Let \(B:\mathbb {C}^I\to \mathbb {C}^n\) be the synthesis map with columns \(v_i\). The frame operator is \(BB^\dagger \), while the vectors span exactly when \(B\) is surjective. A right inverse of \(B\) makes \(B^\dagger \) injective, which gives positive definiteness of \(BB^\dagger \). Conversely, if \(BB^\dagger \) is positive definite, then \(B^\dagger (BB^\dagger )^{-1}\) is a right inverse of \(B\).

Write \(E\) in the Kraus form (1). For each \(i\), let \(d_i\) be the block-diagonal matrix with \(K_i\) on every diagonal block. Then

Each term on the right is positive because conjugation preserves positivity, and a finite sum of positive matrices is positive.

The defining entrywise positivity condition is exactly Theorem 2.2.1.

By the Kraus representation (1), if \(X \ge 0\), then each summand \(K_i X K_i^\dagger \ge 0\) because conjugation preserves positive semidefiniteness. Hence their sum is positive semidefinite.

A quantum channel is completely positive by definition, so Theorem 2.2.3 applies.

Diagonalize \(P\) with non-negative eigenvalues \(\lambda _1,\ldots ,\lambda _D\). Each is at most their sum \(\operatorname{tr}(P)\), so every eigenvalue of \(\operatorname{tr}(P)\mathbb {1}-P\) is non-negative.

Regard \(E\) as the associated positive linear map (Definition 2.4.1). For positive linear maps between matrix \(C^*\)-algebras, \(E(Y^\ast )=E(Y)^\ast \). Since \(X=X^\dagger \) means \(X=X^\ast \), one obtains \(E(X)^\dagger =E(X)^\ast =E(X^\ast )=E(X)\).

The spectral decomposition expresses every density matrix as a convex combination of its unit rank-one eigenprojectors, with the eigenvalues as nonnegative weights summing to one. Conversely, positivity forces every positive summand dominated by a rank-one projector to lie on the same nonnegative ray; the trace-one condition fixes its coefficient to one.

Concatenating Kraus families for the two summands gives a Kraus family for their sum.

If \((A_i)_i\) is a Kraus family, trace preservation and cyclicity give \(\operatorname{tr}((\sum _i A_i^\dagger A_i)X)=\operatorname{tr}(X)\) for every \(X\). Nondegeneracy of the trace pairing implies \(\sum _i A_i^\dagger A_i=\mathbb {1}\).

If \(\mathcal S(X)=\sum _a A_aXA_a^\dagger \) and \(\sum _a A_a^\dagger A_a=I\), cyclicity gives

For every \(X\in M_{D}(\mathbb {C})\), trace duality and trace preservation give

Nondegeneracy of the trace pairing proves the identity.

Each entry \(\tau _{(i_1,i_2),(j_1,j_2)}\) is the value of \(T\) at a fixed matrix, read off at a fixed pair of indices, so (4) and (5) hold entrywise. They make \(T\mapsto \tau _T\) a complex linear map, and (6) is the image of a finite sum under it.

The \((i_2,j_2)\)-slice of \(|\Omega _d\rangle \! \langle \Omega _d|\) is the matrix unit \(E_{i_2 j_2}\) scaled by \(1/d\); substituting it into the defining entries and using linearity gives both compositions.

Expand \(\operatorname{tr}(A\, T(B))\) entrywise, insert the inverse assignment of Theorem 2.6.3, and identify the resulting four-fold sum with \(d\, \operatorname{tr}(\tau \, (A\otimes B^{T}))\).

The entries of \(\tau \) along the diagonal slices are the entries of \(T(E_{i_2 j_2})\) scaled by \(1/d\), so Hermiticity of \(\tau \) is the relation \(T(E_{j_2 i_2})=T(E_{i_2 j_2})^\dagger \) on matrix units; both directions then follow by linearity.

If \(T(X)=\sum _j K_jXK_j^\dagger \), then \(\tau =\sum _j|\psi _j\rangle \! \langle \psi _j|\) with \((\psi _j)_{(i_1,i_2)}=(K_j)_{i_1 i_2}/\sqrt{d}\), so \(\tau \geq 0\). Conversely, a positive semidefinite \(\tau \) decomposes into rank-one outer products \(\sum _j|\psi _j\rangle \! \langle \psi _j|\), and rescaling the vectors \(\psi _j\) by \(\sqrt{d}\) gives Kraus operators via the inverse assignment.

The diagonal slices of \(|\Omega _d\rangle \! \langle \Omega _d|\) sum to \(\mathbb {1}_d/d\), so linearity gives \(\operatorname{tr}_B(\tau )=T(\mathbb {1}_d)/d\); cancelling the factor \(1/d\) gives the equivalence.

The output-factor partial trace has entries \(\operatorname{tr}(T(E_{i_2 j_2}))/d\), which are the entries of \((T^*(\mathbb {1}_{d'}))^{T}/d\) by the defining trace pairing of the adjoint. The equivalence \(\operatorname{tr}(T(X))=\operatorname{tr}(X)\Leftrightarrow T^*(\mathbb {1}_{d'})=\mathbb {1}_d\) is the nondegeneracy of the trace pairing, and transposition and the factor \(1/d\) cancel.

Take the trace of the identity \(\operatorname{tr}_A(\tau )=(T^*(\mathbb {1}_{d'}))^{T}/d\) of Theorem 2.6.8.

Immediate from the partial-trace identities \(\operatorname{tr}_B(\tau )=T(\mathbb {1}_d)/d\) and \(\operatorname{tr}_A(\tau )=(T^*(\mathbb {1}_{d'}))^{T}/d\) of Theorems 2.6.7 and 2.6.8.

The assignment \(T\leftrightarrow \tau \) is linear by Lemma 2.6.2 and one-to-one by Theorem 2.6.3, so it suffices to decompose \(\tau \). The identity

presents \(\tau \) as a complex linear combination of two Hermitian operators, and the spectral decomposition writes each of them as the difference of its positive and negative parts, both positive semidefinite. Every positive semidefinite operator on \(\mathbb {C}^{d'}\otimes \mathbb {C}^{d}\) is the Choi matrix of a completely positive map, so carrying the four positive semidefinite pieces back through the correspondence produces the four completely positive maps, with coefficients \(1,-1,i,-i\). If \(T\) is Hermitian then \(\tau \) is Hermitian by Theorem 2.6.5, so the second summand of (7) vanishes and the positive and negative parts of \(\tau \) alone give two completely positive maps, with coefficients \(1\) and \(-1\).

Cyclicity of the trace gives \(\operatorname{tr}\bigl(\bigl(\sum _j K_j^\dagger K_j\bigr)X\bigr)=\operatorname{tr}(T(X))\) for every \(X\), so the trace-preserving equivalence follows from nondegeneracy of the trace pairing; the unital equivalence is the evaluation at \(X=\mathbb {1}_d\).

Every \(r\)-operator Kraus family gives \(\tau =\sum _{j=1}^{r}|\psi _j\rangle \! \langle \psi _j|\), so \(\operatorname {rank}(\tau )\leq r\). Conversely, the spectral decomposition of \(\tau \geq 0\) into \(\operatorname {rank}(\tau )\) rank-one outer products yields a Kraus family of exactly \(\operatorname {rank}(\tau )\) operators by Theorem 2.6.6, and \(\operatorname {rank}(\tau )\leq dd'\) since \(\tau \) acts on a \(dd'\)-dimensional space.

Choose the \(\psi _j\) in the spectral decomposition of Theorem 2.7.3 to be orthogonal eigenvectors; the corresponding Kraus operators then satisfy \(\operatorname{tr}[K_i^\dagger K_j]=d\, \lambda _i\, \delta _{ij}\), where \(\lambda _i\) is the \(i\)-th nonzero eigenvalue of \(\tau \).

An isometric combination preserves the map by the orthogonality relations of the mixing matrix. Conversely, equal maps have equal trace-pairing adjoints, hence equal Gram matrices for the vectorized Kraus operators \((K_j)_{ab}\); equal Gram matrices are related by a unitary, whose rectangular submatrix is the claimed isometry after zero-padding.

If \((K_{i,j})_j\) is the Kraus family chosen for \(\operatorname{tr}_{B_i}\), then the diagonal-block formula gives

Choose Kraus operators \((V_{i,a})_a\) for \(\operatorname{tr}_{B_i}\) and extend each \(V_{i,a}\) by zero outside the \(i\)th summand. Their controlled family satisfies

Its Kraus form gives complete positivity, and the displayed resolution of the identity gives trace preservation.

Since the involution on \(M_{D}(\mathbb {C})\) is conjugate transpose, \(P^*=P^\dagger \). Therefore \(P^\dagger =P\), \(P^2=P\) if and only if \(P^*=P\), \(P^2=P\).

Self-adjointness follows from \((U^\dagger Q U)^\dagger =U^\dagger Q^\dagger U=U^\dagger Q U\). Idempotence follows from

Expanding column stacking gives

Multiplying the resolution of the identity on both sides by \(Q_k\) gives \(\sum _j Q_kQ_jQ_k=Q_k\). Each summand is positive semidefinite because \(Q_kQ_jQ_k=(Q_jQ_k)^*(Q_jQ_k)\), while the \(j=k\) summand is already \(Q_k\). Subtracting that term and taking traces gives

Hence \(Q_jQ_k=0\) for \(j\ne k\), and taking adjoints gives \(Q_kQ_j=0\).

In the involutive algebra \(M_{D}(\mathbb {C})\), \(P\) satisfies \(P^\dagger =P\) and \(P^2=P\). Hence \((\mathbb {1}-P)^\dagger =\mathbb {1}-P\) and \((\mathbb {1}-P)^2=\mathbb {1}-2P+P^2=\mathbb {1}-P\). Thus \(\mathbb {1}-P\) is an orthogonal projection.

If \(P\) is an invariant projection for \(\mathcal{E}_A\), then \((\mathbb {1}-P)A^iP=0\) for all \(i\). Since the \(\{ A^i\} \) span \(M_{D}(\mathbb {C})\), \((\mathbb {1}-P)MP=0\) for all \(M\), so \(P=0\) or \(P=\mathbb {1}\).

It is already written in the Kraus form (1), with Kraus operators \(\{ A^i\} \).

Apply Lemma 6.13.1 to the Kraus map determined by \((K_i)_i\).

Apply the Kraus-map trace identity directly, since transfer-map notation abbreviates the finite Kraus maps of the two matrix families.

Cyclicity of trace gives

since left canonicity makes the sum in parentheses equal to the identity.

Complete positivity is Theorem 2.10.1. Trace preservation follows by cyclicity and the normalization hypothesis:

If \(D=0\), then the matrix space and its endomorphism algebra contain only zero, so every endomorphism has spectral radius zero.

Put \(c=\| T(\mathbb {1})\| _\infty \). If \(c{\gt}0\), normalize the map by \(S=c^{-1}T\). Positivity gives \(T(\mathbb {1})\geq 0\), and the operator-norm order bound gives \(T(\mathbb {1})\leq c\mathbb {1}\); hence \(S(\mathbb {1})\leq \mathbb {1}\). For a contraction \(A\), one has \(A^\dagger A\leq \mathbb {1}\). Apply Theorem 7.1.2.2 to \(S\), with dominant operator \(\mathbb {1}\), to obtain

The C\(^*\)-identity therefore gives \(\| S(A)\| _\infty \leq 1\). If \(c=0\), the same argument applied directly to \(T\) gives \(T(A)^\dagger T(A)\leq T(\mathbb {1})=0\), and thus \(T(A)=0\). Finally, rescale an arbitrary nonzero \(X\) to the contraction \(A=\| X\| _\infty ^{-1}X\). This proves (12) with coefficient one.

If \(T(X)=\mu X\) for a nonzero \(X\), then Theorem 2.11.5 gives

Division by \(\| X\| _\infty {\gt}0\) proves (6.3). In finite dimension the spectral values are precisely the eigenvalues, so taking the supremum of their moduli proves (6.2).

Unitality gives \(T(\mathbb {1})=\mathbb {1}\), so the nonzero identity matrix is an eigenvector with eigenvalue \(1\). Theorem 2.11.6 gives \(|\mu |\leq \| \mathbb {1}\| _\infty =1\) for every eigenvalue \(\mu \) and \(\varrho (T)\leq 1\). The eigenvalue \(1\) gives the reverse inequality.

Theorem 3.16.5 places every eigenvalue in the closed unit disk. Theorem 8.9.8 supplies a nonzero positive semidefinite fixed point, so \(1\) is an eigenvalue. Hence \(\varrho (T)\leq 1\) and \(\varrho (T)\geq 1\), respectively.

Trace preservation of \(\mathcal K_K\) is the cyclicity computation

and \(\mathcal K_K\) is completely positive by Definition 6.1.1. For \(D\geq 1\), Theorem 3.16.5 therefore gives \(|\mu |\leq 1\). For \(D=0\) the algebra \(M_{0}(\mathbb {C})\) is trivial, so \(\mathcal K_K\) has no eigenvector and the bound is vacuous.

The spectral radius is the supremum of the moduli of the spectral values,

and on the finite-dimensional space \(M_{D}(\mathbb {C})\) the spectral values are exactly the eigenvalues. Theorem 2.11.10 bounds each of them by \(1\).

For each \(i\), set \(C_i=\require{mathtools}\begin{psmallmatrix} A_i& 0\\ 0& B_i\end{psmallmatrix}\) and embed a rectangular matrix \(X\) as \(J(X)=\require{mathtools}\begin{psmallmatrix} 0& X\\ 0& 0\end{psmallmatrix}\). Then

Hence a nonzero eigenvector \(X\) of \(\mathcal M_{A,B}\) with eigenvalue \(\mu \) gives the nonzero eigenvector \(J(X)\) of the trace-preserving Kraus map \(\mathcal K_C\) with the same eigenvalue. Theorem 2.11.10 gives \(|\mu |\leq 1\).

On the finite-dimensional space \(M_{D_1\times D_2}(\mathbb {C})\), every spectral value is an eigenvalue. Theorem 2.11.13 bounds the modulus of each spectral value by \(1\), so the spectral radius is at most \(1\).

By (15), \(P\) is idempotent. The fixed-point relation \(E(\rho )=\rho \) gives \(E\circ P=P\), and trace preservation gives \(P\circ E=P\). These imply \((E-P)\circ P=0\) and \(P\circ (E-P)=0\), so in the binomial expansion of \((P+(E-P))^n\) all cross terms vanish, yielding (16).

\(\rho \) is an eigenvector with eigenvalue \(1\), and \(|1|=1\).

Peripheral eigenvalues are a subset of all eigenvalues, which is a finite set in finite dimensions.

Since \(\mu ^n\) is an eigenvalue for every positive \(n\), the pigeonhole principle on the finite eigenvalue set gives \(\mu ^a=\mu ^b\) for some \(a\neq b\), hence \(\mu ^{|a-b|}=1\).

The closure hypothesis provides \(\mu ^n\in \mathrm{peripheral}(E)\) for all positive \(n\), which in particular means \(\mu ^n\) is an eigenvalue. Theorem 2.13.3 then gives \(\mu ^p=1\) for some \(p{\gt}0\).

Take \(p\) to be the least common multiple of the exponents \(p_\mu \).

The fixed point \(\rho \) gives \(1\in \mathrm{peripheral}(E^p)\). For the reverse inclusion, let \(\nu \in \mathrm{peripheral}(E^p)\). Spectral mapping gives \(\nu =\mu ^p\) for some \(\mu \in \operatorname{spec}(E)\). From \(|\nu |=1\) and \(p{\gt}0\) we obtain \(|\mu |=1\), hence \(\mu \in \mathrm{peripheral}(E)\). The hypothesis \(\mu ^p=1\) then forces \(\nu =1\).

Let \(E(X)=\mu X\) with \(|\mu |=1\), and set \(G:=E(X^\dagger X)-E(X)^\dagger E(X)\). By (3), \(G\geq 0\). Since \(E^*(\rho )=\rho \),

Because \(\rho {\gt}0\) and \(G\geq 0\), this forces \(G=0\), so \(E(X^\dagger X)=E(X)^\dagger E(X)=X^\dagger X\). Hence \(X^\dagger X\) is a positive semidefinite fixed point, which is positive definite by irreducibility (Theorem 8.3.11). Thus \(X\) is invertible. Theorem 6.2.2 gives the corresponding intertwining relation with each Kraus operator. Iterating the multiplicative identity of Theorem 6.2.5 gives \(E(X^n)=\mu ^nX^n\), so \(\mu ^n\) is peripheral.

Combine the closure-under-powers result (Lemma 2.13.2.1) with Theorem 2.13.4.

Choose a normalized Kraus representation \(E(X)=\sum _iK_iXK_i^\dagger \) using Theorem 3.7.2, and set \(L_i=K_i^\dagger \). The normalization \(\sum _iK_i^\dagger K_i=\mathbb {1}\) makes the Kraus map associated with \((L_i)_i\) unital. Theorem 8.1.20 gives a nonzero positive-semidefinite fixed point \(\rho \) of \(E\), and irreducibility makes \(\rho \) positive definite by Theorem 8.3.11.

The adjoint of the Kraus map associated with \((L_i)_i\) is \(E\), so \(\rho \) is a positive-definite fixed point of that adjoint map. Irreducibility is preserved on passing from \((K_i)_i\) to \((K_i^\dagger )_i\), while a peripheral eigenvalue \(\mu \) of \(E\) becomes the peripheral eigenvalue \(\overline\mu \) of the latter Kraus map. Theorem 2.13.2.2 gives \(\overline\mu ^{\, p}=1\) for some \(p{\gt}0\). Taking complex conjugates yields \(\mu ^p=1\).

If \(\mathrm{peripheral}(E)=\{ 1\} \), then \(|\mathrm{peripheral}(E)|=1\). Conversely, if \(|\mathrm{peripheral}(E)|=1\), then \(1\in \mathrm{peripheral}(E)\) by Theorem 2.13.1, so the only element is \(1\).

Let \((E-P)(X)=\nu X\) with \(X\neq 0\). Then \(E(X)=\nu X+P(X)\). If \(\operatorname{tr}(X)=0\), then \(P(X)=0\), so \(E(X)=\nu X\) and hence \(\nu \) is an eigenvalue of \(E\) with \(|\nu |\le 1\). If \(|\nu |=1\), primitivity forces \(\nu =1\), so \(X\) is a trace-zero fixed point of \(E\); by assumption this implies \(X=0\), a contradiction. Thus \(|\nu |{\lt}1\).

If instead \(\operatorname{tr}(X)\neq 0\), trace preservation and \(\operatorname{tr}(P(X))=\operatorname{tr}(X)\), which follows from (15), give

Hence \(\nu \operatorname{tr}(X)=0\) and therefore \(\nu =0\). In either case, \(|\nu |{\lt}1\).