Continuity follows from continuity of the matrix exponential. The exponentials of \(isH\) and \(itH\) multiply to \(e^{i(s+t)H}\) because the two exponents commute. For Hermitian \(H\), \((itH)^\dagger =-itH\), so \(U_H(t)^\dagger =e^{-itH}\) and \(U_H(t)U_H(t)^\dagger =e^0=\mathbb {1}\).

Since the common commutant of the matrix units consists of scalar matrices, there is a matrix unit \(B\) with \(BK-KB\ne 0\). Define \(f(t)=Be^{itK}-e^{itK}B\). Then \(f(0)=0\) and \(f'(0)=i(BK-KB)\ne 0\). Hence \(f(t)/t\to f'(0)\) as \(t\to 0\) through nonzero values, and \(f(t)\ne 0\) throughout a sufficiently small punctured neighborhood. A scalar \(e^{itK}\) would instead give \(f(t)=0\).

The continuous homomorphisms \(X\mapsto X\otimes \mathbb {1}\) and \(Y\mapsto \mathbb {1}\otimes Y\) preserve the exponential series. The two tensor-factor actions commute, so the exponential addition formula gives the first identity. Transposition and adjunction preserve the exponential, and their composition is entrywise conjugation. Finally, \(\overline{itH}=-it\overline H\); applying the tensor-sum identity gives (1).

Since \((YE_{ja}Y^{-1}X_0)_{ia}=Y_{ij}(Y^{-1}X_0)_{aa}\), the formula follows entrywise. A scalar multiple of \(Y\) obeys the same intertwining relation as \(Y\).

Fix an index \(a\) and set \(X(t)_{ij}=\alpha _t(E_{ja})_{ia}\). This matrix is differentiable and \(X(0)=\mathbb {1}\). Write \(Z=X'(0)\) and \(\delta (M)=\alpha '_0(M)\). Differentiating \(\alpha _t(M)X(t)=X(t)M\) gives \(\delta (M)=ZM-MZ\). Preservation of adjoints gives \(\delta (M^\dagger )=\delta (M)^\dagger \), hence \([Z+Z^\dagger ,M]=0\) for all \(M\). Therefore \(H=-\frac{i}{2}(Z-Z^\dagger )\) is Hermitian and \(i[H,M]=[Z,M]\), which proves (2).

Since \((t \cdot L)\) commutes with \((s \cdot L)\), this follows from the exponential addition formula for commuting elements in a Banach algebra.

The exponential of the zero endomorphism is the identity.

The exponential is analytic (convergence radius \(= \infty \)), hence continuous.

Apply the chain rule to the composition \(\mathbb {R}\xrightarrow {t \mapsto t \cdot L} \operatorname{End}(M_{D}(\mathbb {C})) \xrightarrow {\exp } \operatorname{End}(M_{D}(\mathbb {C}))\).

This is Theorem 12.3.7 at \(t=0\) together with Theorem 12.3.5.

Both semigroups agree on \([0,\infty )\), so their derivatives within \([0,\infty )\) at \(t=0\) agree. Since \([0,\infty )\) has unique differentials at \(0\), \(L = L'\).

This is a direct unpacking of the definition of the linear-map exponential semigroup.

Evaluate the derivative of the continuous-linear-map semigroup at \(X\) and transport the statement through Theorem 12.3.10.

Transport Theorem 12.3.5 through Theorem 12.3.10.

Transport the continuous-linear-map semigroup law from Theorem 12.3.4 through Theorem 12.3.10; the initial condition is Theorem 12.3.12.

Since \(T_0 = \mathbb {1}\) and \(T\) is continuous, the integral \(M_\varepsilon = \int _0^\varepsilon T_s\, ds\) is invertible for small \(\varepsilon {\gt} 0\). Using the semigroup property, \(T_t = M_\varepsilon ^{-1}(M_{t+\varepsilon } - M_t)\) is differentiable for \(t \ge 0\). ODE uniqueness then gives \(T_t = e^{tL}\) for all \(t \ge 0\). Since scalar multiples of \(L\) commute, the exponential addition formula gives the complex group law in (4); substituting \(-z\) gives the two-sided inverses. Restricting the complex parameter to \(\mathbb {R}\subset \mathbb {C}\) gives the real group extension.

Define \(f(s) = e^{(t-s)L}e^{sL'}\). Then \(f'(s) = e^{(t-s)L}\Delta \, e^{sL'}\) by Theorem 12.3.7, and \(e^{tL'} - e^{tL} = f(t) - f(0) = \int _0^t f'(s)\, ds\), which gives (5).

Apply the triangle inequality and submultiplicativity of the operator norm to the Duhamel integral (5).

Apply (6) and bound both suprema by \(1\).

This is the Weierstrass M-test step, using the scalar convergence of \(\sum _n a^n/n!\) and lifting via norm-bounded summability.

Induction on \(n\). The base case uses the semigroup supremum bound. The inductive step bounds the integrand pointwise by \(M\| \Delta \| \cdot M(u\| \Delta \| M)^n/n!\) and integrates \(\int _0^s u^n\, du = s^{n+1}/(n+1)\), recovering the factorial.

The bound (8) gives a factorial majorant. Lemma 12.4.4 then applies the Weierstrass M-test.

This is the special case \(s=t\) of (8).

By induction on \(n\). The base case is continuity of the matrix exponential. For the inductive step, the parametric integral \(t\mapsto \int _0^t e^{(t-s)L}\Delta \, \widetilde{T}^{(n)}_s\, ds\) is continuous by the continuous-parameter primitive theorem; the pasting lemma extends this to all of \(\mathbb {R}\) since \(\widetilde{T}^{(n+1)}_t = 0\) for \(t \le 0\).

Induction on \(N\). The base case is the semigroup supremum bound. The inductive step uses the telescoping remainder identity \(R_{N+1}(s) = \int _0^s e^{(s-u)L}\Delta \, R_N(u)\, du\) derived from the Duhamel formula (5), bounds the integrand pointwise, and integrates to recover the factorial.

By (10), the partial-sum remainder \(\| e^{tL'}-\sum _{n{\lt}N}\widetilde{T}^{(n)}_t\| \le M'(t\| \Delta \| M)^N/N!\) tends to zero as \(N \to \infty \), since \(c^N/N! \to 0\) for any constant \(c\). Together with the summability of the series (Corollary 12.4.6), this identifies the sum via uniqueness of limits.

The claim follows from (11).

Let \(\phi (\rho ) = \sum _i K_i\rho K_i^\dagger \). By the cyclic property of trace, \(\operatorname{tr}(L(\rho )) = \operatorname{tr}((\sum _i K_i^\dagger K_i)\rho ) - \operatorname{tr}(\kappa \rho ) - \operatorname{tr}(\kappa ^\dagger \rho ) = \operatorname{tr}((\kappa +\kappa ^\dagger )\rho ) - \operatorname{tr}(\kappa \rho )-\operatorname{tr}(\kappa ^\dagger \rho )=0\).

The Hamiltonian part satisfies \(\operatorname{tr}(i[\rho ,H]) = 0\) by the cyclic property. Each dissipator term satisfies \(\operatorname{tr}(L_j\rho L_j^\dagger ) = \operatorname{tr}(L_j^\dagger L_j\rho ) = \operatorname{tr}(\rho L_j^\dagger L_j)\), so the three terms sum to zero.

Compute \(\kappa ^\dagger = -iH+\frac{1}{2}\sum _j L_j^\dagger L_j\) using \(H^\dagger = H\) and \((A^\dagger A)^\dagger = A^\dagger A\). Setting \(S = \sum _j L_j^\dagger L_j\), expand \(\phi (\rho )-\kappa \rho -\rho \kappa ^\dagger = \sum _j L_j\rho L_j^\dagger -iH\rho -\tfrac {1}{2}S\rho +i\rho H-\tfrac {1}{2}\rho S\), which is \(i[\rho ,H]+\sum _j(L_j\rho L_j^\dagger -\tfrac {1}{2}L_j^\dagger L_j\rho -\tfrac {1}{2}\rho L_j^\dagger L_j)\).

By Theorem 12.5.2.3, the Lindblad form equals a generator decomposition with \(\phi \) CP.

Expanding the commutator brackets \([L_j,\rho L_j^\dagger ]\) and \([L_j\rho ,L_j^\dagger ]\) in (14) yields the standard dissipator terms in (13).

Write \(L(\rho ) = \phi (\rho )-\kappa \rho -\rho \kappa ^\dagger \). The left- and right-multiplication terms are annihilated by \(P\), so the projected Choi matrix reduces to the projected Choi matrix of \(\phi \), which is positive.

Decompose the Hermitian Choi matrix as \(\tau _L = Q-|\psi \rangle \langle \Omega |-|\Omega \rangle \langle \psi |\) with \(Q \ge 0\) supported on \(|\Omega \rangle ^\perp \). The Choi–Jamiolkowski isomorphism applied to \(Q\) gives the CP map \(\phi \), and \((\kappa \otimes \mathbb {1})|\Omega \rangle = |\psi \rangle \) defines \(\kappa \).

From \((e^{tL}\otimes \operatorname{id})(|\Omega \rangle \! \langle \Omega |) \ge 0\), expand at infinitesimal \(t\) and project onto \(P\) to get \(P(L\otimes \operatorname{id})(|\Omega \rangle \! \langle \Omega |)P \ge 0\). Then Theorem 12.5.3.2 gives CCP.

Approximate \(e^{tL}\) by the completely positive Euler steps

Each step is CP, hence so are its powers. These powers converge in operator norm to \(e^{tL}\), and the cone of CP maps is closed.

Combine Theorems 12.5.4.1 and 12.5.4.2.

Set \(c_i = -\operatorname{tr}(L_i)/d\).

Direct expansion of (16) and (17) cancels the linear, quadratic, and purely imaginary scalar terms.

Compare projected Choi matrices and use Choi injectivity.

After identifying \(\phi = \phi '\) (Theorem 12.5.6.2), the generator equality forces \(\Delta \kappa \cdot \rho +\rho \cdot (\Delta \kappa )^\dagger =0\) for all \(\rho \). Setting \(\rho =\mathbb {1}\) gives \(\Delta \kappa +(\Delta \kappa )^\dagger =0\) (skew-Hermiticity), which converts the equation to \([\Delta \kappa ,\rho ]=0\) for all \(\rho \). The scalar commutant lemma gives \(\Delta \kappa =c\cdot \mathbb {1}\), and skew-Hermiticity forces \(c=i\lambda \) for some \(\lambda \in \mathbb {R}\).

Tracelessness makes the Choi matrices of the CP parts vanish on the maximally entangled vector. Projecting the Choi matrix of the common generator onto its orthogonal complement therefore gives \(\phi =\phi '\). The remaining generator equality and matrix units show that \(\kappa -\kappa '\) is a purely imaginary scalar matrix, proving (18). Theorem 2.7.5 applied to the zero-padded Kraus families gives (19).

For fixed \(\rho \), the function \(f(t)=\operatorname{tr}(e^{tL}(\rho ))\) has derivative \(f'(t)=\operatorname{tr}(L(e^{tL}(\rho )))=0\), so \(f\) is constant and \(\operatorname{tr}(e^{tL}(\rho ))=f(t)=f(0)=\operatorname{tr}(\rho )\).

Differentiate the identity \(\operatorname{tr}(e^{tL}(\rho ))=\operatorname{tr}(\rho )\) at \(t=0\) from the right.

CP follows from Theorem 12.5.4.2. The trace constraint implies trace-annihilation by Theorem 12.5.1.5, so Theorem 12.5.7.1 gives trace preservation.

Apply Theorem 12.5.4.1 to obtain a CCP decomposition. Then apply Theorem 12.5.7.2 to the trace-preserving part of the semigroup and rewrite the resulting trace-annihilation condition as (21).

Forward: apply Theorem 12.5.4.1 to the CP part and Theorem 12.5.7.2 to the TP part. Reverse: combine Theorem 12.5.4.2 with Theorem 12.5.7.1.

Forward: apply Theorem 12.5.8.3 to write \(L(\rho )=\phi (\rho )-\kappa \rho -\rho \kappa ^\dagger \) with trace constraint. Choose Kraus operators \(\{ L_j\} \) for \(\phi \), set \(H=\frac{i}{2}(\kappa ^\dagger -\kappa )\), and use \(\sum _jL_j^\dagger L_j=\kappa +\kappa ^\dagger \) to obtain \(\kappa =iH+\frac{1}{2}\sum _jL_j^\dagger L_j\). Theorem 12.5.2.3 gives the Lindblad formula (22). Reverse: Theorems 12.5.2.4 and 12.5.2.2 show that every Lindblad form satisfies the right-hand side of Theorem 12.5.8.4.

Forward: diagonalize \(C=B^\dagger B\) and set \(L_j=\sum _k B_{jk}F_k\). Reverse: keep the same Hamiltonian and family \(F_k=L_k\), and take \(C=\mathbb {1}\).

By Theorem 12.5.8.5, it suffices to compare the Lindblad and basis-level Kossakowski forms. From a Kossakowski form, factor \(C=B^\dagger B\) and set \(L_j=\sum _kB_{jk}F_k\). Conversely, shift the Lindblad operators to be traceless, expand them in the fixed basis as \(L_j=\sum _kM_{jk}F_k\), and set \(C=M^\dagger M\). Substitution into (23) gives the commutator form (14).

Direct multiplication of the three Pauli matrices proves the identities; substituting \(\sigma _a^\dagger =\sigma _a\) and \(\sigma _a^2=\mathbb {1}\) into the Lindblad dissipator gives the final formula.

Apply Theorem 12.6.1 separately to the two dissipators in (26) and collect the two scalar coefficients.

Cyclicity of the trace makes each map \(X\mapsto \sigma _aX\sigma _a-X\) symmetric for the Hilbert–Schmidt pairing. Real linear combinations preserve this symmetry.

Expand \(X\) in the basis \(\{ \mathbb {1},\sigma _x,\sigma _y,\sigma _z\} \). The three Pauli coefficients vanish because the three eigenvalues in (27) are nonzero, while the identity coefficient is unrestricted.

The defining formula gives \(\mathcal L_{a,b}(\tau _1)=0\). Theorem 12.6.5 identifies its kernel, and exponentiating a kernel vector leaves it fixed. Conversely, a vector fixed for every non-negative time lies in the generator kernel.

Entropy decay and Pinsker first give

The time hypothesis (28) implies \(e^{-\gamma _*t/2}\le N^{-2p}/(2N\log 2)\), so the squared trace norm is at most \(N^{-2p}\). Both sides are non-negative, and taking square roots gives the result.

The two root relations give \(p_1\theta =2\pi k_1\) and \(p_2\alpha \theta =2\pi k_2\). If \(\theta \ne 0\), their quotient expresses \(\alpha \) as a rational number. The specialization \(\alpha =\pi \) is the irrational rescaling used by Wolf.

Decompose \(E^n\) into the fixed-point projection and its complementary part. On the complement, every eigenvalue has modulus \({\lt}1\), so the complementary powers tend to \(0\); the trace-zero hypothesis removes the fixed-point component.

Subtract the trace multiple \(\operatorname{tr}(V)\sigma \) and apply the vanishing theorem for trace-zero fixed points of an irreducible channel.

Otherwise the preceding theorem would force the fixed point to vanish.

Peripheral powers remain peripheral for an irreducible channel with faithful fixed point. The bounded-order lemma then gives \(\mu ^p=1\), and the semigroup power identity turns the eigenvector equation for \(T_t\) into a fixed-point equation for \(T_{pt}\).

Apply Theorem 12.7.1.14 and then use the previous corollary at the irreducible time \(pt\).

Take traces in \(E(V)=\mu V\) and use trace preservation.

Combine Theorem 12.7.1.15 with Theorem 12.7.1.16.

The unique faithful fixed density gives the one-dimensional peripheral spectrum criterion, and Theorem 12.7.1.17 collapses the peripheral set to \(\{ 1\} \).

By semigroup commutativity, \(T_{t_0}(T_u(\sigma ))=T_u(T_{t_0}(\sigma ))=T_u(\sigma )\), and uniqueness of the fixed density at time \(t_0\) identifies \(T_u(\sigma )\) with \(\sigma \).

This is the fixed-point trace-scalar theorem for the irreducible channel \(T_{t_0}\).

Factor \(e^{tL}-\mathbb {1}=(\int _0^t e^{sL}\, ds)L\). After division by \(t\), the integral factor tends to the identity as \(t\downarrow 0\), and is therefore invertible throughout a sufficiently small punctured interval.

The \(\lambda \)-eigenspace of \(e^{tL}\) is nonzero and invariant under \(L\). The restriction of \(L\) to this finite-dimensional complex space has an eigenvector, which is a common eigenvector of \(L\) and \(e^{tL}\).

Put \(P_t=\int _0^t e^{sL}\, ds\). The factorization \(e^{tL}-\mathbb {1}=P_tL\) puts the nonzero vector \(L(X)\) in \(\ker P_t\). Since \(P_t\) commutes with \(L\), this kernel contains an eigenvector of \(L\). Its eigenvalue is nonzero because \(P_t\) acts as multiplication by \(t\) on \(\ker L\).

Let \(\sigma \) be the unique faithful fixed density of \(T_{t_0}\). By Theorem 12.7.1.20, it is fixed for all times, and Theorem 12.7.1.21 makes the fixed-point space of \(T_{t_0}\) one-dimensional; hence \(\ker L=\mathbb {C}\sigma \). Theorem 12.7.1.22 makes \(T_u\) irreducible for every sufficiently small \(u{\gt}0\). Choose such a \(u\) for which \(\pi u\) is also in this interval. A purely imaginary generator eigenvalue \(\mu \) produces peripheral eigenvalues at both times. Their roots-of-unity relations and Theorem 12.7.1.6 force \(\mu =0\). Finally, if an arbitrary \(T_s\) had a fixed point outside \(\ker L\), Theorem 12.7.1.24 would produce a nonzero eigenvalue \(\mu \) with \(e^{s\mu }=1\). Such a \(\mu \) is purely imaginary and has just been excluded. Thus every fixed point is a scalar multiple of \(\sigma \), which gives irreducibility of every \(T_s\), \(s{\gt}0\).

First apply Theorem 12.7.1.25 to propagate irreducibility from the single time \(t_0\) to every positive time. Then apply Theorem 12.7.1.18 to conclude primitivity of every positive-time slice.

Forward: apply Theorem 12.7.1.26. Reverse: take \(t_0=1\) (or any positive time); the right-hand side already includes irreducibility of \(T_t\) for every \(t{\gt}0\).

Conditions 1–3 follow from Theorem 12.7.1.27. For condition 4, fix the faithful stationary density \(\rho _\infty \) of an irreducible slice. The powers of the primitive unit-time slice send every trace-zero matrix to zero. Writing \(t=\lfloor t\rfloor +(t-\lfloor t\rfloor )\) and using the uniform bound on the residual interval \([0,1]\) gives the continuous-time limit.

Relaxation excludes a proper invariant compression: such a compression would contain a density matrix fixed at every integer time, which would have to equal the positive-definite limit.

For the implication from condition 5 to condition 1, choose \(t_0{\gt}0\) so that the fixed-point space of \(e^{t_0L}\) is precisely \(\ker L\). Such a time exists because

and the integral factor is invertible for all sufficiently small positive \(t\). Thus \(e^{t_0L}\) has the unique fixed direction \(\mathbb {C}\rho _\infty \), with \(\rho _\infty {\gt}0\), and is irreducible by Theorem 6.4. Conversely, \(\ker L\) is contained in the fixed-point space of every \(e^{tL}\), so an irreducible slice forces condition 5.

Expand the Schrödinger and Heisenberg formulas and apply cyclicity of the trace term by term.

The Hamiltonian commutator term vanishes, and each adjoint dissipator \(\mathcal{D}_{L_j}^*(A)\) is zero under the commutation hypotheses.

From \(L^*(A)=0\) and \(L^*(A^\dagger )=0\), compute \(L^*(A^\dagger A)=\sum _j[A,L_j]^\dagger [A,L_j]\). Faithfulness of \(\rho _0\) and \(\operatorname{tr}(\rho _0L^*(A^\dagger A))=0\) force \([A,L_j]=0\) for all \(j\). Similarly, \([A,L_j^\dagger ]=0\), and then \(L^*(A)=0\) directly gives \([A,H]=0\).

The inclusion \(\supseteq \) is the easy direction (commutant elements are in the kernel by a direct calculation). The reverse inclusion \(\subseteq \) is Theorem 12.8.8.

Differentiate the identity \(Pe^{tL}(PXP)P=e^{tL}(PXP)\) at \(t=0\).

Apply the compression map termwise to the exponential series. Every iterate of \(L\) preserves the compression, so the whole series does as well.

Combine Theorems 12.9.8 and 12.9.9.

The equivalence \(L(\rho )=0\iff e^{tL}(\rho )=\rho \) for all \(t\ge 0\) follows from the generator–semigroup correspondence: exponentiate for the forward direction, and differentiate at \(t=0\) for the reverse.

Let \(\rho _0\) be a rank-deficient fixed density. Let \(P\) be the support projection of \(\rho _0\). Since \(\rho _0\) is rank-deficient, \(P\neq \mathbb {1}\); since \(\rho _0\neq 0\), \(P\neq 0\). Channel positivity and trace preservation force \(e^{tL}\) to preserve \(PM_{D}(\mathbb {C})P\).

First pass from semigroup invariance to generator invariance by Theorem 12.9.8. The generator-level vanishing condition forces \((\mathbb {1}-P)L_jP=0\) and \((\mathbb {1}-P)\kappa P=0\) through the sum-of-squares identity.

The block-upper-triangular relations imply generator-level compression invariance; then apply Theorem 12.9.9.

Combine Theorems 12.9.14 and 12.9.13.

The block-upper-triangular structure implies \(L\) preserves the compression \(PM_{D}(\mathbb {C})P\). Therefore the semigroup \(e^{tL}\) also preserves \(PM_{D}(\mathbb {C})P\) for all \(t\ge 0\). Choose \(t_0{\gt}0\) so that the fixed-point space of \(e^{t_0L}\) is \(\ker L\). The invariant compression contains a density matrix \(\rho _0\) fixed by \(e^{t_0L}\), obtained from the Cesàro means of any density matrix supported in \(PM_{D}(\mathbb {C})P\). Hence \(L(\rho _0)=0\) and \(P\rho _0P=\rho _0\). Since \(P\neq \mathbb {1}\), the matrix \(\rho _0\) has nontrivial kernel.

The conclusion records the adjacent equivalences (1)\(\leftrightarrow \)(2), (2)\(\leftrightarrow \)(3), and (3)\(\leftrightarrow \)(4). They follow from (1)\(\leftrightarrow \)(2), (1)\(\to \)(3), (3)\(\leftrightarrow \)(4), and (4)\(\to \)(2).

Expand the Lindblad operators in a basis of \(V\). The Gram matrix of their coefficient rows is positive semidefinite; its square root gives a family indexed by a basis of \(V\) with the same completely positive and anticommutator sums.

The chosen traceless basis transports the row span of \(\sqrt C\) to the Lindblad span. Since this transport is injective and \(\operatorname{rank}((\sqrt C)^\dagger \sqrt C)=\operatorname{rank}(\sqrt C)\), the two dimensions agree.

Factor \(C=(\sqrt C)^\dagger \sqrt C\). The resulting Lindblad operators are the images of the rows of \(\sqrt C\) under the chosen basis equivalence, so their span has dimension \(\operatorname{rank}(C)\). This gives the upper bound after compressing the family to a basis of its span. For the reverse bound, shift an arbitrary minimizing representation to a traceless representation without changing its cardinality. Wolf’s Proposition 7.4(2) identifies its Lindblad span with the square-root family’s span.

From \(L=\varphi -\kappa (\cdot )-(\cdot )\kappa ^\dagger \) and \((\mathbb {1}-P)L(P)=0\): the \((\mathbb {1}-P)\varphi (P)\) term vanishes because each \((\mathbb {1}-P)L_jP=0\), and the \((\mathbb {1}-P)(P\kappa ^\dagger )\) term vanishes because \((\mathbb {1}-P)P=0\). What remains is \((\mathbb {1}-P)\kappa P=0\).

Assume for contradiction that a block-upper-triangular decomposition exists with nontrivial projection \(P\). By Theorem 12.9.1.8, \((\mathbb {1}-P)\kappa P=0\). Induction on elements of the subalgebra shows \((\mathbb {1}-P)AP=0\) for every generator. Since the generated algebra is all of \(M_{D}(\mathbb {C})\), every matrix satisfies \((\mathbb {1}-P)AP=0\), forcing \(P=0\) or \(P=\mathbb {1}\)—a contradiction.

Assume for contradiction that a block-upper-triangular decomposition exists with nontrivial projection \(P\). The block vanishing \((\mathbb {1}-P)L_jP=0\) and Hermitian closure give \(PL_j(\mathbb {1}-P)=0\) as well, so \([P,L_j]=0\) for all \(j\). But then \(P\) lies in the commutant of the Lindblad span, contradicting triviality.

A block-upper-triangular traceless Lindblad family lies in a proper subspace whose dimension is at most \(d^2-d\). The formal rank minimization argument then contradicts the large-rank bound.

A reducible semigroup would have an invariant compression, and Theorem 12.9.13 would then produce a block-upper-triangular Lindblad form.

Choose \(t_0{\gt}0\) for which the fixed-point space of \(e^{t_0L}\) equals \(\ker L\). If this channel had a nontrivial invariant compression, it would have a density fixed point supported in that compression. This would be a rank-deficient element of \(\ker L\), hence would make the semigroup reducible by Proposition 7.6. Thus \(e^{t_0L}\) is irreducible, and Proposition 7.5 supplies the faithful stationary density and global convergence.

By Theorem 12.9.1.9, no block-upper-triangular decomposition exists. Then apply Theorem 12.9.1.12.

By Theorem 12.9.1.10, no block-upper-triangular decomposition exists. Then apply Theorem 12.9.1.12.

The rank hypothesis rules out block-upper-triangular Lindblad data by Theorem 12.9.1.11. Then apply Theorem 12.9.1.12.

Each of the three hypotheses excludes a block-upper-triangular Lindblad representation and hence excludes reducibility. Apply Theorem 12.9.1.13. In the third case, Theorem 12.9.1.7 identifies the minimum Lindblad-family size with the literal matrix rank \(\operatorname{rank}(C)\).