(\Delta _F(X)=0\). Polarization of this equality gives the first identity in (??). Positivity gives preservation of the adjoint, so the mutually inverse maps are star-algebra equivalences. This supplies an algebraic route to the block-classification step for which [ CPGSV16 , Appendix C.4, line 1997 ] cites [ WPG10 , Theorem 8 ] . The latter instead uses extreme states and the positive inverse to identify the blocks, then the Schwarz condition to exclude transposition.

A star-algebra isomorphism permutes the minimal nonzero central idempotents, inducing an equivalence \(\sigma :I\simeq J\) between the summands. Its restriction to matched ideals is a complex-linear bijection between full matrix algebras. Comparing complex dimensions gives \(D_i^2=E_{\sigma (i)}^2\), and positivity of the block dimensions gives \(D_i=E_{\sigma (i)}\).

Theorem 25.7.3 gives the star-algebra isomorphism \(\Phi :\mathcal A\simeq \mathcal B\). Applying Theorem 25.7.4 to \(\Phi \) gives the equivalence \(\sigma \), the matched block ideals, and the dimension equalities in (??).

By Skolem–Noether, there is an invertible matrix \(P\) such that \(\Phi (X)=PXP^{-1}\). Preservation of the adjoint implies that \(P^*P\) commutes with every matrix, hence \(P^*P=c\mathbb {1}\) for a scalar \(c\). Positivity and invertibility give \(c{\gt}0\). Therefore \(U=c^{-1/2}P\) is unitary and implements the same automorphism.

The restriction of \(\Phi \) to a matched block is a star-algebra isomorphism. After reindexing by \(D_i=E_{\sigma (i)}\), it is an automorphism of a full complex matrix algebra. Skolem–Noether writes this automorphism as conjugation by an invertible matrix \(P\). Preservation of the adjoint implies that \(P^*P\) commutes with every matrix and is therefore a positive scalar multiple of the identity. Dividing \(P\) by the positive square root of this scalar produces the unitary in (??).

The image of a matrix supported on the \(i\)-th summand is supported on the \(\sigma (i)\)-th summand. Unitary conjugation and reindexing preserve the trace, so

Summing this identity over the source summands proves the result.

Let \(\Phi =S^*:\mathcal A\simeq \mathcal B\) be the star-algebra isomorphism obtained from the inverse pair. Theorem 25.7.7 supplies \(\sigma \), the dimension equalities, and the unitaries for \(\Phi \). Lemma 25.7.8 shows that \(\Phi \) preserves the total trace, hence Lemma 25.6.23 gives \(\Phi ^*=\Phi ^{-1}\). On the other hand, involutivity of the trace adjoint gives \(\Phi ^*=S\). Therefore \(S=\Phi ^{-1}\), and the two inverse laws imply \(T=\Phi \). The unitary formula for \(\Phi \) gives (??), while the inverse formula for \(S=\Phi ^{-1}\) gives (??) with the same summand matching and the same unitaries.

The map \(P_T^*\) is a positive unital idempotent map whose range is \(\operatorname{Fix}(T^*)\). For \(A\in \operatorname{Fix}(T^*)\), the Schwarz inequality gives \(T^*(A^*A)\geq T^*(A)^*T^*(A)=A^*A\), while

Since \(\rho \) is positive definite, these relations imply \(T^*(A^*A)=A^*A\). Hence \(\operatorname{Fix}(T^*)\) is a \(*\)-subalgebra, and \(P_T^*\) is a positive retraction onto it. The blockwise classification of positive retractions gives the unitary, the matrix dimensions, the density matrices \(\sigma _k\), and (??)–(??).

For one summand, cyclicity of the trace and the defining property of the partial trace give

Summing over the diagonal summands and applying the unitary change of basis proves (??) by nondegeneracy of the trace pairing. Its range is (??) because \(\operatorname{tr}_{m_k}(\sigma _k\otimes X)=X\).

Extend \(T\) by diagonal compression and block-diagonal embedding. The canonical extension is positive and trace preserving, its trace adjoint satisfies the Schwarz inequality, and the embedded family \(\iota (\rho )\) is a positive-definite fixed point. After choosing a numbered basis of the finite-dimensional ambient space, apply the full-support density-block theorem. Simultaneous reindexing preserves positivity, trace preservation, the Schwarz inequality, the trace adjoint, and positive definiteness. Reindexing the resulting unitary and block decomposition back to the original space gives (??).

Since \(\rho \) is a fixed point of \(E\), the support projection satisfies \((\mathbb {1}-Q)K_iQ=0\) by Theorem 25.4.2, so the compressed family \(QK_iQ\) is supported on the corner and trace-preserving there. Compress it to the support sector along the isometry \(V:\mathbb {C}^r\to \mathbb {C}^D\) of Theorem E.5.1, which satisfies \(V^\dagger V=\mathbb {1}_r\) and \(VV^\dagger =Q\). The compressed family \(C\) is trace-preserving on \(\mathbb {C}^r\), and \(\sigma =V^\dagger \rho V\) is a positive definite fixed point of the compressed Schrödinger map by Theorem 25.4.1.4. The block form of the Heisenberg-picture fixed points of \(C\) from Theorem 25.6.19 yields a unitary \(U\in M_{r}(\mathbb {C})\), dimensions \(d_k\), and multiplicities \(m_k\) with \(\sum _kd_km_k=r\).

Write \(\Phi (X):=VXV^\dagger \) for the compression isomorphism onto the corner and \(C^*\) for the Heisenberg-picture adjoint of the compressed family. The corner-restricted fixed-point condition corresponds to the fixed-point equation of the compressed map:

With \(W:=VU\), which satisfies \(W^\dagger W=\mathbb {1}_r\) and \(WW^\dagger =Q\), this equivalence gives (??) and (??). Finally, \(r\leq D\) because the isometry \(V\) embeds \(\mathbb {C}^r\) into \(\mathbb {C}^D\).

Set \(B_i=\rho ^{-1/2}K_i\rho ^{1/2}\). The equation \(T(\rho )=\rho \) implies that the family \(\{ B_i\} \) is unital, and trace preservation of \(T\) implies that \(\rho \) is a positive-definite fixed point of the adjoint map of the gauged family. Apply Theorem 25.2.4 to \(B\). The equivalence

identifies the fixed points of the gauged map with \(\rho ^{-1/2}F_T\rho ^{-1/2}\).

Closure under addition, scalars, and conjugation is linearity together with the stability of fixed points under conjugate transposition. The corner unit \(Q\) belongs to the set because \(\sqrt{\rho }\, Q\sqrt{\rho }=\rho \) is a fixed point. The support projection absorbs the square root, \(Q\sqrt{\rho }=\sqrt{\rho }=\sqrt{\rho }\, Q\), since

For closure under multiplication, compress to the support sector. Since \(\rho \) is a fixed point of \(T\), the support projection satisfies \((\mathbb {1}-Q)K_iQ=0\) by Theorem 25.4.2. The compressed family \(QK_iQ\) is trace-preserving on the corner, and compression along the isometry \(V:\mathbb {C}^r\to \mathbb {C}^D\) of Theorem E.5.1, with \(V^\dagger V=\mathbb {1}_r\) and \(VV^\dagger =Q\), produces a trace-preserving family \(C\) on \(\mathbb {C}^r\) whose Schrödinger map has the positive definite fixed point \(\sigma =V^\dagger \rho V\) by Theorem 25.4.1.4. The square root compresses along the isometry:

by uniqueness of the positive square root, since \((V^\dagger \sqrt{\rho }V)^2 =V^\dagger \sqrt{\rho }\, Q\sqrt{\rho }V=\sigma \). Hence, for corner-supported \(Y\),

For corner-supported \(W\), the compressed map satisfies

since \(T(W)\) is again corner-supported. Thus \(T(W)=W\) exactly when the compressed map fixes \(V^\dagger WV\). The set therefore corresponds, element by element, to the weighted fixed points of the compressed family with respect to the positive definite fixed point \(\sigma \), which form a \(*\)-subalgebra by Theorem 25.7.14.

Closure under multiplication transports back through the correspondence. A corner-supported \(Y_j\) satisfies \(Y_jQ=Y_j\), so

The product of two members therefore corresponds to the product of their compressed images, which lies in the compressed \(*\)-subalgebra.

Compress to the support sector as in the proof of Theorem 25.7.15 and set

where \(\sigma =V^\dagger \rho V\) is positive definite, so \(\sqrt{\sigma }\) is invertible. Then \(Y\) is corner-supported. Writing \(Z=\sqrt{\sigma }^{-1}(V^\dagger XV)\sqrt{\sigma }^{-1}\) for the compressed middle factor, (??) gives

Hence \(\sqrt{\rho }\, Y\sqrt{\rho }=QXQ=X\), since both sides are corner-supported. In particular, \(\sqrt{\rho }\, Y\sqrt{\rho }\) is fixed by \(T\).

The trace-preservation hypothesis is the normalization \(\sum _iK_i^\dagger K_i=\mathbb {1}\) of the channel form of the transfer map (Theorem 4.7.2).

The support projection absorbs \(\rho \), so \((\mathbb {1}-Q)\rho =0\) and

because conjugation by \(\mathbb {1}-Q\) preserves the Loewner order. Hence

so \((\mathbb {1}-Q)\sqrt{P}=0\) and \((\mathbb {1}-Q)P=((\mathbb {1}-Q)\sqrt{P})\sqrt{P}=0\). Thus \(QP=P\); taking conjugate transposes gives \(PQ=P\) since \(P\) and \(Q\) are Hermitian.

Positivity of the mean-ergodic projection gives \(\rho _0\succeq 0\), while its range and idempotence give \(T(\rho _0)=\rho _0\). If \(A\succeq 0\), then \(A\preceq \operatorname{tr}(A)\mathbb {1}\); applying the positive map \(T_\infty \) gives

Scalar domination therefore gives \(Q_0T_\infty (A)Q_0=T_\infty (A)\). For a fixed Hermitian matrix \(H\), write \(H=H^+-H^-\). Since \(T_\infty (H)=H\), linearity gives

so \(Q_0HQ_0=H\). Finally, decompose an arbitrary fixed point into two Hermitian fixed matrices, using preservation of conjugate transpose, and conclude that \(Q_0XQ_0=X\).

Theorem 25.1.7 gives bounded orbits. Apply Theorem 25.7.19.

First suppose \(A\succeq 0\) and \(QAQ=A\). On the support of \(\rho \), the matrix \(\rho \) is positive definite, so \(Q\preceq c\rho \) for some positive scalar \(c\). Moreover, \(A\preceq \operatorname{tr}(A)Q\), because this follows by compressing \(\operatorname{tr}(A)\mathbb {1}-A\succeq 0\) with \(Q\). Positivity and \(T(\rho )=\rho \) therefore give

Scalar domination transfers the support, hence \(QT(A)Q=T(A)\).

Write \(H_1=X+X^*\) and \(H_2=\mathrm{i}(X-X^*)\), so that both matrices are Hermitian and \(X=\tfrac 12H_1-\tfrac {\mathrm{i}}2H_2\). Setting \(A_j^\pm =QH_j^\pm Q\) gives positive matrices with \(QA_j^\pm Q=A_j^\pm \) and

Applying the positive-input conclusion to these four matrices and using linearity of \(T\) gives (??).

Theorem 25.7.20 gives \(\rho _0\) and shows that \(Q_0XQ_0=X\) for every fixed point \(X\) of \(T\). Choose an isometry onto the range of \(Q_0\). Positivity follows from positivity of conjugation and of \(T\). The matrix \(VYV^*\) is supported on \(Q_0\), so Theorem 25.7.21 shows that its image is also supported there. Inserting \(Q_0=VV^*\) gives

which proves (??). By cyclicity of the trace, the isometry identity, and trace preservation of \(T\),

Thus \(\widetilde T\) is trace preserving. The compressed point \(\sigma _0=V^*\rho _0V\) is positive definite. Since \(V\sigma _0V^*=Q_0\rho _0Q_0=\rho _0\),

If \(T(X)=X\), then \(Q_0XQ_0=X\), and hence \(X=V(V^*XV)V^*\). Moreover,

Conversely, if \(\widetilde T(Y)=Y\), then (??) gives \(T(VYV^*)=V\widetilde T(Y)V^*=VYV^*\). These two implications prove (??).

The right partial trace is the sum, over the nonempty set \(J\), of the positive-definite principal submatrices of \(X\) indexed by \(I\times \{ j\} \). A nonempty finite sum of positive-definite matrices is positive definite.

The left partial trace is the sum, over the nonempty set \(I\), of the positive-definite principal submatrices of \(X\) indexed by \(\{ i\} \times J\). A nonempty finite sum of positive-definite matrices is positive definite.

By Theorem 25.8.2, \(\operatorname{tr}_I(A\otimes B)=\operatorname{tr}(A)B=B{\gt}0\), and hence \(\operatorname{tr}(B){\gt}0\). Theorem 25.8.1 gives \(\operatorname{tr}_J(A\otimes B)=\operatorname{tr}(B)A{\gt}0\). Rescaling by the positive number \(\operatorname{tr}(B)^{-1}\) proves \(A{\gt}0\).

Theorem 25.7.22 gives positivity, trace preservation, a positive definite fixed point for \(\widetilde T\), and the stated correspondence between the original and compressed fixed points. Directly from the trace pairing, \(\widetilde T^*(A)=V^*T^*(VAV^*)V\). Compressing the Schwarz defect for \(T^*\) and adding the positive term through \(\mathbb {1}-VV^*\) proves the Schwarz inequality for \(\widetilde T^*\). Theorem 25.7.11 gives (??) with positive semidefinite trace-one matrices \(\sigma _k\).

Apply this description to a positive definite fixed point \(\rho _{\mathrm{full}}\) of \(\widetilde T\). Its unitary coordinate matrix is positive definite and has the form

Hence every principal block \(\sigma _k\otimes X_k\) is positive definite. Both tensor factors have positive dimension, and \(\operatorname{tr}(\sigma _k)=1\). Theorem 25.8.3, applied to \(\sigma _k\otimes X_k\), therefore gives \(\sigma _k{\gt}0\).

Let \(V:\mathbb {C}^n\to \mathbb {C}^D\) be the maximal-support isometry, and let \(U_c\) be the unitary in the density-block description of the compressed fixed points. The fixed-space equivalence and (??) give

Put \(W=VU_c\). Then \(W^*W=\mathbb {1}_n\). Compare \(W\) with the canonical inclusion of \(\mathbb {C}^n\) as the second summand of \(\mathbb {C}^{D-n}\oplus \mathbb {C}^n\). The two inclusions have the same Gram matrix, so Theorem J.4.2 gives an ambient unitary \(U\) carrying the canonical inclusion to \(W\). Hence, for every \(A\in M_{n}(\mathbb {C})\),

Substituting the compressed density-block description into (??) proves (??). The first block is the whole orthogonal complement of the maximal stationary support, and is therefore one zero summand. Positive definiteness and trace one for every \(\sigma _k\) are supplied by Theorem 25.8.4.

Take the fixed point \(\rho _0\) of Theorem 25.7.20. Every fixed point \(X\) of \(T\) satisfies \(Q_0XQ_0=X\) for the support projection \(Q_0\) of \(\rho _0\), so Theorem 25.7.16 applies to \(X\) and produces the corner-supported \(Y\).

In the spectral decomposition \(\rho =U\operatorname{diag}(\lambda )U^\dagger \), the support projection is \(U\operatorname{diag}(\mathbf1_{\lambda {\gt}0})U^\dagger \). Multiplication by the invertible matrices \(U\) and \(U^\dagger \) preserves the rank, and

where the middle equality follows because the eigenvalues of \(\rho \succeq 0\) are non-negative.

With \(\rho =U\operatorname{diag}(\lambda )U^\dagger \) as above,

where the last equality follows because the rank of a positive semidefinite matrix is the number of its positive eigenvalues.

Let \(\rho _0\) be the fixed point of maximal support of Theorem 25.7.20, with support projection \(Q_0\), and let \(P\) be the support projection of \(\rho \). If \(\rho _0=0\), then every fixed point vanishes because \(Q_0=0\) annihilates it, and the claim is trivial. Suppose that \(\rho _0\neq 0\). Its trace is then positive, and \(\sigma =(\operatorname{tr}\rho _0)^{-1}\rho _0\) is a fixed-point density matrix with \(\operatorname{rank}\sigma =\operatorname{rank}\rho _0\). The rank hypothesis gives \(\operatorname{rank}\rho _0\leq \operatorname{rank}\rho \). Conversely, \(Q_0\rho Q_0=\rho \), so

where the last equality is Lemma 25.8.7. Hence the ranks agree.

From \(\rho Q_0=\rho \), one has \(\rho (\mathbb {1}-Q_0)=0\), so the range of \(\mathbb {1}-Q_0\) lies in the kernel of \(\rho \). The support projection \(P\) of \(\rho \) vanishes on that kernel, and therefore

The last identity follows by taking conjugate transposes. The difference \(R=Q_0-P\) is Hermitian and idempotent, since

Its trace vanishes by Lemma 25.8.8 and the equality of the ranks, so \(\operatorname{tr}(R^\dagger R)=\operatorname{tr}(R)=0\). Hence \(R=0\) and \(P=Q_0\). The claim now follows from the maximal-support property of \(\rho _0\).

By Theorem 25.8.9, every fixed point \(X\) of \(T\) satisfies \(QXQ=X\) for the support projection \(Q\) of \(\rho \), so Theorem 25.7.16 applies to \(X\) and produces the corner-supported \(Y\).

By induction on \(n\), the \(n\)-th iterate \(T^n\) sends \(P_{c,k+n}XP_{c,k+n}\) to \(P_{c,k}T^n(X)P_{c,k}\), using the multiplicative-domain factorizations and the cyclic action \(T(P_{c,j+1})=P_{c,j}\) at each step. Specializing to \(n=m(c)\) and using \(k+m(c)=k\) in \(\{ 0,\ldots ,m(c){-}1\} \) yields corner preservation of \(T^{m(c)}\).

Write \(N=m(c)q\). Corner preservation is stable under taking powers, so \(T^N=(T^{m(c)})^q\) preserves each corner by Lemma 25.10.3.