5 Schwarz Inequalities and Multiplicative Domains

This chapter develops the Schwarz-inequality theory of [ Wol12 , Chapter 5 ] : the Kadison–Schwarz inequality for Kraus maps and the multiplicative domain as a \(*\)-subalgebra on which the map restricts to a \(*\)-homomorphism. Trace adjoints, positive retractions, and the faithful-fixed-point form of peripheral Schwarz equality supply the operator-theoretic results used by the Fundamental Theorem of Chapter 11. The \(k\)-positive and Schmidt-rank theory appears later in the full blueprint.

5.1 Kadison–Schwarz inequality

A linear map is completely positive (Definition 4.1.4) if and only if it has a Kraus form. The Kadison–Schwarz inequality and the multiplicative-domain characterisation are proved first for Kraus maps. The transfer map defined in (??) is a Kraus map (Theorem 4.7.1), so these results apply directly to transfer maps.

Definition 5.1.1 Kraus map

✓

Given operators \(\{ K_i\} _{i=0}^{d-1}\) with \(K_i \in M_{D}(\mathbb {C})\), the Kraus map is

\begin{align} E(X) & = \sum _{i=0}^{d-1} K_i X K_i^\dagger . \label{eq:schwarz_kraus_map} \end{align}

A linear map is completely positive (Definition 4.1.4) if and only if it can be written in this form, as in (??).

Definition 5.1.2 Adjoint Kraus map

✓

The adjoint Kraus map is

\begin{align} E^*(X) & = \sum _{i=0}^{d-1} K_i^\dagger X K_i. \label{eq:schwarz_kraus_adjoint} \end{align}

When the \(\{ K_i\} \) are the matrices of an MPS tensor \(A\), this is the transfer map of the conjugate-transposed family \(i \mapsto (A^i)^\dagger \).

Definition 5.1.3 Unital Kraus map

✓

A Kraus map is unital if \(\sum _{i=0}^{d-1} K_i K_i^\dagger = \mathbb {1}\). Equivalently, \(E(\mathbb {1}) = \mathbb {1}\).

Definition 5.1.4 Trace-preserving Kraus map

✓

A Kraus map is trace-preserving if \(\sum _{i=0}^{d-1} K_i^\dagger K_i = \mathbb {1}\). Equivalently, the adjoint Kraus map is unital. This is the standard MPS normalization condition. In the later gauge language it is the left-canonical condition, so Kadison–Schwarz arguments are often applied to the adjoint map.

Theorem 5.1.5 Kadison–Schwarz inequality

✓

Let \(E(X) = \sum _i K_i X K_i^\dagger \) be a unital Kraus map, so that \(\sum _i K_i K_i^\dagger = \mathbb {1}\). Then, for every \(X \in M_{D}(\mathbb {C})\),

\begin{align} E(X^\dagger X) & \geq E(X)^\dagger E(X). \label{eq:schwarz_kadison} \end{align}

This is [ Wol12 , Equation (5.2) ] .

Proof ▶

Set

\begin{align} v & = \begin{bmatrix} X^\dagger \\ \mathbb {1} \end{bmatrix}, \notag \\ vv^\dagger & = \begin{bmatrix} X^\dagger X & X^\dagger \\ X & \mathbb {1} \end{bmatrix} \geq 0. \notag \end{align}

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 (??).

Theorem 5.1.6 Kadison–Schwarz for the adjoint channel

✓

If \(\sum _i K_i^\dagger K_i = \mathbb {1}\), then the adjoint Kraus map satisfies, for every \(X \in M_{D}(\mathbb {C})\),

\begin{align} E^*(X^\dagger X) & \geq E^*(X)^\dagger E^*(X). \label{eq:schwarz_adjoint_kadison} \end{align}

Proof ▶

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 5.1.5 to \(\{ K_i^\dagger \} \).

Theorem 5.1.7 Hilbert–Schmidt contraction

✓

If a Kraus map is both unital and trace-preserving, then \(\operatorname{tr}(E(X)^\dagger E(X)) \le \operatorname{tr}(X^\dagger X)\) for all \(X \in M_{D}(\mathbb {C})\).

Proof ▶

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

5.2 Multiplicative domain

Theorem 5.2.1 KS gap decomposition

✓

For a unital Kraus map \(E\), the Kadison–Schwarz gap decomposes as

\begin{align} E(X^\dagger X)-E(X)^\dagger E(X) & = \sum _{i=0}^{d-1} R_i^\dagger R_i, \label{eq:schwarz_ks_gap}\\ R_i & := X K_i^\dagger -K_i^\dagger E(X). \label{eq:schwarz_ks_remainder} \end{align}

Proof ▶

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

Theorem 5.2.2 KS equality implies Kraus commutation

✓

Let \(E\) be a unital Kraus map. If \(E(X^\dagger X) = E(X)^\dagger E(X)\) for some \(X\), then, for every \(i\),

\begin{align} X K_i^\dagger & = K_i^\dagger E(X). \label{eq:schwarz_kraus_relation} \end{align}

Proof ▶

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

Theorem 5.2.3 KS equality for peripheral eigenvectors

✓

Let \(E\) be a Kraus map that is both unital and trace-preserving. If \(E(X) = \mu X\) with \(|\mu | = 1\), then the Kadison–Schwarz gap vanishes:

\begin{align} E(X^\dagger X) & = E(X)^\dagger E(X). \label{eq:schwarz_peripheral_equality} \end{align}

Proof ▶

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

\begin{align} \operatorname{tr}(G) & = \operatorname{tr}(E(X^\dagger X))-\operatorname{tr}(E(X)^\dagger E(X)) \notag \\ & = \operatorname{tr}(X^\dagger X)-|\mu |^2\operatorname{tr}(X^\dagger X) =0. \notag \end{align}

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

Theorem 5.2.4 Left multiplicative identity from KS equality

✓

Let \(E\) be a unital Kraus map. If \(E(X^\dagger X) = E(X)^\dagger E(X)\), then, for every \(Y \in M_{D}(\mathbb {C})\),

\begin{align} E(X^\dagger Y) & = E(X)^\dagger E(Y). \label{eq:schwarz_left_multiplicative} \end{align}

Proof ▶

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

\begin{align} E(X^\dagger Y) & = \sum _i K_iX^\dagger YK_i^\dagger = \sum _i E(X)^\dagger K_iYK_i^\dagger = E(X)^\dagger E(Y). \notag \end{align}

Theorem 5.2.5 Right multiplicative identity from KS equality

✓

Let \(E\) be a unital Kraus map. If \(E(X^\dagger X) = E(X)^\dagger E(X)\), then, for every \(Y \in M_{D}(\mathbb {C})\),

\begin{align} E(YX) & = E(Y)E(X). \label{eq:schwarz_right_multiplicative} \end{align}

Proof ▶

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

\begin{align} E(YX) & = \sum _i K_iYXK_i^\dagger = \sum _i K_iYK_i^\dagger E(X) = E(Y)E(X). \notag \end{align}

5.3 Trace adjoints and positive retractions

Definition 5.3.1 Trace-pairing adjoint

✓

The trace-pairing adjoint \(E^* : M_{D'}(\mathbb {C}) \to M_{D}(\mathbb {C})\) of a linear map \(E : M_{D}(\mathbb {C}) \to M_{D'}(\mathbb {C})\) between matrix algebras of possibly different dimensions is the adjoint for the bilinear pairing \((A,B)\mapsto \operatorname{tr}(AB)\).

The trace-duality identities and positivity results behind the applications below are proved in Section A.1. The density and rank-one support arguments for the retraction theorem are collected in Section A.2.

Theorem 5.3.2 Positive retractions onto one matrix factor

✓

Let \(m,d{\gt}0\), and let \(E:M_{md}(\mathbb {C})\to M_{md}(\mathbb {C})\) be a positive complex-linear map. Suppose that the image of \(E\) is contained in \(\mathbb {1}_m\otimes M_{d}(\mathbb {C})\) and that \(E(\mathbb {1}_m\otimes X)=\mathbb {1}_m\otimes X\) for every \(X\in M_{d}(\mathbb {C})\). Then there is a density matrix \(\rho \in M_{m}(\mathbb {C})\) such that, for every \(A\in M_{md}(\mathbb {C})\),

\begin{align} E(A) & =\mathbb {1}_m\otimes \operatorname{tr}_m\! \left((\rho \otimes \mathbb {1}_d)A\right). \label{eq:schwarz_positive_retraction_factor} \end{align}

This is the one-factor case of [ Wol12 , Proposition 1.5 and Equation (1.40) ] .

Proof ▶

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

\begin{align} 0 & \leq R(Y\otimes P_\psi ) \leq \operatorname{tr}(Y)P_\psi . \notag \end{align}

By Lemma A.2.2, 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 A.2.3, 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 A.2.1 therefore supplies a density matrix \(\rho \) satisfying \(f(Y)=\operatorname{tr}(\rho Y)\).

Finally, on simple tensors,

\begin{align} \operatorname{tr}_m\! \left((\rho \otimes \mathbb {1}_d)(Y\otimes X)\right) & =\operatorname{tr}(\rho Y)X =R(Y\otimes X). \notag \end{align}

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

Theorem 5.3.3 Positive retractions onto finite-dimensional star-algebras

✓

Let \(S\subseteq M_{D}(\mathbb {C})\) be a unital \(*\)-subalgebra, and let \(E:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})\) be a positive complex-linear map whose image lies in \(S\) and whose restriction to \(S\) is the identity. There are positive integers \(d_k,m_k\), a unitary \(U\), and density matrices \(\rho _k\in M_{m_k}(\mathbb {C})\) such that

\begin{align} U^*SU & =\bigoplus _k \mathbb {1}_{m_k}\otimes M_{d_k}(\mathbb {C}), \label{eq:schwarz_star_algebra_blocks} \end{align}

and, for every \(A\in M_{D}(\mathbb {C})\),

\begin{align} E(A) & =U\left(\bigoplus _k \mathbb {1}_{m_k}\otimes \operatorname{tr}_{m_k}\! \left((\rho _k\otimes \mathbb {1}_{d_k})(U^*AU)_{kk}\right)\right)U^*. \label{eq:schwarz_star_algebra_retraction} \end{align}

Equivalently, cyclicity of the partial trace permits the factor \(\rho _k\otimes \mathbb {1}_{d_k}\) to be placed on the right of \((U^*AU)_{kk}\). This is [ Wol12 , Proposition 1.5 and Equation (1.40) ] .

Proof ▶

The standard finite-dimensional Wedderburn theorem in [ Wol12 , Proposition 1.5 ] supplies \(U\) and the block form (??). 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 (??). Entrywise expansion of the partial trace gives

\begin{align} \operatorname{tr}_{m_k}((\rho _k\otimes \mathbb {1})A_{kk}) & =\operatorname{tr}_{m_k}(A_{kk}(\rho _k\otimes \mathbb {1})). \notag \end{align}

5.4 Douglas-type factorization

Theorem 5.4.1 Range inclusion implies right factorization

If \(\operatorname{ran}(A) \subseteq \operatorname{ran}(B)\), then there exists \(C\) such that \(A = B C\).

Proof ▶

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 \(C=(B^\dagger B)^\dagger B^\dagger A\).

Theorem 5.4.2 Vectorwise range criterion implies factorization

✓

If \(A v \in \operatorname{ran}(B)\) for every vector \(v\), then there exists \(C\) with \(A = B C\).

Proof ▶

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

5.5 Peripheral Schwarz equality with a faithful fixed point

Theorem 5.5.1 Fixed-point peripheral eigenvectors satisfy KS equality

✓

Let \(E\) be unital and let \(E^*\) be trace-preserving with a positive definite fixed point \(\rho {\gt} 0\). If \(E(X) = \mu X\) with \(|\mu | = 1\), then \(E(X^\dagger X) = E(X)^\dagger E(X)\).

Proof ▶

Define the Kadison–Schwarz gap \(G=E(X^\dagger X)-E(X)^\dagger E(X)\). By Theorem 5.1.5, \(G\succeq 0\). The adjointness identity of Lemma A.3.1, the fixed-point equation \(E^*(\rho )=\rho \), the eigenvalue equation \(E(X)=\mu X\), and \(|\mu |=1\) give

\begin{align} \operatorname{tr}(\rho G) & =\operatorname{tr}(\rho E(X^\dagger X)) -\operatorname{tr}(\rho E(X)^\dagger E(X)) \notag \\ & =\operatorname{tr}(E^*(\rho )X^\dagger X) -\operatorname{tr}(\rho (\mu X)^\dagger (\mu X)) \notag \\ & =\operatorname{tr}(\rho X^\dagger X) -|\mu |^2\operatorname{tr}(\rho X^\dagger X) =0. \notag \end{align}

Since \(\rho {\gt}0\) and \(G\succeq 0\), Lemma A.3.2 yields \(G=0\), which is the asserted equality.

5.6 Additional results on multiplicative domains and order

5.6.1 Abstract Schwarz maps and their multiplicative domains

Definition 5.6.1.1 Schwarz inequality for a linear map

✓

A complex-linear map \(E:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})\) satisfies the Schwarz inequality if, for every \(A\in M_{D}(\mathbb {C})\),

\begin{align} E(A^\dagger A)-E(A^\dagger )E(A) & \succeq 0. \label{eq:schwarz_abstract_inequality} \end{align}

This is [ Wol12 , Equation (5.2) ] .

Definition 5.6.1.2 Abstract one-sided multiplicative domains

✓

For a complex-linear map \(E:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})\), set

\begin{align} \mathcal{A}_R(E) & =\{ A:E(XA)=E(X)E(A)\text{ for every }X\} , \notag \\ \mathcal{A}_L(E) & =\{ A:E(AX)=E(A)E(X)\text{ for every }X\} , \notag \\ \mathcal{A}(E) & =\mathcal{A}_R(E)\cap \mathcal{A}_L(E). \notag \end{align}

These are [ Wol12 , Equations (5.11)–(5.12) ] , with the source’s convention that the subscript records the side on which \(A\) multiplies the varying matrix.

Theorem 5.6.1.3 Abstract multiplicative-domain characterization

✓

If \(E\) satisfies the Schwarz inequality, then

\begin{align} A\in \mathcal{A}_R(E) & \quad \Longleftrightarrow \quad E(A^\dagger A)=E(A^\dagger )E(A), \notag \\ A\in \mathcal{A}_L(E) & \quad \Longleftrightarrow \quad E(AA^\dagger )=E(A)E(A^\dagger ). \notag \end{align}

Consequently, \(A\in \mathcal{A}(E)\) exactly when both equalities hold. This is [ Wol12 , Equations (5.13)–(5.14) ] .

Proof ▶

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

\begin{align} L & =E(A^\dagger Y)-E(A^\dagger )E(Y), \notag \\ R & =E(Y^\dagger A)-E(Y^\dagger )E(A), \notag \\ Q & =E(Y^\dagger Y)-E(Y^\dagger )E(Y)\succeq 0. \notag \end{align}

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

\begin{align} t(L+R)+t^2Q & \succeq 0, \notag \\ it(L-R)+t^2Q & \succeq 0. \notag \end{align}

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

\begin{align} E(A^\dagger Y) & =E(A^\dagger )E(Y), \notag \\ E(Y^\dagger A) & =E(Y^\dagger )E(A). \notag \end{align}

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 \(A^\dagger \).

Theorem 5.6.1.4 Kraus maps satisfy the abstract Schwarz inequality

✓

Every unital Kraus map satisfies the abstract Schwarz inequality of Definition 5.6.1.1.

Proof ▶

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

\begin{align} E(A^\dagger A)-E(A^\dagger )E(A) & =E(A^\dagger A)-E(A)^\dagger E(A)\succeq 0. \notag \end{align}

5.6.2 Kraus specialization and full algebraic structure

Definition 5.6.2.1 Right and left multiplicative domains

✓

For a Kraus map \(E\), define

\begin{align} \mathcal{A}_R(E) & :=\{ X\in M_{D}(\mathbb {C}):\forall Y\in M_{D}(\mathbb {C}), E(XY)=E(X)E(Y)\} , \notag \\ \mathcal{A}_L(E) & :=\{ X\in M_{D}(\mathbb {C}):\forall Y\in M_{D}(\mathbb {C}), E(YX)=E(Y)E(X)\} . \notag \end{align}

We follow the convention that \(\mathcal{A}_R(E)\) controls right multiplication and \(\mathcal{A}_L(E)\) controls left multiplication. These are Kraus-map specializations of the preceding abstract domains, with the names interchanged: here the subscript records the side on which the varying factor is appended, whereas the preceding convention records the side occupied by the fixed element.

Definition 5.6.2.2 Full multiplicative domain

✓

The multiplicative domain of \(E\) is \(\mathcal{A}(E):=\mathcal{A}_R(E)\cap \mathcal{A}_L(E)\).

Remark 5.6.2.3 Comparison of the two naming conventions

✓

Write \(\mathcal{A}_R^{\mathrm W},\mathcal{A}_L^{\mathrm W}\) for Wolf’s convention and \(\mathcal{A}_R^{\mathrm K},\mathcal{A}_L^{\mathrm K}\) for the Kraus convention of this subsection. For the linear map associated with a Kraus family,

\begin{align} \mathcal{A}_R^{\mathrm W} & =\mathcal{A}_L^{\mathrm K}, \notag \\ \mathcal{A}_L^{\mathrm W} & =\mathcal{A}_R^{\mathrm K}, \notag \\ \mathcal{A}^{\mathrm W} & =\mathcal{A}^{\mathrm K}. \notag \end{align}

Theorem 5.6.2.4 Multiplicative-domain characterization

✓

Let \(E\) be a unital Kraus map. Then

\begin{align} X\in \mathcal{A}_R(E) & \quad \Longleftrightarrow \quad E(XX^\dagger )=E(X)E(X)^\dagger , \notag \\ X\in \mathcal{A}_L(E) & \quad \Longleftrightarrow \quad E(X^\dagger X)=E(X)^\dagger E(X). \notag \end{align}

Together these are the Kraus-map specialization of Theorem 5.6.1.3.

Proof ▶

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 5.2.4, and the right-domain implication follows by applying Theorem 5.2.4 to \(X^\dagger \).

Theorem 5.6.2.5 One-sided multiplicative domains are subalgebras

✓

For a unital Kraus map \(E\), both \(\mathcal{A}_R(E)\) and \(\mathcal{A}_L(E)\) are unital subalgebras of \(M_{D}(\mathbb {C})\).

Proof ▶

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 5.6.2.4 iteratively; the \(\mathcal{A}_L(E)\) case is analogous. Containment of \(\mathbb {1}\) follows from unitality of \(E\).

Theorem 5.6.2.6 Multiplicative domain is a \(*\)-subalgebra

✓

For a unital Kraus map \(E\), the full multiplicative domain \(\mathcal{A}(E)\) is a \(*\)-subalgebra of \(M_{D}(\mathbb {C})\). This is the algebraic content of [ Wol12 , Theorem 5.7 ] .

Proof ▶

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

\begin{align} E(X^\dagger X) & =E(X)^\dagger E(X), \notag \\ E(XX^\dagger ) & =E(X)E(X)^\dagger . \notag \end{align}

The characterization of Theorem 5.6.2.4 shows that \(X^\dagger \in \mathcal{A}(E)\). Combined with Theorem 5.6.2.5, this gives a \(*\)-subalgebra.

Theorem 5.6.2.7 Restriction to multiplicative domain is a \(*\)-homomorphism

✓

If \(E\) is unital, then the restricted map \(E|_{\mathcal{A}(E)}:\mathcal{A}(E)\to M_{D}(\mathbb {C})\) is a \(*\)-algebra homomorphism. This is the homomorphism conclusion of [ Wol12 , Theorem 5.7 ] .

Proof ▶

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

5.6.3 Positive maps preserve order and spectral intervals

The elementary order-preservation results used below are proved in Section A.1.

Theorem 5.6.3.1 Order intervals are preserved

✓

Let \(T:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})\) be positive with \(T(\mathbb {1})\le \mathbb {1}\). If \(a\le 0\le b\) and \(a\mathbb {1}\le A\le b\mathbb {1}\), then

\begin{align} a\mathbb {1}& \le T(A)\le b\mathbb {1}. \label{eq:schwarz_order_interval} \end{align}

This is the matrix form of [ Wol12 , Equation (5.21) ] .

Proof ▶

By Theorem A.1.1, \(T(a\mathbb {1})\le T(A)\le T(b\mathbb {1})\). 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 (??).

Theorem 5.6.3.2 Spectral interval contractivity

✓

Let \(T:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})\) be positive with \(T(\mathbb {1})\le \mathbb {1}\), and let \(A\in M_{D}(\mathbb {C})\) be Hermitian. If \(\operatorname{spec}(A)\subseteq [a,b]\) with \(a\le 0\le b\), then

\begin{align} \operatorname{spec}(T(A)) & \subseteq [a,b]. \notag \end{align}

This is [ Wol12 , Equation (5.21) ] .

Proof ▶

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 5.6.3.1 and translate (??) back into spectral bounds.