Take \(X=(T^*(\mathbb {1}))^{-1/2}\). It is invertible by Lemma I.1.5, and Theorem I.1.6 gives directly that \(\rho \mapsto T(X\rho X^\dagger )\) is positive and trace-preserving. The normalization behind that theorem is Lemma I.1.4, which gives \(X^\dagger T^*(\mathbb {1})X=\mathbb {1}\). Indeed, the trace-pairing identity and cyclicity then give, for every \(\rho \),

Diagonalize \(A\) with non-negative eigenvalues \(\lambda _i\). Then \(\operatorname{tr}(A^2)=\sum _i\lambda _i^2\) and \(\operatorname{tr}(A)=\sum _i\lambda _i\), so the desired inequality is \(\sum _i\lambda _i^2\le (\sum _i\lambda _i)^2\), which follows because all mixed products are non-negative.

Diagonalize \(A\) with real eigenvalues \(\lambda _i\). Suppose that some eigenvalue \(\lambda _0\) is negative. Set \(T=\sum _i\lambda _i\) and \(R=\sum _{i\ne 0}\lambda _i\). Since \(T\ge 0\) and \(T=\lambda _0+R\), one has \(T{\lt}R\), and hence \(T^2{\lt}R^2\). Cauchy’s inequality gives

contradicting (??). Thus all eigenvalues are non-negative.

Let \(E(X)=\sum _i K_iXK_i^\dagger \) be a Kraus representation. For every \(Y\geq 0\) in \(M_{D}(\mathbb {C})\otimes M_{2}(\mathbb {C})\),

because each summand is positive semidefinite. Thus \(E\otimes \operatorname{id}_2\) is positive, so \(E\) is two-positive.

If \(X\geq 0\), then \(X\otimes E_{11}\geq 0\) in \(M_{D}(\mathbb {C})\otimes M_{2}(\mathbb {C})\). Two-positivity gives

Its \((1,1)\) corner is \(E(X)\), hence \(E(X)\geq 0\). Therefore \(E\) is positive.

Form the \(2\times 2\) block matrix \((\begin{smallmatrix} \mathbb {1} & X \\ X^\dagger & X^\dagger X \end{smallmatrix})\ge 0\). Apply \(E\otimes \operatorname{id}_{2}\) (positive by \(2\)-positivity) and take the Schur complement of the top-left identity block: \(E(X^\dagger X)-E(X)^\dagger E(X)\ge 0\).

This is precisely the preceding definition of the ampliation.

The zero-vector claim is \(\operatorname{rank}(0)=0\). For the dimension bounds, \(\operatorname{SR}(\psi )=\operatorname{rank}(C_\psi )\le D\) and \(\operatorname{SR}(\psi )=\operatorname{rank}(C_\psi )\le k\) follow from the row-rank and column-rank bounds on a \(D\times k\) matrix, hence \(\operatorname{SR}(\psi )\le \min (D,k)\). For a product vector, each column of \(C_{u\otimes v}\) is a scalar multiple of \(u\), so the column span has dimension at most one.

For a finite-dimensional linear map, the support of the singular-value sequence is the initial interval whose length is the dimension of the range. The coefficient matrix \(C_\psi \) represents the corresponding map \(\mathbb {C}^k\to \mathbb {C}^D\), and this range dimension is \(\operatorname{rank}(C_\psi )\).

Theorem I.3.1 provides a rank-\(k\) projection whose trace against \(A\) equals \(S_k(A)\), so the value is attained. Theorem I.3.2 shows that no rank-\(k\) projection exceeds it. Hence \(S_k(A)\) is the maximum.

Write \(C\) for the coefficient matrix of \(\phi \), so that \(\rho =CC^\dagger \) and the overlap with a vector \(\psi \) of coefficient matrix \(B\) is the Frobenius pairing \(\langle \phi | \psi \rangle =\operatorname{tr}(C^\dagger B)\); normalization reads \(\operatorname{tr}(B^\dagger B)=1\) and the Schmidt-rank constraint reads \(\operatorname{rank}(B)\le n\).

For the upper bound, let \(P\) be the orthogonal projection onto the column space of \(B\), so \(\operatorname{rank}(P)\le n\) and \(PB=B\). The Cauchy–Schwarz inequality for the Frobenius pairing gives

Since \(P\) has rank at most \(n\) and \(\rho \) is positive semidefinite, the positive-semidefinite form of Ky Fan’s maximum principle (Theorem 22.2.2.2) gives \(\operatorname{tr}(P\rho )\le \| \rho \| _{(n)}\). Thus \(|\langle \phi | \psi \rangle |^2\le \| \rho \| _{(n)}\) for every admissible \(\psi \).

For the converse, let \(P_n\) be the eigenprojection of \(\rho \) onto its \(n\) largest eigenvalues, so \(\operatorname{tr}(P_n\rho )=\| \rho \| _{(n)}\), and set

Then \(\operatorname{rank}(B)\le \operatorname{rank}(P_n)=n\) and \(\operatorname{tr}(B^\dagger B)=\| \rho \| _{(n)}^{-1}\operatorname{tr}(P_n\rho )=1\), while

Thus the bound is attained. For a normalized \(\phi \) with \(1\le n\), one has \(\| \rho \| _{(n)}\ge \| \rho \| _{(1)}{\gt}0\), since \(\operatorname{tr}\rho =1\), which justifies the normalization in (??).

The Ky-Fan \(D\)-norm sums all eigenvalues of \(\rho \), so \(\| \rho \| _{(D)}=\operatorname{tr}\rho =\| \phi \| ^2=1\). The bound \(1\) is attained at \(\psi =\phi \), whose Schmidt rank is at most \(D\); for any normalized \(\psi \), the Cauchy–Schwarz inequality gives \(|\langle \phi | \psi \rangle |^2\le \| \phi \| ^2\| \psi \| ^2=1\).

By (??), \(\langle \psi |\tau |\psi \rangle =\sum _i\nu _i|\langle \phi _i | \psi \rangle |^2\), while (??) gives \(\sum _i|\langle \phi _i | \psi \rangle |^2=\| \psi \| ^2=1\). Subtracting \(\nu _0\) times the latter from the former gives

For an eigenvalue \(\nu _i{\gt}0\), the factor \(\nu _i-\nu _0\) is non-negative, so those terms only increase the sum and may be dropped. For an eigenvalue \(\nu _i\le 0\), the factor \(\nu _i-\nu _0\) is nonpositive, and Theorem 22.2.2.3 gives \(|\langle \phi _i | \psi \rangle |^2\le \| \rho _i\| _{(n)}\). Multiplication by the nonpositive factor reverses the inequality, which gives (??).

Replacing \(\nu _0\) by \(\nu \) in (??) gives

Every factor \(\nu _i-\nu \) is nonpositive, so all terms are nonpositive and keeping only the \(i=j\) term gives \(\langle \psi |\tau |\psi \rangle \le \nu +(\nu _j-\nu )|\langle \phi _j | \psi \rangle |^2\). Theorem 22.2.2.3 supplies a normalized \(\psi \) of Schmidt rank at most \(n\) attaining \(|\langle \phi _j | \psi \rangle |^2=\| \rho _j\| _{(n)}\); at that vector the right-hand side is the expression in (??).

The eigenbasis expansion (??), together with Parseval’s identity (??), bounds each weight \(\nu _i\) below by \(\nu _{\min }\), giving \(\nu _{\min }\le \langle \psi |\tau |\psi \rangle \). Evaluating at the eigenvector \(\phi _j\) for a least-eigenvalue index \(j\) gives \(\langle \phi _j|\tau |\phi _j\rangle =\nu _j=\nu _{\min }\), so the lower bound is attained and the infimum is (??).

The forward implication is the definition of \(k\)-positivity, since \(|\phi \rangle \! \langle \phi |\) is positive semidefinite. Conversely, every positive semidefinite matrix on \(\mathbb {C}^D\otimes \mathbb {C}^k\) is a finite sum \(\sum _i|\phi _i\rangle \! \langle \phi _i|\), and

by linearity of \(E^{(k)}\), the hypothesis, and closure of the positive semidefinite cone under finite sums.

Write \(\psi =R_X^\dagger \eta \) using Lemma I.4.7. By Theorem 22.2.3.6, the compression \(C_T(X)\) is positive semidefinite. Its quadratic form at \(\eta \) is therefore non-negative, and (??) identifies this number with \(\langle \psi ,\tau _T\psi \rangle \).

By Theorem 22.2.3.6, it is enough to prove that \(C_T(X)\) is positive semidefinite for each \(X\in M_{D,k}(\mathbb {C})(\mathbb {C})\). For any \(\eta \), (??) identifies \(\langle \eta ,C_T(X)\eta \rangle \) with \(\langle R_X^\dagger \eta ,\tau _T R_X^\dagger \eta \rangle \), and Lemma I.4.6 gives \(\operatorname{SR}(R_X^\dagger \eta )\le k\). Thus every quadratic form of \(C_T(X)\) is non-negative. Over a complex finite-dimensional space, polarization recovers Hermiticity from this real non-negative quadratic-form condition, so \(C_T(X)\) is positive semidefinite.

Combine Theorem 22.2.3.3 with Theorem 22.2.3.4.

The forward direction applies the rank-one \(k\)-positivity test to the vector \(\psi _X\) in (??), then uses Lemma I.4.9 in the form

Conversely, given a vector \(\psi \) on the ampliated space, take \(X_{i,p}=D^{1/2}\psi _{(i,p)}\). Since \(D{\gt}0\), this choice satisfies \(D^{-1/2}X_{i,p}=\psi _{(i,p)}\). Substituting this vector in (??) gives \((T\otimes \operatorname{id}_k)(|\psi \rangle \! \langle \psi |)=C_T(X)\). Positivity of \(C_T(X)\) is therefore positivity of the ampliation on the rank-one matrix \(|\psi \rangle \! \langle \psi |\), and the rank-one test gives \(k\)-positivity.

By Theorem 22.2.3.6, \(C_T(X)\ge 0\) for every \(X\in M_{D,k}(\mathbb {C})(\mathbb {C})\). The sandwich formula (??) identifies \(R_X\tau R_X^\dagger \) with \(C_T(X)\).

By Lemma I.4.16, write \(P=VV^\dagger \) with \(V\in M_{D,k}(\mathbb {C})(\mathbb {C})\). The conclusion is then exactly Theorem I.4.15.

By Theorem 22.2.3.6, it is enough to prove positivity of the right-factor compression by an arbitrary \(X\in M_{D,k}(\mathbb {C})(\mathbb {C})\). Choose a rank-\(k\) Hermitian projection \(P\) with \(PX=X\) using Lemma I.4.12. The assumed positivity of \(R_P\tau R_P^\dagger \), together with Lemma I.4.11, gives positivity of the rectangular compression by \(X\).

The forward direction is Theorem 22.2.3.8. The reverse direction is Theorem 22.2.3.9.

For block matrices write \(\rho _{pq}\) and \(X_{pq}\) for their \(M_{D}(\mathbb {C})\)-valued blocks. Expanding the trace by blocks and applying (??) gives

Nondegeneracy of the trace pairing identifies the two adjoints, proving (??).

Since \(E\) is \(k\)-positive, its ampliation \(E^{(k)}\) is positive. The trace-pairing adjoint of this positive map is positive, and (??) identifies that adjoint with \((E^*)^{(k)}\).

One direction is Theorem 22.3.2. The other follows by applying the same theorem to \(E^*\) and using Theorem A.1.4.

Choose a Kraus representation \(E(X)=\sum _i K_iXK_i^\dagger \). For every \(Y\succeq 0\) in \(M_{D}(\mathbb {C})\otimes M_{k}(\mathbb {C})\),

Thus every ampliation \(E\otimes \operatorname{id}_k\) is positive.

The ampliation by \(M_{1}(\mathbb {C})\) is identified with the original matrix algebra: all block indices have the unique value, so positivity of \(E\otimes \operatorname{id}_1\) is exactly positivity of \(E\).

Complete positivity implies \(k\)-positivity for every \(k\). Conversely, assume \(D\ge 1\). If \(E\) is \(D\)-positive, apply \(E\otimes \operatorname{id}_D\) to the maximally entangled projector \(|\Omega \rangle \! \langle \Omega |\). The result is the Choi matrix of \(E\), hence it is positive semidefinite. Choi’s theorem then gives complete positivity.

If \(X\ge 0\), diagonalize \(X\). Its eigenvalues \(\lambda _1,\ldots ,\lambda _D\) are non-negative, and each satisfies \(\lambda _j\le \sum _{i=1}^D\lambda _i=\operatorname{tr}(X)\). Therefore all eigenvalues of \(\operatorname{tr}(X)\mathbb {1}-X\) are non-negative.

Expanding both sides of (??) using (??) and bilinearity of the trace gives \(\operatorname{tr}(\rho )\operatorname{tr}(X)-k^{-1}\operatorname{tr}(\rho X)\) in each case.

The trace term sends the \((a,b)\) slice of \(|\Omega \rangle \langle \Omega |\) to its trace times the identity, giving the contribution \(D^{-1}\mathbb {1}\). The second term is the Choi matrix of \(k^{-1}\) times the identity map, namely \(k^{-1}|\Omega \rangle \langle \Omega |\).

Conjugation by any matrix preserves positive semidefiniteness. Moreover, if \(W=C^\dagger C\), then \(W^T=C^T\overline C=\overline C^\dagger \overline C\geq 0\); hence transposition also preserves positive semidefiniteness, and both maps send the cone into itself. For surjectivity onto the cone, given a positive semidefinite \(A\), set \(W=Y^{-1}A(Y^{-1})^\dagger \), which is again positive semidefinite. Then \(YWY^\dagger =A\) proves the conjugation case, and \(Y(W^T)^TY^\dagger =A\) proves the transpose case, with \(W\) and \(W^T\) positive semidefinite.

The Choi matrix is linear in the map, so it splits over the two terms of \(T_\eta \). The trace term contributes \(\tau (\rho \mapsto \operatorname{tr}(\rho ) \mathbb {1})=D^{-1}\mathbb {1}\), since each elementary slice of \(|\Omega \rangle \langle \Omega |\) contributes its trace times the identity. The identity map contributes \(\tau (\operatorname{id})=|\Omega \rangle \langle \Omega |\). Subtracting \(\eta ^{-1}\) times the second from the first gives \(\tau _\eta =D^{-1}\mathbb {1}-\eta ^{-1}|\Omega \rangle \langle \Omega |\).

By the Schmidt-rank Choi criterion, \(k\)-positivity is the nonnegativity of the Choi quadratic form on all vectors of Schmidt rank at most \(k\). For a normalized such vector \(\psi \), the Choi matrix \(\tau _\eta =D^{-1}\mathbb {1}-\eta ^{-1}|\Omega \rangle \langle \Omega |\) gives \(\langle \psi |\tau _\eta |\psi \rangle =D^{-1}-\eta ^{-1}|\langle \Omega | \psi \rangle |^2\). The reduced density of \(|\Omega \rangle \) on the first factor is \(\mathbb {1}/D\), so by the maximal-overlap principle the supremum of \(|\langle \Omega | \psi \rangle |^2\) over Schmidt-rank-\(k\) normalized vectors is \(\| \mathbb {1}/D\| _{(k)}=k/D\). The infimum of the quadratic form is therefore \(D^{-1}-\eta ^{-1}(k/D)=D^{-1}(1-k/\eta )\), which is non-negative precisely when \(\eta \ge k\). Degree-two homogeneity extends the bound from normalized vectors to all vectors of Schmidt rank at most \(k\).

On \(M_{D}(\mathbb {C})\), \(D\)-positivity is complete positivity, which is positive semidefiniteness of the Choi operator \(\tau _\eta =D^{-1}\mathbb {1}-\eta ^{-1}|\Omega \rangle \langle \Omega |\). Its quadratic form on a vector \(\psi \) is \(D^{-1}\| \psi \| ^2-\eta ^{-1}|\langle \Omega | \psi \rangle |^2\); since \(|\Omega \rangle \) is normalized, Cauchy–Schwarz gives \(|\langle \Omega | \psi \rangle |^2\le \| \psi \| ^2\) with equality at \(\psi =\Omega \), so the form is non-negative for all \(\psi \) precisely when \(\eta \ge D\).

Take \(T_\eta \) at \(\eta =k\). By the threshold equivalence it is \(k\)-positive, since \(\eta =k\ge k\), but not \((k+1)\)-positive, since \(\eta =k{\lt}k+1\).

The two maps differ only in the scalar coefficient of \(X\), namely \(k^{-1}\) against \(\eta ^{-1}\) at \(\eta =k\). The natural-number and real casts of \(k\) into the complex scalars coincide, so the coefficients agree and the maps are equal.

The reduction map equals the family map \(T_\eta \) at \(\eta =k\), and \(T_\eta \) is \(k\)-positive exactly when \(\eta \ge k\); the threshold is met with equality at \(\eta =k\). For \(k{\lt}D\) this is the Schmidt-rank threshold, and the top index \(k=D\) is the completely positive endpoint supplied by the dimension-indexed threshold.

The reduction map equals \(T_\eta \) at \(\eta =k\). For \(k+1{\lt}D\) the threshold gives that \(T_\eta \) is \((k+1)\)-positive iff \(\eta \ge k+1\), so \(T_k\) would require \(k\ge k+1\), which is false. For \(k+1=D\) the dimension-indexed threshold gives that \(T_\eta \) is \(D\)-positive iff \(\eta \ge D\), so \(T_k\) would require \(k\ge D\), contradicting \(k{\lt}D\).

The reduction map is \(T_\eta \) at \(\eta =n\), whose Choi matrix is \(D^{-1}\mathbb {1}-n^{-1}|\Omega \rangle \langle \Omega |\), which is exactly \(W_n\).

Fix block indices \(i_2,j_2\) on the second factor and write \(\rho ^{(i_2,j_2)}\) for the \(D\times D\) slice \(\rho ^{(i_2,j_2)}_{i_1 j_1}=\rho _{(i_1,i_2),(j_1,j_2)}\) on the first factor. Its trace is the corresponding entry of \(\rho _2\):

Applying \(T_\eta \) entry by entry gives

and the first term is exactly the \((i_1,i_2),(j_1,j_2)\) entry of \(\mathbb {1}\otimes \rho _2\).

By the reduction-criterion identity, \((T_n\otimes \operatorname{id})(\rho )\) equals \(\mathbb {1}\otimes \rho _2-n^{-1}\rho \). Scaling a positive semidefinite operator by \(n{\gt}0\) preserves positivity, so \(n\, \mathbb {1}\otimes \rho _2-\rho \ge 0\), which is the first inequality. For the second, the factor swap \(F\) satisfies \(F^2=\operatorname{id}\) and is a unitary reindexing, so it preserves positive semidefiniteness and the order, and it exchanges the reduced densities and the two tensor slots: \((\rho ^{F})_2=\rho _1\) and \((\mathbb {1}\otimes (\rho ^{F})_2-n^{-1}\rho ^{F})^{F} =\rho _1\otimes \mathbb {1}-n^{-1}\rho \). Applying the first form to \(\rho ^{F}\) gives \(n\, \mathbb {1}\otimes (\rho ^{F})_2-\rho ^{F}\ge 0\); conjugating by \(F\) turns this into \(n\, \rho _1\otimes \mathbb {1}-\rho \ge 0\).

On \(|v\rangle \langle v|\) the image is

with \(\overline{v}\) the entrywise conjugate of \(v\). The two vectors \(v\) and \(U\overline{v}\) are orthogonal: antisymmetry of \(U\) gives

hence \(\langle v | U\overline{v} \rangle =0\). The contraction bound gives \(\lVert U\overline{v}\rVert \le \lVert v\rVert \). For any \(w\) the quadratic form is \(\lVert v\rVert ^{2}\lVert w\rVert ^{2} -\lvert \langle v | w \rangle \rvert ^{2} -\lvert \langle U\overline{v} | w \rangle \rvert ^{2}\), which is non-negative because the two-vector Bessel inequality for the orthogonal pair \(v\), \(U\overline{v}\) with \(\lVert U\overline{v}\rVert \le \lVert v\rVert \) gives \(\lvert \langle v | w \rangle \rvert ^{2} +\lvert \langle U\overline{v} | w \rangle \rvert ^{2} \le \lVert v\rVert ^{2}\lVert w\rVert ^{2}\).

A positive semidefinite matrix is a sum of rank-one operators \(|v\rangle \langle v|\). By Lemma 22.7.2 the Breuer–Hall map sends each such operator to a positive semidefinite matrix, and a sum of positive semidefinite matrices is positive semidefinite, so \(T_{\mathrm{BH}}\) is positive.

The diagonal projection of a shifted rank-one projector records the shifted squared moduli: \(D(U_{k0}|v\rangle \! \langle v|U_{k0}^{\dagger })_{ii}=|v_{i-k}|^2\). Substituting this into the definition of \(T_C\) gives the displayed diagonal matrix minus the original rank-one projector.

For every vector \(w\), \(\langle w|(\mathbb {1}-|p\rangle \! \langle p|)|w\rangle =\| w\| ^2-|\langle p | w \rangle |^2\). Since \(\| p\| ^2=\sum _i |p_i|^2\le 1\), Cauchy’s inequality gives \(|\langle p | w \rangle |^2\le \| p\| ^2\| w\| ^2\le \| w\| ^2\), so the quadratic form is non-negative.

Put \(p_i=0\) when \(a_i=0\) and \(p_i=v_i/\sqrt{a_i}\) otherwise. Then

under the same zero-term convention. Lemma 22.8.4 gives \(\mathbb {1}-|p\rangle \! \langle p|\ge 0\). If \(S=\operatorname{diag}(\sqrt{a_i})\), then the support condition gives \(S(\mathbb {1}-|p\rangle \! \langle p|)S^\dagger =\operatorname{diag}(a_i)-|v\rangle \! \langle v|\). Conjugation by \(S\) preserves positive semidefiniteness, which proves the claim.

The matrix in Lemma 22.8.2 has the form \(\operatorname{diag}(a_i)-|v\rangle \! \langle v|\). Each \(a_i\) is non-negative. Since \(n\le d-2\), the coefficient \(d-n\) is positive; hence \(a_i=0\) forces \(v_i=0\). The hypotheses of Lemma 22.8.5 therefore apply and give the stated positivity.

First assume that the denominators are nonzero. After multiplying by their product, the inequality is equivalent to \(x^2y+y^2z+z^2x\ge 3xyz\). This follows from AM–GM applied to the three non-negative numbers \(x^2y\), \(y^2z\), and \(z^2x\). If one of the denominators vanishes, the corresponding two variables vanish. For example, if \(2x+z=0\), then \(x=z=0\), and the remaining term is \(y/(2y)\le 1/2\le 1\) when \(y{\gt}0\), while all terms vanish when \(y=0\). The rank-one positivity then follows from Lemma 22.8.6.

Let \(\Phi \) be the map and let \(X\ge 0\). Scaling the eigenvectors of \(X\) by the square roots of its non-negative eigenvalues produces vectors \(w_\alpha \) with \(X=\sum _\alpha |w_\alpha \rangle \! \langle w_\alpha |\). Linearity gives \(\Phi (X)=\sum _\alpha \Phi (|w_\alpha \rangle \! \langle w_\alpha |)\), each summand is positive semidefinite by hypothesis, and a sum of positive semidefinite matrices is positive semidefinite.

Combine Lemma 22.8.7 with Lemma 22.8.8.

If \(T=0\), every summand vanishes. Otherwise write \(\delta _i=x_i-x_{\sigma (i)}\). Each summand obeys

For \(\sigma (i)=i\) both sides equal \(x_i/T\), and for \(\sigma (i)\ne i\) the difference of the two sides is

by the pair bound \(x_i+x_{\sigma (i)}\le T\) and \(T+\delta _i\ge 2x_i\ge 0\); a vanishing denominator forces \(x_i=0\) and leaves a non-negative right-hand side. Because \(\sigma \) permutes the entries, \(\sum _i x_{\sigma (i)}^2=\sum _i x_i^2\), which gives the identity \(\sum _i x_i\delta _i=\tfrac 12\sum _i\delta _i^2\). Summing the term bounds therefore yields

For \(n=d-2\) the backward shifts \(i-1,\ldots ,i-(d-2)\) run over every index of \(\mathbb {Z}/d\mathbb {Z}\) except \(i\) and \(i+1\), so the shifted sum equals \(T-x_i-x_{i+1}\) and \(a_i=2x_i+T-x_i-x_{i+1}\).

By Lemma 22.8.11, the reciprocal weight sum is \(\sum _i x_i/(T+x_i-x_{i+1})\) with \(x_i=|v_i|^2\). Lemma 22.8.10, applied to the cyclic shift \(\sigma (i)=i+1\), bounds this sum by \(1\), and Lemma 22.8.6 turns the bound into rank-one positivity.

Combine Lemma 22.8.12 with Lemma 22.8.8.

If \(X \ge 0\), write \(X=C^\dagger C\). Then \(X^T=C^T\overline C=\overline C^\dagger \overline C\), so \(X^T \ge 0\). Trace preservation is the identity \(\operatorname{tr}(X^T)=\operatorname{tr}(X)\).

If \(\theta \) were completely positive, its Choi matrix would be positive semidefinite. By Theorem 4.5.7, this Choi matrix is \(D^{-1}F\). For the antisymmetric vector \(v=|01\rangle -|10\rangle \),

contradicting positive semidefiniteness.

Let \(X\ge 0\). Transposition preserves positivity, so \(\theta (X)\ge 0\). Since \(\Phi \circ \theta \) is completely positive, \((\Phi \circ \theta )(\theta (X))\ge 0\). Because \(\theta ^2=\operatorname{id}\), this is exactly \(\Phi (X)\ge 0\).

Write \(\Phi =\Phi _{\mathrm{cp}}+\Phi _{\mathrm{ccp}}\), with \(\Phi _{\mathrm{cp}}\) completely positive and \(\Phi _{\mathrm{ccp}}\) completely copositive. For \(X\ge 0\), the two preceding positivity results give \(\Phi _{\mathrm{cp}}(X)\ge 0\) and \(\Phi _{\mathrm{ccp}}(X)\ge 0\). Their sum is positive, hence \(\Phi (X)\ge 0\).

Both entries equal \(\rho _{(i,k),(j,l)}\): the first transposition exchanges the first index pair, the second restores it.

On the diagonal the row and column indices coincide, so the index exchange in the definition is the identity and each diagonal entry is unchanged.

The entry \((\rho ^{T_1})_{(i,k),(j,l)}=\rho _{(j,k),(i,l)}\) has complex conjugate \(\overline{\rho _{(j,k),(i,l)}}=\rho _{(i,l),(j,k)}\) by Hermiticity of \(\rho \), which is the \(((j,l),(i,k))\) entry of \(\rho ^{T_1}\).

Transposing both factors is the full transpose, so transposing the first factor and then the whole operator transposes only the second factor.

The two partial transposes differ by a global transposition, and transposition preserves positive semidefiniteness, so one is positive exactly when the other is.

Expanding the conjugated operator entry by entry as a double sum over the first factor, the partial transpose exchanges the two summation indices and reverses the roles of the left and right conjugating matrices, leaving them transposed; collecting the Kronecker factors gives the stated form.

By the basis-change law with \(V=U^\dagger \), the new-basis partial transpose equals \(((U U^{\mathsf T})\otimes \mathbb {1}) \rho ^{T_1} ((U U^{\mathsf T})\otimes \mathbb {1})^\dagger \), a conjugation of \(\rho ^{T_1}\) that preserves positive semidefiniteness.

The SWAP operator is self-adjoint, \(F^\dagger =F\), so taking the adjoint of \((A\otimes B)\, F\, (A\otimes B)^\dagger \) returns the same operator.

It is the one-term sum \(\sum _i A_i\otimes B_i\) with the single index taking the value \(A\otimes B\).

Concatenating the two index families realizes \(\rho +\sigma \) as a single sum of Kronecker products of positive semidefinite matrices.

Absorbing \(c\) into the first factor of each Kronecker product keeps every factor positive semidefinite, since a non-negative multiple of a positive semidefinite matrix is positive semidefinite.

Each summand \(A_i\otimes B_i\) is a Kronecker product of positive semidefinite matrices, hence positive semidefinite, and a finite sum of positive semidefinite matrices is positive semidefinite.

The first-factor partial transpose is an entrywise reindexing of the matrix, so it commutes with the entrywise finite sum.

The first-factor partial transpose is linear and acts on a Kronecker product by transposing the first factor, \((A\otimes B)^{T_1}=A^{\mathsf T}\otimes B\). Writing \(\rho =\sum _i A_i\otimes B_i\) with each factor positive semidefinite, the partial transpose is \(\sum _i A_i^{\mathsf T}\otimes B_i\). The transpose of a positive semidefinite matrix is positive semidefinite, so each summand is a Kronecker product of positive semidefinite matrices, and the sum is positive semidefinite.

Applying a map to the first factor is linear in the bipartite matrix, so it commutes with the finite sum.

The reduction map \(T_1\) is positive, so it maps each positive semidefinite first factor \(A_i\) of \(\rho =\sum _i A_i\otimes B_i\) to a positive semidefinite matrix. The ampliation acts by \((T_1\otimes \operatorname{id})(A_i\otimes B_i)=T_1(A_i)\otimes B_i\), each summand is a Kronecker product of positive semidefinite matrices, and the sum is positive semidefinite.

Each summand \(|\psi _i\rangle \langle \psi _i|\) is a rank-one positive semidefinite matrix, and a finite sum of positive semidefinite matrices is positive semidefinite. Concatenating two pure-state families realizes a sum as a single finite sum of pure-state projectors of Schmidt rank at most \(n\), and a pure summand of Schmidt rank at most \(n\) also has Schmidt rank at most any larger bound. Multiplication by a non-negative real \(a\) rescales each pure summand \(\psi _i\) to \(\sqrt a\, \psi _i\), which leaves its Schmidt rank unchanged and reproduces \(a\) times the original state.

A pure state of Schmidt rank at most one has a coefficient matrix of rank at most one, which factors as \(\psi (i,j)=u_i v_j\); its projector is the Kronecker product \(|u\rangle \langle u|\otimes |v\rangle \langle v|\) of two rank-one positive semidefinite matrices, hence separable, and separable matrices are closed under addition. Conversely a Kronecker product \(A\otimes B\) of positive semidefinite matrices, after expanding each factor into rank-one projectors, is a double sum of product-vector projectors, each of Schmidt rank at most one, so it has Schmidt number at most one; a separable matrix is a finite sum of such products.

A convex combination \(a x + b y\) with \(a,b\ge 0\) is a sum of two non-negatively scaled states of Schmidt number at most \(n\), and that bound is preserved under non-negative scaling and under addition; rescaling a vector by a non-negative scalar scales its coefficient matrix and so does not increase the Schmidt rank.

A trace-one state of Schmidt number at most \(n\) is a convex combination of pure states \(|\psi \rangle \langle \psi |\) whose vector \(\psi \) has unit norm and Schmidt rank at most \(n\), so \(S_n\) is the convex hull of the set \(P_n\) of those pure states. The set \(P_n\) is the image, under the continuous projector map \(\psi \mapsto |\psi \rangle \langle \psi |\), of the unit vectors of Schmidt rank at most \(n\). In the Euclidean norm those unit vectors form the unit sphere, which is compact in finite dimension, intersected with the closed set on which the Schmidt rank is at most \(n\) (the coefficient matrix depends continuously on the vector and the matrices of rank at most \(n\) form a closed set); hence \(P_n\) is compact. The trace of \(|\psi \rangle \langle \psi |\) is the squared Euclidean norm of \(\psi \), so the trace-one constraint is exactly the unit-norm constraint and the weights of the convex decomposition are the squared norms, summing to the trace. The convex hull of a compact set is compact in a finite-dimensional real normed space.

The map \(T_n\) is \(n\)-positive for \(1\le n{\lt}D\). Write \(\psi \) through the maximally entangled vector as \(\psi =\psi _X\) for a square matrix \(X\) on the right factor with \(\operatorname{rank}(X)=\operatorname{SR}(\psi )\le n\). Then \((T_n\otimes \operatorname{id})(|\psi \rangle \langle \psi |)\) is the right-factor Choi compression of \(T_n\) by \(X\). Its quadratic form on any vector \(\eta \) is the Choi quadratic form of \(T_n\) on \(R_X^\dagger \eta \), whose Schmidt rank is at most \(\operatorname{rank}(X)\le n\). The Schmidt-rank Choi criterion makes that form non-negative, so the compression is positive semidefinite.

Write \(\psi \) through the maximally entangled vector as \(\psi =\psi _X\) for a square matrix \(X\) on the right factor with \(\operatorname{rank}(X)=\operatorname{SR}(\psi )\le n\). Then \((T\otimes \operatorname{id})(|\psi \rangle \langle \psi |)\) is the right-factor Choi compression of \(T\) by \(X\), and its quadratic form on a vector \(\eta \) equals the Choi quadratic form \(C_T\) of \(T\) evaluated on \(R_X^\dagger \eta \), a vector of Schmidt rank at most \(\operatorname{rank}(X)\le n\). Because \(T\) is \(n\)-positive, \(\langle R_X^\dagger \eta ,\, C_T\, R_X^\dagger \eta \rangle \ge 0\), so the compression is positive semidefinite.

The state is a finite sum of pure-state projectors of Schmidt rank at most \(n\). Applying \(T\) to the first factor is linear, so it distributes over the sum; the pure-state step makes each summand positive semidefinite, and a finite sum of positive semidefinite matrices is positive semidefinite.

Pad the bipartite state to the square system \(M_{D}(\mathbb {C})\otimes M_{D}(\mathbb {C})\) with \(D=\max (d,d')\) by zero rows and columns: when \(d'\le d\) the second factor is padded up to \(d\); when \(d\le d'\) the first factor is padded up to \(d'\) and \(T\) is extended to a map on \(M_{d'}(\mathbb {C})\) that preserves \(n\)-positivity. Padding preserves the Schmidt-number bound, the square result makes the padded ampliation positive semidefinite, and the original ampliation is its principal submatrix, hence positive semidefinite.

The state is a finite sum of pure-state projectors of Schmidt rank at most \(n\). Applying \(T_n\) to the first factor is linear, so it distributes over the sum; the pure-state step makes each summand positive semidefinite, and a finite sum of positive semidefinite matrices is positive semidefinite.

The Schmidt-number premise gives \((T_n\otimes \operatorname{id})(\rho )\ge 0\) on the general system (first step), and the dimension-general operator implication then yields \(n\, \mathbb {1}\otimes \rho _2\ge \rho \). The factor swap of a state of Schmidt number at most \(n\) again has Schmidt number at most \(n\), since it sends each pure summand \(\psi \) to \(\psi \circ \mathrm{swap}\), whose Schmidt coefficient matrix is the transpose \(M_{\psi \circ \mathrm{swap}}=M_\psi ^{\mathsf T}\) of that of \(\psi \) and so has the same rank; applying the first step to the swapped state (whose first factor is \(d'\)) and the operator implication in its symmetric form gives \(n\, \rho _1\otimes \mathbb {1}\ge \rho \).

The real bilinear pairing \((X,H)\mapsto \operatorname{Re}\operatorname{tr}(X H)\) is nondegenerate, because on the diagonal pair \((H,H^{\dagger })\) it equals the squared Frobenius norm \(\operatorname{Re}\operatorname{tr}(H H^{\dagger })=\sum _{i,j}\lvert H_{ij}\rvert ^2\), which is positive unless \(H=0\). Hence the map sending \(H\) to the functional \(X\mapsto \operatorname{Re}\operatorname{tr}(X H)\) is an injective real-linear map from the matrix space to its real dual. These two spaces have the same finite real dimension, so the map is also surjective, and every \(g\) is represented by some \(H\). Replacing \(H\) by its Hermitian part \(\tfrac 12(H+H^{\dagger })\) keeps the value on Hermitian inputs, since \(\operatorname{tr}(X H^{\dagger })=\overline{\operatorname{tr}(X H)}\) when \(X\) is Hermitian, and produces a Hermitian representing matrix.

The set \(S_n\) of trace-one states of Schmidt number at most \(n\) is convex and compact, and \(\rho \notin S_n\). The geometric Hahn–Banach theorem separates the closed convex set \(\{ \rho \} \) from the compact convex set \(S_n\) by a continuous real-linear functional \(f\) and a constant \(c\) with \(f(\rho ){\lt}c{\lt}f(\sigma )\) for every \(\sigma \in S_n\). Representing \(f\) as the trace form of a Hermitian \(H_0\) and setting \(W=H_0-c\, \mathbb {1}\) gives \(\operatorname{Re}\operatorname{tr}(W\rho )=f(\rho )-c{\lt}0\). For a vector \(\psi \) of Schmidt rank at most \(n\) the projector \(|\psi \rangle \langle \psi |\), after normalization by \(\lVert \psi \rVert ^2\), is a trace-one state of \(S_n\), so its value under \(f\) is at least \(c\); multiplying back by the squared norm gives \(\operatorname{Re}\langle \psi |W|\psi \rangle \ge 0\), with the zero vector giving the value \(0\) directly.

The forward implication is the separating-hyperplane construction (22.13.13). For the converse, suppose \(\rho \) had Schmidt number at most \(n\), so \(\rho =\sum _i|\psi _i\rangle \langle \psi _i|\) with each \(\psi _i\) of Schmidt rank at most \(n\). By linearity of the trace, \(\operatorname{Re}\operatorname{tr}(W\rho )=\sum _i\operatorname{Re}\operatorname{tr}\! (W|\psi _i\rangle \langle \psi _i|)\), a sum of non-negative terms, contradicting \(\operatorname{Re}\operatorname{tr}(W\rho ){\lt}0\). This converse uses no density-matrix hypotheses on \(\rho \).

The Choi correspondence \(T\mapsto \tau _T\) is complex-linear and injective. Its domain, the space of linear maps on \(M_{D}(\mathbb {C})\), and its codomain, the bipartite matrix space on \(\mathbb {C}^D\otimes \mathbb {C}^D\), both have complex dimension \(D^4\), so the injective map is also surjective. Hence every \(W\) is a Choi matrix.

By the surjectivity of the Choi correspondence (22.13.15) there is a linear map \(T\) with \(\tau _T=W\). Because \(W\) is Hermitian, the Choi quadratic form \(\langle \psi |W|\psi \rangle \) is real, and it equals the witness expectation \(\operatorname{tr}(W|\psi \rangle \langle \psi |)\); the witness condition makes it non-negative on every vector of Schmidt rank at most \(n\). This is exactly the Schmidt-rank Choi criterion (22.2.3.5) for \(n\)-positivity of \(T\).

Since \(\tau _T=(T\otimes \mathbb {1})(|\Omega \rangle \langle \Omega |)\), cyclicity of the trace gives \(\operatorname{tr}(\tau _T\rho )=\operatorname{tr}(\rho \, (T\otimes \mathbb {1})(|\Omega \rangle \langle \Omega |))\). The \(\mathbb {1}\)-ampliation of the trace-pairing adjoint is the trace-pairing adjoint of the ampliation, so moving \(T\) across the trace pairing replaces \((T\otimes \mathbb {1})\) by \((T^{*}\otimes \mathbb {1})\) acting on \(\rho \) and turns the pairing against \(|\Omega \rangle \langle \Omega |\) into the quadratic form \(\langle \Omega | \cdot \, |\Omega \rangle \).

The entanglement witness \(W\) for \(\rho \) (22.13.13) is the Choi matrix of an \(n\)-positive map \(P\) (22.13.16). Its trace-pairing adjoint \(T=P^{*}\) is again \(n\)-positive (22.3.2). By the Choi trace pairing through the trace-pairing adjoint (22.13.17), \(\langle \Omega |(T\otimes \mathbb {1})(\rho )|\Omega \rangle =\operatorname{tr}(W\rho )\), whose real part is negative. A positive semidefinite matrix has non-negative quadratic forms, so \((T\otimes \mathbb {1})(\rho )\) is not positive semidefinite.

It is a non-negative linear combination of the transpose map and the map \(A \mapsto \operatorname{tr}(A) \mathbb {1}\), and both of these maps send positive semidefinite matrices to positive semidefinite matrices.

Write the Schwarz gap as \(F(A)\). One checks that \(F(A + \lambda \mathbb {1}) = F(A)\) for every scalar \(\lambda \), so it is enough to treat the case \(\operatorname{tr}(A) = 0\). In that case a direct \(2 \times 2\) computation shows that \(F(A)\) dominates \(\frac{1}{2} (A^\dagger A)^T\), hence is positive semidefinite.

Compute the Choi matrix of \(T\) and test it on the antisymmetric vector \(|01\rangle - |10\rangle \). The resulting expectation value is negative, so the Choi matrix is not positive semidefinite.