Applying \(E\) to (??) gives \(E(\sigma _N)=N^{-1}\sum _{n=1}^{N}E^n(\rho _0)\). Subtracting \(\sigma _N\) cancels all but the first and last terms, yielding (??).

Start with any \(\rho _0\in \mathcal{D}_D\) (Theorem B.1.3). Lemma B.1.6 keeps every \(\sigma _N\), \(N\geq 1\), in the compact set \(\mathcal{D}_D\), so a subsequence converges to some \(\rho \in \mathcal{D}_D\). The telescope identity (??) gives

because the iterates \(E^N(\rho _0)\) lie in the compact set \(\mathcal{D}_D\) and are therefore uniformly bounded. Along the convergent subsequence, continuity of \(E\) now gives \(E(\rho )=\rho \). In  [ Wol12 , Theorem 6.11 ] , existence is proved via Brouwer’s fixed-point theorem; we use the Cesàro argument instead.

Choose a Kraus decomposition

and set \(T(B)=B+E(B)\) for nonzero \(B\geq 0\). Always \(\ker T(B)\subseteq \ker B\): if \(v\in \ker T(B)\), then

and both terms are non-negative, so \(v^\dagger Bv=0\) and hence \(Bv=0\). If \(B\) is not positive definite, irreducibility forces

Indeed, equality of the kernel dimensions would turn the inclusion above into \(\ker B\subseteq \ker T(B)\). Thus, for \(v\in \ker B\),

Every summand is non-negative, so equality would imply, for every \(i\),

Equivalently, \(\operatorname{supp}B\) would be invariant under every \(K_i\), making the support projection of \(B\) a nontrivial invariant projection.

We prove by induction on \(n\) that every nonzero \(B\geq 0\) with \(\dim \ker B\leq n\) satisfies \(T^n(B){\gt}0\). For \(n=0\), the kernel is trivial, so \(B{\gt}0\). Suppose the claim holds for \(n\) and let \(\dim \ker B\leq n+1\). If \(B{\gt}0\), then \(T(B)=B+E(B){\gt}0\), and every subsequent iterate remains positive definite. Otherwise the strict kernel decrease gives \(\dim \ker T(B)\leq n\); the induction hypothesis applied to \(T(B)\) yields \(T^n(T(B))=T^{n+1}(B){\gt}0\). Since \(A\neq 0\), rank–nullity gives \(\dim \ker A\leq D-1\), which proves (??).

Write \(E(X)=\sum _i K_iXK_i^\dagger \) and let \(Q\) be the support projection of \(\rho \). For \(v\in \ker \rho \), the kernel inclusion gives

Every summand is non-negative, so \(K_i^\dagger (\ker \rho )\subseteq \ker \rho \) for every \(i\). Equivalently, for every \(X\),

Irreducibility forces \(Q\in \{ 0,I\} \), while \(\rho \neq 0\) rules out \(Q=0\). Thus \(Q=I\), so \(\ker \rho =\{ 0\} \) and \(\rho {\gt}0\).

The eigenvector equation gives

Hence \(\ker \rho \subseteq \ker E(\rho )\). Theorem 6.2.2 then forces \(\rho \) to be positive definite.

Suppose \(\rho \) is not positive definite, and let \(Q\) be its support projection. Since \(\rho =Q\rho Q\) and \(\rho =\sum _iA^i\rho (A^i)^\dagger \), the \((\mathbb {1}-Q)\) corner satisfies

Every summand is positive semidefinite, so each vanishes. The restriction of \(\rho \) to its support is positive definite; hence, for every \(i\),

Thus \(A^iQ=QA^iQ\), and for every \(X\) each summand \(A^iQXQ(A^i)^\dagger \) lies in the \(Q\) corner. Therefore

so \(Q\) is invariant. Irreducibility forces \(Q=0\) or \(Q=\mathbb {1}\). The first case contradicts \(\rho \neq 0\), while the second says that \(\rho \) is positive definite. This is the support-projection mechanism of [ Wol12 , Theorem 6.2(1) ] .

Injectivity implies irreducibility of the transfer map. Apply Theorem 6.2.4.

Theorem 6.2.1 gives \((\mathbb {1}+ E)^{D-1}(A) {\gt} 0\). Since \(B \ge 0\) is nonzero, \(\operatorname{tr}(B(\mathbb {1}+ E)^{D-1}(A)) {\gt} 0\). Expanding by the binomial theorem gives

Each term \(\operatorname{tr}(BE^k(A))\) is non-negative since both factors are positive semidefinite. The \(k = 0\) term vanishes by the hypothesis \(\operatorname{tr}(BA) = 0\). Hence at least one \(k \in \{ 1,\ldots ,D-1\} \) contributes a strictly positive term. This proves the implication (2)\(\Rightarrow \)(4) in [ Wol12 , Theorem 6.2(4) ] by a binomial expansion and positivity.

If \(\sigma =0\), take \(c=0\). Otherwise, Theorem 6.2.4 makes both \(\rho \) and \(\sigma \) positive definite. Write \(\rho =SS^\dagger \) and set

If \(c_0{\gt}0\) is the smallest eigenvalue of \(H\), then \(H-c_0\mathbb {1}\geq 0\) and \(H-c_0\mathbb {1}\) is singular. Since \(S\) is invertible, congruence by \(S\) preserves positivity and singularity and gives

so \(\tau \) is positive semidefinite and singular. Linearity makes \(\tau \) a fixed point. If \(\tau \neq 0\), Theorem 6.2.4 would make it positive definite, a contradiction. Thus \(\tau =0\) and \(\sigma =c_0\rho \). This is the minimal-eigenvalue mechanism of [ Wol12 , Theorem 6.3(2) ] .

Injectivity implies irreducibility of the transfer map. Apply Theorem 6.3.1.

Extract Kraus operators \(K\) for \(E\). Irreducibility of \(E\) passes to the adjoint transfer map \(E^\dagger (X) = \sum _i K_i^\dagger X K_i\), which is a positive CP map. The Perron–Frobenius existence theorem (Theorem 6.7.2) together with the irreducible positive-definiteness upgrade (Theorem 6.2.3) gives a positive definite eigenvector \(\tau {\gt} 0\) with eigenvalue \(t {\gt} 0\): \(E^\dagger (\tau ) = t\tau \). For any nonzero positive semidefinite \(X\) with \(E(X) = sX\), the trace-pairing identity (??) yields

Since \(\tau {\gt} 0\) and \(X \ge 0\), \(X \neq 0\), the scalar \(\operatorname{tr}(\tau X) {\gt} 0\), so \(s = t\). Applying this to both \((\rho ,r_1)\) and \((\sigma ,r_2)\) gives \(r_1 = t = r_2\). This is the completely positive specialization of [ Wol12 , Theorem 6.3(3) ] : uniqueness of a positive eigenvalue admitting a nonzero positive semidefinite eigenvector. The proof follows Wolf’s dual-map trace argument.

Choose Kraus operators for \(E\) and rescale each by \(r^{-1/2}\). Their transfer map is \(r^{-1}E\), so both \(\rho \) and \(\sigma \) are fixed points. The rescaled map remains irreducible, and Theorem 6.3.1 gives the claimed proportionality.

Under the normalization \(\sum _i (A^i)^\dagger A^i = \mathbb {1}\), the transfer map is a quantum channel (Theorem 4.7.2). The Cesàro fixed-point theorem (Theorem 6.1.3) then provides the fixed point; injectivity is not needed for this step.

Combine Theorems 6.4.1, 6.2.5, and 6.3.2.

Existence is the channel fixed-point theorem (Theorem 6.4.1). Irreducibility upgrades the fixed point to positive definite, and Theorem 6.3.1 gives uniqueness up to scalar multiple.

Conjugating each summand gives \((S^{-1}A^iS)(S^{-1}A^iS)^\dagger =S^{-1}A^i\rho (A^i)^\dagger (S^\dagger )^{-1}\). The substitution \(\mathcal{E}_A(\rho )=\rho \) leaves \(S^{-1}\rho (S^\dagger )^{-1}\), and \(\rho =SS^\dagger \) cancels the outer factors. The term-by-term calculation is in Section B.2.

Conjugating each summand gives \((SA^iS^{-1})^\dagger (SA^iS^{-1}) =(S^\dagger )^{-1}(A^i)^\dagger \sigma A^iS^{-1}\). The adjoint fixed-point substitution leaves \((S^\dagger )^{-1}\sigma S^{-1}\), and \(\sigma =S^\dagger S\) cancels the outer factors. See Section B.2 for the full calculation.

Each conjugated summand is \((B^i)^\dagger B^i =\sigma ^{-1/2}(A^i)^\dagger \sigma A^i\sigma ^{-1/2}\). Substituting \(\sum _i(A^i)^\dagger \sigma A^i=\sigma \) and cancelling the outer inverse square roots gives \(\mathbb {1}\). The complete calculation appears in Section B.2.

Suppose \(Q\neq 0,\mathbb {1}\) is an invariant projection for \(E'\). Thus \((\mathbb {1}-Q)E'(QXQ)Q=0\) for every \(X\). Setting \(R = CQC^\dagger \), the support projection \(P\) of \(R\) is an invariant projection for \(E\). Irreducibility of \(E\) forces \(P = 0\) or \(P = \mathbb {1}\). The invertibility of \(C\) then forces \(Q = 0\) or \(Q = \mathbb {1}\), a contradiction. This is the completely positive specialization of the similarity result in  [ Wol12 , Proposition 6.6 ] ; the scalar case \(C=\mathbb {1}\) gives the scaling result stated next.

An invariant projection for \(cE\) is an invariant projection for \(E\), since the nonzero constant may be removed from the invariance relation.

If \(E\) annihilates a nonzero positive semidefinite matrix \(\rho \), take \(r=0\). Otherwise \(E(\sigma )\neq 0\) for every nonzero \(\sigma \geq 0\), and Theorem 6.7.2 gives (??) with \(r{\gt}0\).

Normalize \(E\) on the density matrices by

Positivity and the nonvanishing hypothesis make this a continuous self-map. Theorem B.1.5 gives a fixed point \(\rho \). Clearing the positive denominator yields \(E(\rho )=r\rho \) with \(r=\operatorname{tr}(E(\rho )){\gt}0\). Thus this theorem is the positive-eigenvalue specialization of Theorem 6.7.1.

By Theorem B.4.1, \(\mathcal{E}_A^\dagger \neq 0\). Since \(\mathcal{E}_A^\dagger (X) = \sum _i (A^i)^\dagger X A^i\) is a Kraus map, it is positive. Irreducibility of \(A\) transfers to the adjoint transfer map. More precisely, if \(P\) is invariant for \(\mathcal{E}_A\), so that \((\mathbb {1}-P)K_iP=0\) for all Kraus operators \(K_i\), then taking adjoints gives \(PK_i^\dagger (\mathbb {1}-P)=0\), which means \(\mathbb {1}-P\) is invariant for \(\mathcal{E}_A^\dagger \). Hence \(\mathcal{E}_A\) is irreducible if and only if \(\mathcal{E}_A^\dagger \) is irreducible, and together with \(\mathcal{E}_A^\dagger \neq 0\) this implies \(\mathcal{E}_A^\dagger (\tau ) \neq 0\) for every nonzero positive semidefinite matrix \(\tau \). Hence Theorem 6.7.2 applies to \(\mathcal{E}_A^\dagger \), giving \(\sigma _0 \ge 0\), \(\sigma _0\neq 0\), and \(r {\gt} 0\) with \(\mathcal{E}_A^\dagger (\sigma _0) = r\sigma _0\). Set \(T^i = r^{-1/2}(A^i)^\dagger \). Then \(\mathcal{E}_T(\sigma _0) = \sigma _0\), and \(\mathcal{E}_T\) is irreducible by Theorem 6.6.2, since irreducibility of \(\mathcal{E}_A\) transfers to the adjoint and is preserved under scaling. Since \(\mathcal{E}_T\) is irreducible with a positive semidefinite fixed point, Theorem 6.2.4 gives that \(\sigma _0\) is positive definite.

By Theorem 6.7.3, obtain \(\sigma \) positive definite and \(r {\gt} 0\) with \(\mathcal{E}_A^\dagger (\sigma ) = r\sigma \). Set \(c = r^{-1/2}\) and \(A'^i = cA^i\). Then \(\mathcal{E}_{A'}^\dagger (\sigma ) = \sigma \). By the square-root trace-preserving gauge construction (Theorem 6.5.4), the tensor defined by (??) satisfies \(\sum _i (B^i)^\dagger B^i = \mathbb {1}\). Lemma B.2.1 gives the required gauge equivalence to \(r^{-1/2}A\).

Apply Theorem 6.7.3 to the conjugate-transposed Kraus family. Irreducibility transfers to that family, and the nonzero Kraus hypothesis is preserved by taking adjoints. Translating the resulting adjoint eigenvector equation back gives \(\mathcal{E}_A(\rho )=r\rho \) with \(\rho \) positive definite and \(r{\gt}0\). The unital gauge theorem then gives the representative (??), and the same square-root similarity gives gauge equivalence to \(r^{-1/2}A\).

By Theorem B.5.1, the first \(D\) terms in (??) already form a positive definite matrix. Every remaining term is positive semidefinite, so the full exponential series is the sum of a positive definite matrix and a positive semidefinite tail.

The forward implication is Theorem 6.8.1. Conversely, let \(P\) be an invariant orthogonal projection. Invariance gives \(E(PM_{D}(\mathbb {C})P)\subseteq PM_{D}(\mathbb {C})P\), so induction yields \(E^k(P)\in PM_{D}(\mathbb {C})P\) for every \(k\geq 0\). Therefore every term, and hence the convergent series

lies in the same corner. If \(P\neq 0\), then at \(t=1\) the assumed positive definiteness is impossible unless \(P=\mathbb {1}\), because a positive definite matrix cannot be supported on a proper corner. Thus every invariant projection is \(0\) or \(\mathbb {1}\), so \(E\) is irreducible.

Every Cesàro mean (??) of a density matrix stays inside the compact convex set of density matrices, so a subsequence converges to a density matrix \(\sigma \). The telescope identity (??) gives \(E(\sigma )=\sigma \). Theorem 6.2.3 makes \(\sigma \) positive definite. If \(\tau \) is another density-matrix fixed point, Theorem 6.3.4 gives \(\tau =c\sigma \). Taking traces yields \(1=c\), hence \(\tau =\sigma \).

Every subsequential limit of the Cesàro means is again a density-matrix fixed point. By Theorem 6.9.1, there is only one such fixed point, so every convergent subsequence has the same limit. Compactness then forces the whole sequence to converge to that limit.

Choose Kraus operators \(\{ K_i\} \) for \(E\). Regarded as an MPS tensor, they have transfer map \(E\), and irreducibility of \(E\) is precisely irreducibility of this tensor. Theorem 6.7.3 therefore gives a positive definite matrix \(\sigma {\gt}0\) and \(t{\gt}0\) such that \(E^\dagger (\sigma )=t\sigma \). The weighted trace pairing with \(\rho \) gives

Since \(\operatorname{tr}(\sigma \rho ){\gt}0\), one has \(t=r\). Rescale by \(r^{-1}\) and conjugate by \(\sigma ^{1/2}\) as in Theorem 6.5.4; this produces a trace-preserving Kraus map \(F\) similar to \(r^{-1}E\). The transfer-operator contraction (Theorem 7.4.3) gives \(\rho _{\operatorname{spec}}(F)\leq 1\). The conjugate of \(\rho \) is a nonzero fixed point of \(F\), so \(1\) is an eigenvalue and \(\rho _{\operatorname{spec}}(F)\geq 1\). Hence \(\rho _{\operatorname{spec}}(F)=1\). Similarity preserves the spectrum, and undoing the scalar rescaling gives \(\rho _{\operatorname{spec}}(E)=r\).

Combine Perron–Frobenius existence for the map and its adjoint with the uniqueness theorem for positive semidefinite Perron eigenvectors and the spectral-radius identity of Theorem 6.10.1.

Gauge by the positive definite left eigenvector in (??) and rescale by the Perron eigenvalue to obtain a trace-preserving map with a positive definite fixed point. A nontrivial invariant projection would then produce, by Cesàro averaging in its corner algebra, a second positive semidefinite fixed point not proportional to the Perron vector, contradicting the uniqueness clause in Definition 6.11.1.

Combine Theorems 6.11.2 and 6.11.3.