Apply the identity map in the definition of the Choi matrix.

The \((i_2,j_2)\) slice of \(|\Omega \rangle \! \langle \Omega |\) is \(D^{-1}E_{i_2j_2}\), so equality of the Choi matrices gives \(D^{-1}T(E_{ij})=D^{-1}S(E_{ij})\) entrywise for every matrix unit \(E_{ij}\). Cancel the nonzero factor \(D^{-1}\) and extend from the matrix units to all of \(M_{D}(\mathbb {C})\) by linearity.

The \((i_2,j_2)\) slice of \(|\Omega \rangle \! \langle \Omega |\) is \(D^{-1}E_{i_2j_2}\). Transposition sends this matrix unit to \(D^{-1}E_{j_2i_2}\), whose \((i_1,j_1)\) entry is \(D^{-1}\) exactly when \(i_1=j_2\) and \(i_2=j_1\).

The \((i_2,j_2)\) slice of \(|\Omega \rangle \! \langle \Omega |\) is \(D^{-1}E_{i_2j_2}\). Therefore the \(((i_1,i_2),(j_1,j_2))\) entry of the Choi matrix is

which is the corresponding entry of the stated sum of outer products.

(\(\Rightarrow \)) If \(T(X) = \sum _i K_i X K_i^\dagger \), then \(\tau = \sum _i |v_i\rangle \! \langle v_i|\) where \((v_i)_{(a,b)} = c\, K_i(a,b)\) and \(c = 1/\sqrt{D}\). This is a sum of rank-one PSD matrices.

(\(\Leftarrow \)) If \(\tau \ge 0\), write \(\tau = \sum _m |v_m\rangle \! \langle v_m|\) by spectral decomposition, define \(K_m(a,b) = v_m(a,b)/c\), and recover \(T(X) = \sum _m K_m X K_m^\dagger \) by linearity on matrix units.

Substitute the combination formula, expand the triple sum over \(j,\ell ,\ell '\), swap summation order, and use the entry-wise form of \(U^\dagger U = \mathbb {1}\) to collapse to \(\ell = \ell '\) diagonal terms.

Pad the smaller family by zero operators to a family \(A'\) of size \(r_1\). Evaluating equality of the two maps on matrix units and taking matrix entries gives equality of the Gram matrices of the Stinespring vectors, \(M_B^\dagger M_B=M_{A'}^\dagger M_{A'}\). Hence the assignment from each column of \(M_{A'}\) to the corresponding column of \(M_B\) is well-defined and isometric on their common span. Extend this isometry to the ambient space \(\mathbb {C}^{r_1}\) and let \(W\) be its unitary matrix. Restricting \(W\) to the original \(r_2\) coordinates gives a rectangular matrix \(V\) with \(V^\dagger V=\mathbb {1}_{r_2}\). Reading the column identity entrywise yields \(B_\alpha =\sum _jV_{\alpha j}A_j\).

The two traces are the sums of the squared absolute values of their respective matrix entries. The bijection

identifies the two sums by (11).

Apply Theorem 3.2.3 and substitute (10) into the entrywise Frobenius-norm formula for \(C(\mathcal L)\).

Set \(G=C(\mathcal L)^\dagger C(\mathcal L)\). The contraction inequality applied in the matrix-unit bases shows that every eigenvalue of the positive semidefinite matrix \(G\) is at most one. Hence

Theorem 3.2.3 identifies the left-hand side with \(\| J(\mathcal L)\| _F^2\).

The projector \(|\Omega \rangle \! \langle \Omega |\) is the rank-one matrix built from the vector \(|\Omega \rangle \), so this is Lemma 6.12.2 with \(c=1\) and \(\psi =\Omega \), read after transporting along a bijection between the pair index set of the bipartite space \(\mathbb {C}^d\otimes \mathbb {C}^d\) and a single index set of the same cardinality \(d^2\).

Pass to Choi matrices. The Choi matrix of \(\operatorname{id}\) is \(|\Omega \rangle \! \langle \Omega |\) and the Choi correspondence is linear, so the hypothesis reads \(|\Omega \rangle \! \langle \Omega |=\sum _\alpha \tau _\alpha \), where \(\tau _\alpha \) is the Choi matrix of \(T_\alpha \). Each \(\tau _\alpha \) is positive semidefinite because \(T_\alpha \) is completely positive, so every summand is dominated by the whole sum: \(\tau _\alpha \leq |\Omega \rangle \! \langle \Omega |\). Then Lemma 3.3.1 gives \(\tau _\alpha =c_\alpha |\Omega \rangle \! \langle \Omega |\) with \(c_\alpha \geq 0\). The right-hand side is the Choi matrix of \(c_\alpha \operatorname{id}\), so Lemma 3.1.7 gives \(T_\alpha =c_\alpha \operatorname{id}\).

Evaluate \(\sum _\alpha T_\alpha =\operatorname{id}\) at the identity matrix and take the trace: the left-hand side gives \(d\sum _\alpha c_\alpha \) and the right-hand side gives \(d\). Divide by \(d\geq 1\).

By Theorem 3.3.2, \(T_\alpha (\rho )=c_\alpha \rho \), whose trace is \(c_\alpha \operatorname{tr}(\rho )=c_\alpha \).

The component maps of an instrument are completely positive and their sum is the identity by hypothesis, so Theorem 3.3.4 applies.

Both sides equal \(\sum _{i\in I}\sum _{j\in J}\overline{A_{ij}}B_{ij}\).

Insert \(U_{J,J}^{-1}U_{J,J}=\mathbb {1}\) between the two transported maps.

Complete positivity implies that \(E^*\) preserves adjoints. Therefore, for \(X\in \mathbb C^{J\times J}\) and \(Y\in \mathbb C^{I\times I}\),

The preceding lemma transports this identity to the Euclidean inner products and proves the assertion.

Associativity proves the single-Kraus composition identity. The two trace-adjoint assertions follow from cyclicity and non-degeneracy of the trace pairing. The transport assertions follow by inserting the vectorization and its inverse.

The assertion for \(N=0\) is immediate. For the induction step, apply \(E\) to the expansion for \(E^N(X)\) and distribute the finite sums. Prepending one letter to each word of length \(N\) gives every word of length \(N+1\) exactly once.

A unital completely positive map is contractive in the matrix operator norm, so every eigenvalue has modulus at most one. The spectral radius is the supremum of the eigenvalue moduli, so it is at most one as well, and similarity under Frobenius vectorization preserves the spectrum.

Write \(\Gamma _a(X)=aXa\). The completely positive map

is unital because \(\Phi (\sigma )=\tau \) and trace preservation gives \(\Phi ^*(\mathbb {1})=\mathbb {1}\). The adjoint-square of the Frobenius transport of \(L\) is similar, by congruence with \(\sigma ^{1/4}\), to the Frobenius transport of \(E\). Its spectral radius is therefore at most one. Positivity and self-adjointness of the adjoint-square identify its spectral radius with its operator norm, which proves the contraction estimate.

The square-root identity follows from uniqueness of the positive square root. The support inverse identities follow from the corresponding cancellation relations on the positive spectral subspace. The norm identity follows by cyclicity of the trace and \(W^\dagger W=\mathbb {1}\).

Choose Kraus operators \(A_i\) for \(\Phi \). The joint column Gram matrix of \(A_i\sqrt\sigma \) is \(\tau \), so its column support is \(P\). Since \(\sqrt\sigma \) is invertible, \(PA_i=A_i\) for every \(i\), which gives the two support identities. Cyclicity of the trace then shows that compression by \(V\) preserves the trace of every output.

Let \(P\) be the support projection of \(\tau \), and choose an isometry \(V:\operatorname {supp}\tau \to \mathbb C^J\) with \(VV^\dagger =P\). Since \(\sigma \) is positive definite, every Kraus operator of \(\Phi \) has range in \(\operatorname {supp}\tau \). Hence \(\Psi (A)=V^\dagger \Phi (A)V\) is completely positive and trace preserving, while \(\tau _c=V^\dagger \tau V\) is positive definite and \(\Psi (\sigma )=\tau _c\). The full-support estimate applies to \(\Psi \). Functional calculus on the support gives \(V^\dagger \tau ^{-1/4}_{\operatorname {supp}}V=\tau _c^{-1/4}\), and expansion by \(V\) preserves the Hilbert–Schmidt norm. Transporting the compressed estimate along \(V\) proves the result.

By trace preservation,

where \(\Omega ^{(ij)} = \frac{1}{D} E_{ij}\) is the \((i,j)\)-slice of \(|\Omega \rangle \! \langle \Omega |\). Taking the trace gives \(\frac{1}{D}\delta _{ij}\).

The Choi matrix entries satisfy \(\tau _{(a,i),(b,j)} = \frac{1}{D}(T(E_{ij}))_{ab}\). Hermiticity of \(\tau \) says \(\overline{\tau _{(b,j),(a,i)}} = \tau _{(a,i),(b,j)}\), which unravels to \(T(E_{ji})_{ba} = \overline{T(E_{ij})_{ab}}\), i.e., \(T(E_{ij}^\dagger ) = T(E_{ij})^\dagger \). By linearity this extends to all \(B\).

One has \(\operatorname{tr}(\tau ) = \operatorname{tr}(\operatorname{tr}_A(\tau )) = \operatorname{tr}(\frac{1}{D}\mathbb {1}_D) = 1\).

Working entrywise reduces the claim to the scalar polarization identity

in \(\mathbb {C}\), which holds after substituting \(i^2 = -1\).

Apply the polarization identity summand-by-summand.

Rank-one ray scalar rigidity gives \(T=c\, \operatorname{id}\) for some \(c\in \mathbb {C}\). If \(D{\gt}0\), evaluating at the first coordinate projector gives \(P_{e_0}=T(P_{e_0})=cP_{e_0}\), whose \((0,0)\) entry yields \(c=1\). For \(D=0\) the matrix algebra is trivial, so the conclusion also holds.

Polarization expresses every outer product \(|u\rangle \! \langle v|\) as a complex linear combination of four self-outer-products. Thus \(T\) and \(S\) agree on all outer products, in particular on every matrix unit. Linearity then gives \(T=S\) on all of \(M_{D}(\mathbb {C})\).

Expanding and applying the orthogonality relation \(\sum _i V_{ij}\overline{V_{ij'}} = \delta _{j j'}\) from \(V^\dagger V = \mathbb {1}\) gives

Embed each vector \(\psi _i\), \(\phi _j\) as the \(0\)-th column of a \(D \times D\) matrix with zeros elsewhere; call these \(K_i\), \(L_j \in M_{D}(\mathbb {C})\). A direct entry-wise computation shows that for any \(X \in M_{D}(\mathbb {C})\) both \(\sum _i K_i X K_i^\dagger \) and \(\sum _j L_j X L_j^\dagger \) collapse to \(X_{00} \cdot \rho \), so the density equality \(\rho _\psi = \rho _\phi \) forces the two Kraus families to define the same CP map. Theorem 3.7.9 then supplies an isometry \(V\) with \(V^\dagger V = \mathbb {1}\) and \(K_i = \sum _j V_{ij}\, L_j\); reading the equation off at column \(0\) recovers the vector relation \(\psi _i = \sum _j V_{ij} \phi _j\).

Combine the two directions above.

For any \(N\),

Non-degeneracy of the trace pairing forces \(\sum _i K_i^\dagger K_i = \mathbb {1}\).

Complete positivity gives a finite Kraus family satisfying \(E(X)=\sum _i K_i X K_i^\dagger \). Since \(E\) is trace-preserving, Theorem 3.7.1 gives \(\sum _i K_i^\dagger K_i=\mathbb {1}\).

One has

Evaluate the Kraus formula at the identity matrix.

Expand the Kraus sums, interchange the finite summations, and use \(W^\dagger W = \mathbb {1}\) to collapse the coefficient matrix.

Expand \(\operatorname{tr}(K_j^\dagger K_i)\) using the change of basis. Hilbert–Schmidt orthonormality of both Kraus families identifies the Gram matrix with the identity, yielding the matrix identity \(U U^\dagger = \mathbb {1}_r\), hence also \(U^\dagger U = \mathbb {1}_r\).

Use the trace pairing: for all \(X\),

Nondegeneracy of the trace pairing gives the result.

Specialise Theorem 3.7.7 at \(Y = \mathbb {1}\).

Reindex by choosing enumerations of those finite index sets and apply Theorem 3.1.12.

The implication \((1) \Rightarrow (2)\) is Theorem 3.7.9. Conversely, expand \(\sum _\alpha B_\alpha X B_\alpha ^\dagger \) using \(B_\alpha = \sum _j V_{\alpha j} A_j\), interchange the finite sums, and use \(V^\dagger V = \mathbb {1}\) to collapse the coefficient matrix to the diagonal.

Apply Theorem 3.7.10 with \(|\iota | = |\iota |\). In the square case an isometry matrix is unitary, and the converse is immediate.

For the forward implication, expand the Choi matrix using the Kraus formula and write \(\tau _E = \sum _j v_j v_j^\dagger \) with one vector \(v_j\) for each Kraus operator. The column space of this sum lies in the span of the \(r\) vectors \(\{ v_j\} \), so the rank is at most \(r\). For the converse, use the spectral decomposition of the positive semidefinite Choi matrix, discard the zero-eigenvalue terms, and rescale the remaining eigenvectors into Kraus operators.

Choose \(k\) with \(\psi _k\neq 0\). For \(v\in \mathbb {C}^D\), the matrix

satisfies \(X\psi =v\). Hence the map \(X\mapsto X\psi \) from \(M_{D}(\mathbb {C})\) to \(\mathbb {C}^D\) is surjective. If \(\mathcal S_m=M_{D}(\mathbb {C})\), its image is precisely \(H_m(K,\psi )\).

Apply Theorem 3.8.2 at every sufficiently large length.

The Kraus operators of \(E^m\) are the matrices \(K_\sigma \) with \(\sigma \in I^m\). Hence

The matrix indices retain their row–column order. Since \(D^{-1/2}\neq 0\), the vectors \(v_\sigma \) span \(\mathbb {C}^{D\times D}\) exactly when the matrices \(K_\sigma \) span \(M_{D}(\mathbb {C})\). The finite-frame criterion now gives Equation 18.

Apply Equation 18 pointwise for every sufficiently large \(m\).

Irreducibility gives a positive-definite density matrix \(\rho \) fixed by \(E\). Put \(N=E-P_\rho \), where \(P_\rho (X)=\operatorname {tr}(X)\rho \), and define

For every word length \(m\), expansion of the Kraus-map power gives

The trivial peripheral spectrum implies \(\lVert N^m\rVert \to 0\), while \(E^m=P_\rho +N^m\). Compactness of the unit sphere gives a constant \(\delta {\gt}0\) such that

and the operator-norm estimate gives

Hence \(\operatorname {Re}Q_{E^m}(B){\gt}0\) for every nonzero \(B\) and all sufficiently large \(m\). The trace-pairing identity then excludes a nonzero functional annihilating \(\mathcal S_m\), so \(\mathcal S_m=M_{D}(\mathbb {C})\). Finally, Corollary 3.8.5 gives the Choi statement.

Corollary 3.8.5 gives \(\mathcal S_m=M_{D}(\mathbb {C})\) for every sufficiently large \(m\). Apply Theorem 3.8.3.

The first equality drops the terms outside the image of \(e\), which vanish because \(|0\rangle \! \langle 0| = 0\); the second reindexes along \(e\), which is injective; the third substitutes \(\Psi _{e(i)} = \psi _i\).

Apply Lemma 3.9.1 to the two inclusions \(\iota _1 \hookrightarrow \iota _1 \sqcup \iota _2\) and \(\iota _2 \hookrightarrow \iota _1 \sqcup \iota _2\), which are injective and have complementary images.

Apply Lemma 3.9.1 to the inclusion \(\{ k : k {\lt} m\} \hookrightarrow \{ k : k {\lt} n\} \).

Suppose the two density operators agree. The hypotheses on \(\Psi \) and \(\Phi \) then give \(\sum _k |\Psi _k\rangle \! \langle \Psi _k| = \sum _k |\Phi _k\rangle \! \langle \Phi _k|\), and both families are indexed by the same finite set, so the cardinality hypothesis of Theorem 3.6.7 holds with equality. That theorem supplies \(U \in \mathbb {C}^{\iota \times \iota }\) with \(U^\dagger U = \mathbb {1}\) and \(\Psi _k = \sum _\ell U_{k\ell }\Phi _\ell \). A square matrix over a finite index set with \(U^\dagger U = \mathbb {1}\) also satisfies \(U U^\dagger = \mathbb {1}\), so \(U\) is unitary.

Conversely, a unitary \(U\) satisfies \(U^\dagger U = \mathbb {1}\), so Theorem 3.6.6 gives \(\sum _k |\Psi _k\rangle \! \langle \Psi _k| = \sum _k |\Phi _k\rangle \! \langle \Phi _k|\). Together with the two density agreements this reads

Lemmas 3.9.2 and 3.9.4 show that either padding leaves both density operators unchanged, which is exactly the hypothesis of Theorem 3.9.5.

Use \((A\otimes B)(C\otimes D)=AC\otimes BD\) and \(\mathbb {1}\otimes \mathbb {1}=\mathbb {1}\).

Compute \((V^\dagger (A\otimes \mathbb {1}_r)V)_{ab}\) by summing over the product index \((i,j)\); the \(\delta _{jl}\) from \(\mathbb {1}_r\) collapses the sum over \(l\), giving \(\sum _j K_j^\dagger A K_j\). For Kraus operators of \(T\), the trace pairing identifies this sum with \(T^*(A)\), as displayed in the proof of Theorem 3.10.13.

One has \((V^\dagger V)_{ab} =\sum _{(i,j)}\overline{V_{(i,j),a}}V_{(i,j),b} =\sum _j\sum _i\overline{(K_j)_{ia}}(K_j)_{ib} =(\sum _jK_j^\dagger K_j)_{ab}\).

Expand the middle expression in (26) and sum over \(k\) to recover \(\sum _kK_k\rho K_k^\dagger \).

Choose a Kraus representation of \(E\) and apply the explicit Heisenberg-picture Stinespring formula.

Choose Kraus operators for the channel and use trace preservation to obtain the Stinespring isometry condition \(V^\dagger V=\mathbb {1}\).

By minimality of the Kraus rank, \(T\) has a Kraus family of exactly \(\operatorname{rank}(\tau )\) operators; padding it with zeros gives a family \(\{ K_j\} _{j=0}^{r-1}\) for every \(r\geq \operatorname{rank}(\tau )\). Set \(V:=\sum _j K_j\otimes |j\rangle \) for an orthonormal basis \(\{ |j\rangle \} \) of \(\mathbb {C}^{r}\), so that \(V^\dagger (A\otimes \mathbb {1}_r)V=\sum _j K_j^\dagger AK_j\). The trace pairing gives

for every \(X\in M_{d}(\mathbb {C})\), and nondegeneracy of the trace pairing yields (27). Finally \(V^\dagger V=\sum _j K_j^\dagger K_j\), and the latter equals \(\mathbb {1}_d\) exactly when \(T\) is trace preserving.

The Kraus operators are recovered from \(V\) as the blocks \(K_j=(\mathbb {1}_{d'}\otimes \langle j|)V\), which satisfy \(V=\sum _jK_j\otimes |j\rangle \), so that \(T^*(A)=\sum _jK_j^\dagger AK_j\) for every \(A\in M_{d'}(\mathbb {C})\). For every \(X\in M_{d}(\mathbb {C})\), the trace pairing gives

and nondegeneracy of the trace pairing yields \(T(X)=\sum _jK_jXK_j^\dagger \): an \(r\)-operator Kraus family of \(T\). Any \(r\)-operator Kraus family forces \(\operatorname{rank}(\tau )\le r\), the bound in the minimality of the Kraus rank.

The coordinate formula \(W_{j,(i,a)}=d'^{-1/2}V_{(a,j),i}\) is invertible because \(d'{\gt}0\). Equivalently, with the canonical tensor-factor identifications used in the source, \(V=d'(W\otimes \mathbb {1}_{d'})(\mathbb {1}_d\otimes |\Omega \rangle )\). The coefficient \(d'^2\) printed in the source footnote is a normalization typo; see docs/paper-gaps/wolf_lecture_notes_errata.tex. Expanding the Kronecker product shows directly that multiplication by \(\mathbb {1}_{d'}\otimes C\) on \(V\) becomes multiplication by \(C\) on \(W\).

Write \(V=\sum _jK_j\otimes |j\rangle \). Then \(T(A)=\sum _jK_j^\dagger A K_j\), so \(T\) has Kraus operators \(K_j^\dagger \). Both sides of (32) have entries

at the pair of indices \((i,a),(k,b)\).

Equivalently, (33) is \(W_1^\dagger W_1\leq W_2^\dagger W_2\). Take the positive square root \(S\) of \(W_2^\dagger W_2-W_1^\dagger W_1\). After adjoining zero rows, the matrices with row blocks \((W_1,S,0)\) and \((0,0,W_2)\) have the same Gram matrix. A unitary carrying the second matrix to the first has an \(R_1\times R_2\) corner \(C\) satisfying \(W_1=CW_2\); orthonormality of its columns gives \(C^\dagger C\leq \mathbb {1}_{R_2}\).

The Gram identity gives \(\operatorname{rank}(W_2)=\operatorname{rank}(W_2^\dagger W_2)=\operatorname{rank}(\tau _2)=r_2\), so \(W_2\) has full row rank and is surjective. A surjective right factor can be cancelled.

Complete positivity of \(T_2-T_1\) and the Choi correspondence give \(\tau _1\leq \tau _2\). By Lemma 3.11.4,

which is precisely the squared norm comparison (33). The rectangular factorization lemma gives a contraction \(C\) with \(W_1=CW_2\). Inverting (31) gives (34) with the same supplied \(V_i\).

If \(r_2=\operatorname{rank}(\tau _2)\), then \(\operatorname{rank}(W_2)=\operatorname{rank}(W_2^\dagger W_2)=\operatorname{rank}(\tau _2)=r_2\); hence \(W_2\) is surjective. Thus two coefficients satisfying \(W_1=CW_2=C'W_2\) agree, proving uniqueness.

Direct entrywise computation: \((CC^\dagger )_{ii'}=\sum _jC_{ij}\overline{C_{i'j}}\) collapses to \(\delta _{ii'}\).

The matrix \(C^\dagger C\) is diagonal with a \(1\) on the first \(r\) coordinates and a \(0\) on the last \(s\), so \(\mathbb {1}-C^\dagger C\) is positive semidefinite.

Entrywise expansion of both sides of (35): the Kronecker product with \(C_{r,s}\) picks out the first \(r\) coordinates of the dilation space, matching the action of \(V_{K+\! +L}\) on those coordinates, which is precisely \(V_K\).

Choose a Heisenberg-form Kraus family \(K_1\) of \(T_1\) with Stinespring matrix \(V_1=V_{K_1}\), and Kraus operators \(L\) of the CP map \(T_2-T_1\) in Schrödinger orientation. Let \(L^\dagger \) denote the conjugate-transposed family. Set \(m=r_1+s\) and let \(K_2=K_1\mathbin {+\! +}L^\dagger \). Then

This is the Heisenberg form for \(K_2\), yielding \(T_2(A)=V_{K_2}^\dagger (A\otimes \mathbb {1})V_{K_2}\). Taking \(\widetilde C=C_{r_1,s}\), the intertwining \(V_{K_1}=(\mathbb {1}_D\otimes \widetilde C)V_{K_2}\) is Lemma 3.11.11, and \(\widetilde C^\dagger \widetilde C\le \mathbb {1}\) is Lemma 3.11.10.

First suppose that \(d'{\gt}0\). Choose Kraus families for the maps \(T_i\), and enumerate their disjoint union as \(B_\alpha :\mathbb {C}^{d'}\to \mathbb {C}^d\), with outcome label \(\ell (\alpha )\in I\). The aggregate Stinespring matrix \(\widehat V\), whose \(\alpha \)-th ancilla block is \(B_\alpha ^\dagger \), represents the total map \(T\).

Apply Theorem 3.11.7 to the reflexive order relation \(T\le T\), using \(\widehat V\) for the first supplied representation and \(V\) for the second. It gives a contraction \(C:\mathbb {C}^r\to \mathbb {C}^m\) such that

Let \(C_i\) be the matrix obtained from \(C\) by retaining the rows labelled by \(i\) and setting all other rows to zero, and define the preliminary effect

The labelled fibres partition the rows of \(C\), so

For a nonminimal supplied dilation the middle sum need not yet be the identity. Set \(R=\mathbb {1}_r-C^\dagger C\geq 0\). Since \(V\) and \(\widehat V\) represent the same total map, the contraction identity gives

Choose \(i_0\in I\) and put \(P_{i_0}=E_{i_0}+R\) and \(P_i=E_i\) for \(i\ne i_0\). These operators are positive, their sum is \(\mathbb {1}_r\), and (42) shows that their Stinespring compressions still give the maps \(T_i\).

If \(d'=0\), every matrix \(A\in M_{0}(\mathbb {C})\) is zero. Choose \(i_0\in I\), set \(P_{i_0}=\mathbb {1}_r\) and \(P_i=0\) for \(i\ne i_0\), and use linearity. This separate case preserves the statement without an additional positive-dimension assumption.

This is the specialization of Theorem 3.11.14 to \(d'=d=D\).

Take Heisenberg-form Kraus families \(K_1\) for \(T_1\) and Schrödinger-form \(L\) for \(T_2\), and form \(K=K_1\mathbin {+\! +}L^\dagger \) on \(\mathbb {C}^{r_1+s}\). Choose \(P_1=P_{\mathrm{top}}\), \(P_2=P_{\mathrm{bot}}\) as in Definition 3.11.13. The Kronecker identity \(A\otimes (C^\dagger C) =(\mathbb {1}\otimes C)^\dagger (A\otimes \mathbb {1})(\mathbb {1}\otimes C)\) rewrites \(V^\dagger (A\otimes P_{\mathrm{top}})V\) as \(((\mathbb {1}\otimes C)V)^\dagger (A\otimes \mathbb {1})((\mathbb {1}\otimes C)V)\), which equals \(V_{K_1}^\dagger (A\otimes \mathbb {1})V_{K_1}=T_1(A)\) by Lemma 3.11.11. For the complementary block, \(A\otimes P_{\mathrm{bot}} =A\otimes \mathbb {1}-A\otimes P_{\mathrm{top}}\), and sandwiching by \(V\) yields \((T_1+T_2)(A)-T_1(A)=T_2(A)\).

Apply the existence of an isometric Stinespring dilation (Theorem 3.10.12) and the Schrödinger-picture identity (26).

Rewrite the componentwise identity (44) as a partial trace over the ancilla factor.

Take the isometric open-system representation \(T(\rho )=\operatorname{tr}_E(V\rho V^\dagger )\). Since \(V^\dagger V=\mathbb {1}=W_0^\dagger W_0\), Lemma 10.16.1 gives a unitary \(U\) with \(V=UW_0\). Substituting this identity into the partial trace formula gives (45).

Choose the Stinespring dilation space to have dimension \(r=dd'\). The bound \(\operatorname{rank}(\tau _T)\leq dd'\) and the Schrödinger form of Theorem 3.10.13 give an isometry \(V:\mathbb {C}^d\to \mathbb {C}^{d'}\otimes \mathbb {C}^{dd'}\) satisfying \(T(\rho )=\operatorname{tr}_{\mathbb {C}^{dd'}}(V\rho V^\dagger )\). Reorder the codomain as \(\mathbb {C}^d\otimes \mathbb {C}^{d'}\otimes \mathbb {C}^{d'}\), with the retained output factor last. Choose a normalized \(\varphi \in \mathbb {C}^{d'}\otimes \mathbb {C}^{d'}\) and let \(W_\varphi x=x\otimes \varphi \). Since the reordered Stinespring map and \(W_\varphi \) are isometries, their Gram matrices agree. By Lemma 10.16.1, a unitary \(U\) on the common \(d(d')^2\)-dimensional space satisfies \(V=UW_\varphi \). Substitution, together with \(W_\varphi \rho W_\varphi ^\dagger =\rho \otimes |\varphi \rangle \! \langle \varphi |\), gives (46).

Apply the generalized-inverse construction from quantum steering to the positive summands \(\sigma _i\). Their supports lie in the support of \(\rho \). The resulting effects sum to the projection onto the part of \(\mathbb {C}^r\) seen by \(C\); add the complementary projection to one effect. This preserves (48) and makes the effects sum to the identity.

The supplied Stinespring matrix gives a purification \(C\) of the Choi matrix \(\tau _T\). Complete positivity and \(T=\sum _iT_i\) give a positive decomposition \(\tau _T=\sum _i\tau _{T_i}\). Lemma 3.12.2 yields a POVM satisfying \(\tau _{T_i}=CP_i^{\mathsf T}C^\dagger \). The Choi identity for inserting an environment effect gives (50). Finally,

Restrict \(U\) to inputs of the form \(x\otimes \varphi \) and reorder the output factors. This gives a supplied Stinespring matrix \(V:\mathbb {C}^d\to \mathbb {C}^{d'}\otimes \mathbb {C}^{dd'}\) satisfying (49). Apply Theorem 3.12.3 and rewrite its effect-insertion formula in the original environment-first tensor order. The result is (52), with the same rank bound.

Sum the identities \(E_i=M_i^\dagger M_i\) and use the defining resolution of the identity for the POVM.

The Stinespring Gram identity gives \(V^\dagger V=\sum _iM_i^\dagger M_i=\sum _iE_i=\mathbb {1}_D\) by the resolution of identity.

Each identity follows from direct entrywise computation using the delta-function form in (54).

Computing entrywise,

Take \(r=n\) and the explicit witnesses \(V=\sum _iM_i\otimes |i\rangle \) and \(P_i=\mathbb {1}_D\otimes |i\rangle \! \langle i|\).

Combine the previously proved identities \(V^\dagger V=\mathbb {1}_D\), \(P_i^2=P_i\), \(P_i^\dagger =P_i\), \(P_iP_j=0\) for \(i\ne j\), \(\sum _iP_i=\mathbb {1}\), and \(V^\dagger P_iV=E_i\).

For each outcome \(i\), the \(i\)-th block of \(V\) has Gram matrix \(E_i\). Comparing with the canonical square root \(M_i\) gives \(V_i^\dagger V_i=M_i^\dagger M_i\), hence \(V_i=U_iM_i\) for a unitary \(U_i\) on \(\mathbb {C}^D\). Assemble the \(U_i\) block-diagonally to obtain an isometry \(W=\bigoplus _iU_i\) on \(\mathbb {C}^D\otimes \mathbb {C}^n\), and then \(V=W\sum _iM_i\otimes |i\rangle =WV_0\).

Assemble the \(n\times d\) matrix \(\Psi _{i,j}:=\psi _i(j)\). From the resolution of identity one obtains \(\Psi ^\dagger \Psi =\mathbb {1}_d\). Thus \(\Psi \) is an isometry, which implies \(d\le n\). Let \(J\) be the \(n\times d\) inclusion matrix \(J_{i,j}=1\) if \(i\) equals the embedding of \(j\) into \(\{ 0,\ldots ,n{-}1\} \) and \(0\) otherwise; for \(d\le n\), \(J^\dagger J=\mathbb {1}_d\). By Lemma 10.16.1, there exists a unitary \(U\in U(n)\) with \(\Psi =UJ\), hence \(\psi _i(j)=U_{i,\iota (j)}\) where \(\iota :\{ 0,\ldots ,d{-}1\} \to \{ 0,\ldots ,n{-}1\} \) is the natural inclusion. The rows of \(U\) are the required orthonormal basis. The corollary extracts the vectors from a POVM whose effects are given in rank-one form and applies the main theorem.

Complete positivity of the sum follows from closure of CP maps under finite addition; trace preservation is built into the definition.

Linearity of trace and trace preservation of \(\sum _i\Phi _i\) give

Each \(\Phi _i\) is completely positive, hence positive, so \(\Phi _i(\rho )\) is positive semidefinite and its trace is a non-negative real.

Expand \(\rho \) and each \(T(\sigma _j)\) in the chosen basis and substitute. The trace-self-dual identity (55) replaces coordinates by traces, giving the double-sum formula (56).

The spectral theorem for Hermitian matrices provides an orthonormal eigenbasis with real eigenvalues. Positive semidefiniteness forces those eigenvalues to be non-negative, and the decomposition is written in SVD form with \(U\) the eigenvector unitary and \(\sigma \) the eigenvalue sequence.

Apply the spectral theorem to the positive definite matrix \(M^{\dagger }M\) to obtain \(M^{\dagger }M=V\operatorname{diag}(\lambda )V^{\dagger }\) with \(\lambda _i{\gt}0\). Set \(\sigma _i:=\sqrt{\lambda _i}\) and \(\Sigma :=\operatorname{diag}(\sigma )\); since \(\Sigma \) is a real positive diagonal, it is self-adjoint and invertible. Define \(U:=MV\Sigma ^{-1}\). Then

Thus \(U\) is unitary and the stated decomposition holds.

Direct specialisation of Theorem 3.14.2 to the index type \(\{ 0,\ldots ,D-1\} \times \{ 0,\ldots ,D-1\} \) used by the transfer-matrix representation.

For a real-eigenvalue block, reversal changes the first superdiagonal of \(J_n(a)\) into the first subdiagonal of \(J_n(a)^{\mathsf T}\); the scalar sign multiplies both sides. For a conjugate pair, simultaneous reversal of the chain and realification coordinates also changes \(B(a,\tau )\) into \(B(a,\tau )^{\mathsf T}\), giving the unsigned identity.

In the Pauli basis, pre- and postcomposition by \(\Theta \) multiply the transfer matrix by \(D\) on the right and left, respectively. Averaging with \(D\widehat T D\) therefore gives 68. Time reversal is positive, so \(\Theta \circ T\circ \Theta \) is positive whenever \(T\) is positive, and convexity gives positivity of \(T'\). If \(T(X)=\sum _iK_iXK_i^\dagger \), then \(\Theta \circ T\circ \Theta \) has Kraus operators \(\sigma _2\overline{K_i}\sigma _2\). Hence the same averaging also preserves complete positivity.

Choose a \(D\)th root \(c\) of \(\det X\). Since \(X\) is invertible, both \(\det X\) and \(c\) are nonzero. Put \(S=c^{-1}X\). Multiplicativity of the determinant gives \(\det S=1\), while direct expansion of \((cS)A(cS)^\dagger \) gives \(\Phi _X=|c|^2\Phi _S\). The displayed nonzero Kraus operator gives Kraus rank at most one, and invertibility excludes Kraus rank zero.

Substitute \(\Phi _{X_i}=|c_i|^2\Phi _{S_i}\) and use linearity of \(T\) and of the outer filtering operation. The two real scalars combine as \(|c_1|^2|c_2|^2=|c_1c_2|^2\).

The matrix \(M-\lambda _{\min }(M)\mathbb {1}\) is positive semidefinite by the spectral theorem. Since \(X^{\dagger }X\) is also positive semidefinite, the trace product \(\operatorname{tr}\! \left(X^{\dagger }X(M-\lambda _{\min }(M)\mathbb {1})\right)\) is non-negative. Expanding this expression and cycling the trace gives (74).

The matrix \(A^{\dagger }A\) is positive semidefinite and has determinant \(|\det A|^{2}=1\). Thus the product of its Hermitian eigenvalues is one. Denoting the eigenvalues of \(A^{\dagger }A\) by \(\lambda _{1},\ldots ,\lambda _{D}\geq 0\), the arithmetic-geometric mean inequality gives

Therefore \(\operatorname{tr}(A^{\dagger }A)=\lambda _{1}+\cdots +\lambda _{D}\geq D\).

Use \((A\otimes _{k}B)^{\dagger }=A^{\dagger }\otimes _{k}B^{\dagger }\), the mixed product identity for Kronecker products, and \(\operatorname{tr}(C\otimes _{k}D)=\operatorname{tr}(C)\operatorname{tr}(D)\). With \(C=A^{\dagger }A\) and \(D=B^{\dagger }B\), these identities give (75).

Write \(X=S_{2}\otimes _{k}S_{1}\). Lemma 3.15.1.15 gives the trace lower bound below. Lemma 3.15.1.17 factors the Hilbert–Schmidt trace. For each \(i\in \{ 1,2\} \), Lemma 3.15.1.16 supplies the first row. Together,

Hence any point whose value is at most the value at the identity lies in a fixed Frobenius ball \(\{ \, \| S\| \leq C\, \} \). Intersecting with the closed set \(\det S=1\) produces a compact set, on which the continuous trace functional attains its minimum by the extreme-value theorem. A point outside the sublevel set has value larger than the value at the identity, so the minimiser on the compact set is a global minimiser.

Apply the weighted arithmetic–geometric-mean inequality with uniform weights \(1/D\):

Raising both sides to the \(D\)-th power gives (76).

Both \(\det M=\prod _{i}\lambda _{i}\) and \(\operatorname{tr}M=\sum _{i}\lambda _{i}\) are expressed through the non-negative eigenvalues \(\lambda _i\) of \(M\). The bound is the arithmetic–geometric-mean inequality with uniform weights \(1/D\), raised to the \(D\)-th power; for a \(D \times D\) matrix, equality holds exactly when all eigenvalues coincide, that is, when \(M\) is a scalar matrix. Relabelling along a bijection with the index set of a \(D \times D\) matrix carries the inequality to an arbitrary finite index set of \(D\) elements.

Choose a minimizing pair \((S_1,S_2)\) by Lemma 3.15.1.18. Hold \(S_2\) fixed and set

Positive-definiteness of \(\tau \) and invertibility of \(S_2\) imply that \(M_1\) is positive-definite. The defining minimality of \((S_1,S_2)\) says that \(S_1\) minimizes \(\operatorname{tr}(XM_1X^\dagger )\) over \(X\in \mathrm{SL}(d_1,\mathbb {C})\). A determinant-one whitening of \(M_1\) attains the trace–determinant arithmetic–geometric-mean bound, while its equality case in Lemma 3.15.1.20 forces \(S_1M_1S_1^\dagger \) to be a scalar matrix. This matrix is \(\operatorname{tr}_2[\tau ']\). Holding \(S_1\) fixed and interchanging the two tensor factors gives \(\operatorname{tr}_1[\tau ']\propto \mathbb {1}_{d_2}\) in the same way.

Let \(\tau \) be the Choi matrix of \(T\) and take the minimiser \((S_{1},S_{2})\) of

over \(\det S_{1}=\det S_{2}=1\) from Lemma 3.15.1.18. Set \(\Phi _{1},\Phi _{2}\) to be the filterings with matrices \(S_{1}^{\mathsf T}\) and \(S_{2}\), so that the Choi matrix of \(\Phi _{2}\circ T\circ \Phi _{1}\) equals \((S_{2}\otimes _{k}S_{1})\tau (S_{2}\otimes _{k}S_{1})^{\dagger }\). Holding one filtering factor fixed, the minimiser is optimal in the other coordinate, so each partial trace minimises a functional \(\operatorname{tr}(SMS^{\dagger })\) over \(\det S=1\) for a positive-semidefinite \(M\), where \(\operatorname{tr}(SMS^{\dagger })\geq D(\det M)^{1/D}\), with equality at the minimiser; by the equality case of Lemma 3.15.1.20, this forces \(SMS^{\dagger }\propto \mathbb {1}\), hence both partial traces are proportional to the identity, which is exactly the doubly-stochastic condition.

Apply Theorem 3.15.1.21 to the rectangular Choi matrix \(\tau \) of \(T\), and let \(S_1,S_2\) be the resulting determinant-one matrices. Define \(\Phi _1,\Phi _2\) to be the filterings with matrices \(S_1^{\mathsf T}\) and \(S_2\), respectively. The Choi matrix of \(\Phi _2\circ T\circ \Phi _1\) is then the minimizing representative \((S_2\otimes S_1)\tau (S_2\otimes S_1)^\dagger \). Its two partial traces are scalar by Equation 79, and the rectangular Choi identities \(\operatorname{tr}_B(\tau )=T(\mathbb {1})/d_1\) and \(\operatorname{tr}_A(\tau )=(T^{*}(\mathbb {1}))^{\mathsf T}/d_1\) give the doubly-stochastic condition.

The first identity follows by substituting the four Pauli matrices into the determinant of a \(2\times 2\) matrix. For the second, a positive semidefinite Hermitian matrix has nonnegative trace and determinant, which give \(x_0\geq 0\) and \(q(x)\geq 0\). Conversely, the two real eigenvalues of \(M(x)\) have sum \(2x_0\geq 0\) and product \(q(x)\geq 0\), so both are nonnegative.

Congruence by \(X\) preserves the determinant because \(\det X=1\); polarization then gives preservation of the Minkowski bilinear form. It preserves the future cone because congruence by an invertible matrix preserves positive semidefiniteness. Moreover, \(L(X)_{00}=\operatorname{tr}(XX^\dagger )/2{\gt}0\). The determinant of \(L(X)\) is a real multiplicative character of \(\operatorname{SL}(2,\mathbb {C})\) and is therefore trivial because \(\operatorname{SL}(2,\mathbb {C})\) equals its commutator subgroup. Associativity of congruence gives multiplicativity, and the two scalar signs cancel in \((-X)M(-X)^\dagger \).