Take \(X=(T^*(\mathbb {1}))^{-1/2}\). It is invertible by Lemma 4.16.5, and Theorem 4.16.6 gives directly that \(\rho \mapsto T(X\rho X^\dagger )\) is positive and trace-preserving. The normalization behind that theorem is Lemma 4.16.4, 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 (2). 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 4.18.1 provides a rank-\(k\) projection whose trace against \(A\) equals \(S_k(A)\), so the value is attained. Theorem 4.18.2 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 4.2.2.2) 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 (9).

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 (80), \(\langle \psi |\tau |\psi \rangle =\sum _i\nu _i|\langle \phi _i | \psi \rangle |^2\), while (81) 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 4.2.2.3 gives \(|\langle \phi _i | \psi \rangle |^2\le \| \rho _i\| _{(n)}\). Multiplication by the nonpositive factor reverses the inequality, which gives (11).

Replacing \(\nu _0\) by \(\nu \) in (12) 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 4.2.2.3 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 (13).

The eigenbasis expansion (80), together with Parseval’s identity (81), 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 (14).

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 4.19.7. By Theorem 4.2.3.6, the compression \(C_T(X)\) is positive semidefinite. Its quadratic form at \(\eta \) is therefore non-negative, and (85) identifies this number with \(\langle \psi ,\tau _T\psi \rangle \).

By Theorem 4.2.3.6, it is enough to prove that \(C_T(X)\) is positive semidefinite for each \(X\in M_{d\times k}(\mathbb {C})\). For any \(\eta \), (85) identifies \(\langle \eta ,C_T(X)\eta \rangle \) with \(\langle R_X^\dagger \eta ,\tau _T R_X^\dagger \eta \rangle \), and Lemma 4.19.6 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 4.2.3.3 with Theorem 4.2.3.4.

The forward direction applies the rank-one \(k\)-positivity test to the vector \(\psi _X\) in (83), then uses Lemma 4.19.9 in the form

Conversely, given a vector \(\psi \) on the ampliated space, take \(X_{a,p}=d^{1/2}\psi _{(a,p)}\). Since \(d{\gt}0\), this choice satisfies \(d^{-1/2}X_{a,p}=\psi _{(a,p)}\). Substituting this vector in (17) 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 4.2.3.6, \(C_T(X)\ge 0\) for every \(X\in M_{d\times k}(\mathbb {C})\). The sandwich formula (82) identifies \(R_X\tau R_X^\dagger \) with \(C_T(X)\).

By Lemma 4.19.16, write \(P=VV^\dagger \) with \(V\in M_{d\times k}(\mathbb {C})\). The conclusion is then exactly Theorem 4.19.15.

By Theorem 4.2.3.6, it is enough to prove positivity of the right-factor compression by an arbitrary \(X\in M_{d\times k}(\mathbb {C})\). Choose a rank-\(k\) Hermitian projection \(P\) with \(PX=X\) using Lemma 4.19.12. The assumed positivity of \(R_P\tau R_P^\dagger \), together with Lemma 4.19.11, gives positivity of the rectangular compression by \(X\).

The forward direction is Theorem 4.2.3.8. The reverse direction is Theorem 4.2.3.9.

For block matrices write \(\rho _{pq}\) and \(X_{pq}\) for their \(M_{D}(\mathbb {C})\)-valued blocks. Expanding the trace by blocks and applying (20) gives

Nondegeneracy of the trace pairing identifies the two adjoints, proving (19).

Since \(E\) is \(k\)-positive, its ampliation \(E^{(k)}\) is positive. The trace-pairing adjoint of this positive map is positive, and (19) identifies that adjoint with \((E^*)^{(k)}\).

One direction is Theorem 4.3.2. The other follows by applying the same theorem to \(E^*\) and using Theorem 6.11.5.

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.

A normalized product vector has Schmidt rank one, hence Schmidt rank at most \(n\), so \(E_n(\tau )\) is nonempty. For the lower bound, Theorem 4.2.2.8 gives \(\nu _{\min }\le \langle \psi |\tau |\psi \rangle \) for every normalized \(\psi \), without any Schmidt-rank restriction.

Theorem 4.2.2.6 bounds \(\langle \psi |\tau |\psi \rangle \) below by the left-hand side at every normalized \(\psi \) of Schmidt rank at most \(n\), that is, at every element of \(E_n(\tau )\). A lower bound for every element of a nonempty set is a lower bound for its infimum.

Theorem 4.2.2.7 supplies a normalized \(\psi \) of Schmidt rank at most \(n\) whose expectation is at most the right-hand side. Its expectation lies in \(E_n(\tau )\), so the infimum of \(E_n(\tau )\) is at most that expectation, hence at most the right-hand side.

The partial trace preserves the trace, so \(\operatorname{tr}\rho _i=\operatorname{tr}|\phi _i\rangle \! \langle \phi _i|=\langle \phi _i | \phi _i \rangle =1\). The Ky-Fan norm sums the \(n\) largest eigenvalues in descending order, and an index past the dimension \(D'\) contributes zero, so the sum stops growing at \(n=D'\), where it is the sum of all eigenvalues of \(\rho _i\), that is \(\operatorname{tr}\rho _i\).

Every vector has Schmidt rank at most \(D'\), so for \(n\ge D'\) the set \(E_n(\tau )\) is the set of expectations in all normalized vectors. By Theorem 4.2.2.8 that set is bounded below by \(\nu _{\min }\) and contains \(\nu _{\min }\), attained at an eigenvector for the least eigenvalue. This gives (23). For (24), let \(j\) be a minimizing index. If \(\nu _{\min }\le 0\) then \(j\) belongs to the summation range and

every remaining summand being nonpositive. If instead every eigenvalue is positive the summation range is empty and the left-hand side is \(\nu _0\le \nu _{\min }\).

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 (26) using (25) 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 \(k^{-1}|\Omega \rangle \langle \Omega |\).

The operator \(\mathbb {1}-F\) is twice the orthogonal projector onto the antisymmetric subspace, so it is positive semidefinite. The Choi criterion therefore shows that \(T_1\circ \theta \) is completely positive, which means that \(T_1\) is completely copositive. Taking the completely positive summand to be zero gives a decomposable representation of \(T_1\).

If \(P\ge cA\ge 0\) with \(c{\gt}0\), then \(A\ge 0\) and every vector in the kernel of \(P\) lies in the kernel of \(A\). Hence \(A\) is supported on \(Q\), so \(QAQ=A\). Conversely, suppose \(A\ge 0\) and \(QAQ=A\). On the support of \(P\), put \(q=\lVert P^{-1/2}AP^{-1/2}\rVert \), where the inverse square root is taken only on that support, and choose \(c=(q+1)^{-1}\). Then \(c{\gt}0\) and \(cP^{-1/2}AP^{-1/2}\le \mathbb {1}\) on the support of \(P\). Conjugating by the support square root gives \(cA\le P\), and therefore \(A\in C(P)\).

The support-corner characterization gives the inclusion from left to right. For the reverse inclusion, use that the positive semidefinite matrices span \(M_{D}(\mathbb {C})\) over \(\mathbb {C}\). Compress such a spanning expression by \(Q\). Every compressed positive semidefinite matrix \(QBQ\) is positive semidefinite and supported on \(Q\), hence belongs to \(C(P)\) by Lemma 4.4.2.3. Every matrix in the corner is equal to its compression, which proves the reverse inclusion.

By Theorem 4.4.2.4, the span is the matrix corner supported on \(Q=\operatorname{supp}(P)\). This corner is linearly equivalent to \(M_{r}(\mathbb {C})\), where \(r=\operatorname{rank}(Q)\). The support projection has \(\operatorname{rank}(Q)=\operatorname{rank}(P)\), equivalently \(\operatorname{tr}(Q)=\operatorname{rank}(P)\), so the corner has complex dimension \(r^2=\operatorname{rank}(P)^2\).

Put \(A=T(\mathbb {1})\), \(S=\sqrt{A}\), and \(R=S^{-1}\). Positivity and rank preservation imply that \(A\) is positive definite. The normalized map

is therefore positive, unital, and rank preserving on Hermitian matrices. For every Hermitian \(H\) and every real \(\lambda \), unitality and the rank-drop characterization of the spectrum give

Thus \(\widetilde T\) preserves the spectrum of every Hermitian matrix. Theorem 1.10.12 yields a unitary \(U\) such that \(\widetilde T(X)=UXU^\dagger \) for all \(X\), or \(\widetilde T(X)=UX^{\mathsf T}U^\dagger \) for all \(X\). Since \(T(X)=S\widetilde T(X)S\), either formula has the required form with the corrected denormalization factor

This factor, rather than \(U(T(\mathbb {1}))^{-1/2}\) printed in the final line of the cited proof, undoes the displayed normalization.

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.

For (1) \(\Rightarrow \) (2), \(T\) is bijective and \(T^{-1}\) is again positive and maps the positive semidefinite cone onto itself. For \(P\geq 0\), \(T\) maps Wolf’s cone \(C(P)\) onto \(C(T(P))\). The complex spans therefore have equal dimension, so

Hence \(T\) preserves rank on positive semidefinite matrices. Decompose a Hermitian \(H\) as \(H=P_+-P_-\) into its positive and negative parts. Rank subadditivity gives \(\operatorname{rank}(T(H))\leq \operatorname{rank}(H)\); applying the same argument to \(T^{-1}\) gives the converse inequality. Implication (2) \(\Rightarrow \) (3) is Theorem 4.4.2.6, and (3) \(\Rightarrow \) (1) is Theorem 4.4.2.7.

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 4.7.2 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.

Substitute \(U=0\) into the defining formula. The twisted-transpose term vanishes, leaving the first reduction map. Apply Theorem 4.4.1.5.

Positivity is Theorem 4.7.3. For indecomposability, write \(F\) for the swap operator on \(\mathbb {C}^{D}\otimes \mathbb {C}^{D}\) and \(|\Psi \rangle \) for the \(U\)-twisted maximally entangled vector, \(\Psi (i,k)=U(i,k)\), with unnormalized projector \(P_{0}=|\Psi \rangle \langle \Psi |\). The witness is the bipartite state

a sum of positive semidefinite terms (\(\mathbb {1}+F\) is twice the projector onto the permutation-symmetric subspace), hence positive semidefinite. Its first-factor partial transpose is

where \(P_{1}\) is the untwisted maximally entangled projector: the partial transpose of \(\mathbb {1}+F\) is \(\mathbb {1}+P_{1}\), and antisymmetry together with unitarity of \(U\) gives \(P_{0}^{T_{1}}=(U^{\dagger }\otimes \mathbb {1})(\mathbb {1}+F)(U\otimes \mathbb {1})-\mathbb {1}\), so the identity summands cancel and \(\rho ^{T_{1}}\) is again a sum of positive semidefinite terms. Hence \(\rho \) has the PPT property. The carrying identity, valid for every bipartite \(\sigma \) and proved by expanding all tensor indices and contracting against \(U^{\dagger }U=\mathbb {1}\), is

At \(\sigma =\rho \) the three traces evaluate to \(\operatorname{tr}(\rho )=D^{2}+2D\), \(\operatorname{tr}(\rho P_{0})=D^{2}\), and \(\operatorname{tr}(F\rho )=D^{2}\) — the last two because antisymmetry gives \(F|\Psi \rangle =-|\Psi \rangle \) — and therefore

So the quadratic form of \((T_{\mathrm{BH}}\otimes \operatorname{id})(\rho )\) at \(|\Psi \rangle \) is negative and \((T_{\mathrm{BH}}\otimes \operatorname{id})(\rho )\) is not positive semidefinite. By Theorem 4.10.17, \(T_{\mathrm{BH}}\) is not decomposable.

The space \(A\), equipped with the induced norm, is the finite-dimensional \(L^2\) space on \(I\). Transport \(T\) to this \(L^2\) space and use continuity of linear maps in finite dimension. This is the boundedness hypothesis in Nowosad’s Theorem 1.8, not a bound for the supremum norm on a function space.

Lagrange interpolation gives, for each value \(w_i\), the idempotent

and places it in \(P(w)\). Since \(e_{[w_i]}^2=e_{[w_i]}\), the Leibniz identity gives \(D(e_{[w_i]})=0\). For arbitrary \(q\in P(w)\), \(qe_{[w_i]}=q_i e_{[w_i]}\), and a second application gives \(D(q)=0\). Thus the idempotent is attached to the whole value class, rather than to a singleton coordinate which need not belong to \(P(w)\). This is the finite-coordinate form of the point-derivation step following Nowosad’s equations (1.28)–(1.29).

The first and second variations at the two genuine local minima give Nowosad’s normalized variation data. The comparison of the two positive quadratic forms yields the power recursion (1.19)–(1.20), then the Laurent-polynomial identity (1.23). Finite value-class interpolation is the coordinate analogue of the analytic extension (1.27), and gives the product rule (1.28). The null ideal in (1.29) is zero because \(\varphi \) is faithful. Coordinate evaluation therefore gives the point derivations of Theorem 4.8.1.3, which vanish. Undoing the normalization proves the multiplication and constancy conclusions. This is the finite specialization of Nowosad  [ Now68 , pp. 409–418 ] . Finally, \(\lambda _{-T}=-\lambda _T\), so negation exchanges local maxima and local minima exactly as used by Yamagami  [ Yam93 , p. 522 ] .

The row and column sum identities together with \(H\succeq 0\) give Nowosad’s first normalized variation package at \(\mathbf1\) directly. A local maximum at \(a\) transfers under the invertible change of variables to the normalized point \(b=(Sa)/s\); applying the sign change \(T\mapsto -T\) gives the second package. The finite Laurent theorem applied to \(\log b\in P(b)\) makes the normalized operator vanish there. Hence the pulled-back curve \(x(t)=sS^{-1}(b^t)\) has tangent \(sS^{-1}(\log b)=\log b\), which lies in \(\ker H\). If \(a\) is non-scalar, so is this tangent, contradicting \(\ker H=\mathbb {R}\mathbf1\).

This proves only the uniqueness assertion of Yamagami’s Lemma 3 on p. 522. It does not prove that \(\mathbf1\) is itself a local maximum, and it does not prove the cyclic reciprocal inequality.

Yamagami’s printed Lemma 6 permits \(1\le m\le d-l\) and \(s\ge d\) [ Yam93 , pp. 523–524 ] . At \(l=1\), however, its endpoint \(m=d-1\), \(s=d\) is false: the cyclic matrix is \(I+C+\cdots +C^{d-1}\), so every nontrivial-character eigenvalue is zero and the displayed strict disk inequality becomes an equality. The statement above is the valid range \(m\le d-2\) needed by Wolf; the omitted endpoint is valid when \(s{\gt}d\). The source correction is recorded separately in [ con26n ] .

In this corrected range, the source argument at \(s=d\) splits according to \(2m\le d\) or \(d{\lt}2m\). In the first case strict convexity is applied to the \(m\) window characters together with the nonnegative real coefficient \(d/2-m\); in the second, the vanishing sum of the nontrivial character replaces the window by its complement. Translation of the eigenvalue by \(s-d\) gives the result for \(s\ge d\).

Formula (47) gives nonnegativity and both row-sum identities directly. Translation is diagonal in the standard character basis; expanding \(H\) therefore gives (49). The corrected open-disk inequality makes \(q_j\) strictly positive off the constant character and also makes every Fourier multiplier of \(S\) nonzero. Thus \(S\) is invertible. Parseval’s identity proves \(H\succeq 0\) and identifies its kernel with the constant vectors. Matrix inversion then reduces \(\lambda _{S^{-1}}(Sx)\) term by term to \(\sum _i x_i/(Sx)_i\).

The negative-Hessian matrix \(H\) is the one in Yamagami’s Lemmas 2–3 on p. 522; the cyclic diagonalization and disk criterion are Lemma 4 and Corollary 5 on p. 523, followed by Lemma 6 on pp. 523–524  [ Yam93 ] .

The coefficient of the distinguished coordinate is unchanged, since \((s-1)-(m-1)=s-m\). For every retained coordinate, the reduced \((m-1)\)-step forward window embeds in the original \(m\)-step window while preserving cyclic order. Nonnegativity of \(x\) therefore makes the reduced denominator no larger than the original denominator. If the common numerator is positive, then \(s-m{\gt}0\) makes the reduced denominator positive, so the original summand is at most the reduced one. If the numerator is zero, both summands have totalized value zero. Summing these comparisons and removing the zero summand at \(N-1\) proves (51). This is the regular-boundary reduction on p. 525 of [ Yam93 ] .

The comparison cannot in general be made strict. For \((N,m,s)=(5,2,5)\) and \(x=(1,0,0,1,0)\), the original denominators are \((3,1,1,4,1)\), whereas the denominators of the reduced vector \(x'=(1,0,0,1)\) are \((3,0,1,4)\). Both sides of (51) equal \(7/12\). Thus a regular original boundary point can become singular after deletion, and the totalized comparison must be non-strict. It is nevertheless sufficient for induction, because an upper bound for the reduced functional is preserved under equality.

A zero denominator gives a forward window of \(m+1\) zero coordinates. Starting immediately after this window, take the first positive coordinate encountered cyclically; it exists because \(a\) is nonzero. Minimality gives the displayed zero string. Translation by \(j\) is injective on the natural indices \(1,\ldots ,m\) when \(m{\lt}N\), which gives the cardinality statement. No decomposition into maximal zero blocks is used.

Choose \(j\) from Lemma 4.8.1.11. The \(m\) summands immediately preceding \(j\) tend to zero: their numerators tend to zero, while their limiting denominators contain \(a_j{\gt}0\). Every remaining positive summand is nonnegative and at most \(1/(s-m)\). The complement has \(N-m\) elements, which gives the first bound; the second is elementary. This one global count covers any number of simultaneous vanishing denominators. Yamagami prints the single-singularity calculation and only indicates the simultaneous case; the completed step and its precise attribution are recorded in [ con26o ] .

Approximate \(x\ge 0\) by \(y_i^{(r)}=x_i+1/(r+1)\). At every nonsingular limiting denominator the summand converges to its direct boundary value. The omitted singular summands of the positive approximants are nonnegative, while their totalized boundary values are zero. Passing to the limit gives the claim. This lemma isolates a general passage from the positive cone to the totalized boundary. The proof of the cardinal inequality below follows Yamagami’s source order more closely by combining strong induction with separate regular and singular boundary arguments.

Argue by strong induction on \(N\). For \(m=1\), the product-one argument of Lemma 4.8.1.22 gives (54). Suppose next that \(m\ge 2\) and that some coordinate of \(x\) vanishes. Rotate that coordinate to \(N-1\), apply Theorem 4.8.1.10, and use the induction hypothesis for \((N-1,m-1,N-1)\).

It remains to consider a strictly positive \(x\). Suppose that \(f_{N,m,N}(x){\gt}1\), normalize \(x\) to the standard simplex, and take the closure \(K\) of its positive superlevel set. This set is nonempty and compact. It cannot contain a boundary point. Indeed, along a sequence approaching a boundary point with a singular denominator, Lemma 4.8.1.12 gives upper limit at most \(1\). At a boundary point with only nonzero denominators, continuity and the preceding inductive boundary estimate again give at most \(1\). Both alternatives contradict the defining superlevel bound.

The functional is continuous on \(K\), so it has a maximizer \(b\) there. Homogeneity makes \(b\) a global, hence local, maximum on the strictly positive cone. Theorem 4.8.1.9 verifies the cyclic matrix hypotheses of Theorem 4.8.1.6, so \(b\) is scalar. Every positive scalar vector has functional value \(1\), contradicting \(f_{N,m,N}(b){\gt}1\). This proves (54) and follows Yamagami’s argument on pp. 522–525  [ Yam93 ] .

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 4.8.1.18 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 4.8.1.16 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 4.8.1.19 therefore apply and give the stated positivity.

Negating the cyclic index turns the backward window in (55) into Yamagami’s forward window. Apply Theorem 4.8.1.14 with \((N,m,s)=(d,n,d)\) and \(x_i=|v_i|^2\). The resulting reciprocal bound and Lemma 4.8.1.20 give the rank-one statement.

Following Tanahashi–Tomiyama, Lemma 3, first suppose that every \(x_i\) is positive and set \(y_i=x_{i-1}/x_i\). Then \(\prod _i y_i=1\), and the sum becomes \(\sum _i 1/(d-1+y_i)\). After clearing its positive common denominator, multiply the resulting non-negative difference by the positive factor \(d-1\). The desired estimate then reduces to

where \(e_k(y)\) is the \(k\)th elementary symmetric sum. The product of the \(k\)-fold monomials occurring in \(e_k(y)\) is \(1\), so AM–GM gives \(e_k(y)\ge \binom {d}{k}\). The coefficients are non-negative for \(k\ge 2\), while the \(k=0,1\) terms are fixed, and substitution of \(e_k(y)=\binom {d}{k}\) gives equality by the binomial identity obtained at \(y_i=1\).

Tanahashi–Tomiyama assume positive coordinates at this step. To obtain the source’s unrestricted positive-map statement, suppose instead that some \(x_i\) vanishes. Then at most \(d-1\) summands are nonzero. Every nonzero summand is at most \(1/(d-1)\) because \((d-1)x_i+x_{i-1}\ge (d-1)x_i\), so their sum is again at most \(1\).

Put \(x_i=|v_i|^2\). Lemma 4.8.1.22 bounds \(\sum _i x_i/a_i\) by \(1\), and Lemma 4.8.1.20 turns this bound into rank-one positivity.

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 4.8.1.23 with Lemma 4.8.1.24.

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 4.8.1.27, the reciprocal weight sum is \(\sum _i x_i/(T+x_i-x_{i+1})\) with \(x_i=|v_i|^2\). Lemma 4.8.1.26, applied to the cyclic shift \(\sigma (i)=i+1\), bounds this sum by \(1\), and Lemma 4.8.1.20 turns the bound into rank-one positivity.

Combine Lemma 4.8.1.28 with Lemma 4.8.1.24.

The cases \(n=1\) and \(n=d-2\) are Theorems 4.8.1.25 and 4.8.1.29. In the remaining case, \(2\le n\le d-3\); Theorem 4.8.1.21 gives positivity on every rank-one projector, and Lemma 4.8.1.24 extends this to all positive semidefinite matrices.

For fixed \(r\) and \(i\), every column of \(z_{r,i}\) except column \(r\) is, by (56), a scalar multiple of \(a_{i,0}\). Column \(r\) is the single exceptional vector \(c_r\circ a_{i,r}\). Hence all columns lie in the span of two vectors, so the Schmidt rank is at most two. The two averages in (61) have non-negative coefficients, which proves the cone membership and positive semidefiniteness. Ha’s second decomposition is proved next; the later factor-swap and pairing steps are recorded in [ con26e ] .

The complete root sum and Ha’s exponent-collision identity give the entry formula for each \(A_r\), hence for \(A_\gamma ^{T_2}\). Evaluating the projectors in (66) gives the same entry formula. This proves the displayed identity. Each of \(u_i\), \(v_i\), \(\alpha _i\), and \(\beta _{i,j}\) has Schmidt rank at most two, so every summand belongs to \(V_2\) and therefore so does the finite sum.

Expanding \(A^\sigma \) in matrix units and applying \(\Phi \) to its first tensor factor gives (69) term by term. Since the coefficient of \(e_i\otimes e_i\) in \(\Omega _d\) is \(d^{-1/2}\), contraction on both sides contributes \(d^{-1}\); multiplying by \(d\) gives (70). The reindexing assertion follows by reindexing the four finite sums in (68).

Tensor-factor exchange sends each projector onto a vector of Schmidt rank at most two to another such projector. Apply this observation first to \(A_\gamma \in V_2\). The identity \((A^\sigma )^{T_1}=(A^{T_2})^\sigma \), followed by Theorem 4.8.1.34, gives the second assertion in (71).

In cyclic coordinates, expansion of \(T_C\) reduces the pairing to

The correlated entries are all \(1\). In the last sum, the column \(k=1\) contributes \(\gamma ^2+d-1\), while each of the remaining \(n-1\) columns contributes \(d\). Consequently,

Simultaneously reindexing the cyclic coordinates by the equivalence \(\mathbb Z/d\mathbb Z\simeq \{ 0,\ldots ,d-1\} \) gives the standard-coordinate identity, and (70) gives its normalized maximally entangled form. Finally \(0{\lt}\gamma {\lt}1\) implies \(\gamma ^2-1{\lt}0\).

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 3.1.8, 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 \(\Phi (X)\ge 0\).

If \(W=P_1+P_2^{T_1}\) and \(W'=Q_1+Q_2^{T_1}\), then for \(a,b\geq 0\) with \(a+b=1\),

Both matrices in parentheses are positive semidefinite.

Let \(\mathcal D\) be the set of positive semidefinite trace-one operators on \(\mathbb {C}^d\otimes \mathbb {C}^{d'}\). Every positive semidefinite operator on this space has Schmidt number at most \(d\), because every vector in a rank-one decomposition has Schmidt rank at most \(d\). Thus \(\mathcal D\) is the compact Schmidt-number section \(S_d\). First-factor partial transposition is a continuous coordinate permutation and commutes with scalar multiplication. The normalized decomposable set is therefore the image of the compact set \([0,1]\times \mathcal D\times \mathcal D\) under the continuous map

This formula always gives a decomposable operator of trace one. Conversely, write a trace-one decomposable operator as \(W=P_1+P_2^{T_1}\). Positivity gives \(\operatorname{tr}(P_1),\operatorname{tr}(P_2)\geq 0\), while \(\operatorname{tr}(P_1)+\operatorname{tr}(P_2)=1\). Normalize each nonzero summand and use \(a=\operatorname{tr}(P_1)\). A positive summand of trace zero vanishes, so in that case it may be replaced by any fixed density operator. This proves the claimed identification. Convexity now follows by intersecting the convex cone of Theorem 4.10.5 with the affine trace-one hyperplane.

Expand the trace in product indices. Both sides are the same finite fourfold sum after interchanging the two first-factor indices. Taking real parts gives the assertion for the real trace pairing.

Write a decomposable map in Wolf’s form \(T=T_1+T_2\circ \theta \), with \(T_1,T_2\) completely positive. The trace adjoint reverses composition and transposition is self-adjoint, so

Rectangular Choi matrices place the output factor first. Consequently, postcomposition by \(\theta \) becomes first-factor partial transposition:

The Choi matrices of \(T_1^*\) and \(T_2^*\) are positive semidefinite by complete positivity, which proves the forward implication. Conversely, reconstruct completely positive maps \(Q_1,Q_2\colon M_{d'}(\mathbb {C})\to M_{d}(\mathbb {C})\) from \(P_1,P_2\geq 0\). Choi injectivity gives \(T^*=Q_1+\theta \circ Q_2\); taking trace adjoints again yields \(T=Q_1^*+Q_2^*\circ \theta \), a decomposable map.

Write \(\Phi =\Phi _{\mathrm{cp}}+\Phi _{\mathrm{ccp}}\) with \(\Phi _{\mathrm{cp}}\) completely positive and \(\Phi _{\mathrm{ccp}}\) completely copositive. Complete positivity implies positivity of every finite ampliation, so \((\Phi _{\mathrm{cp}}\otimes \operatorname{id})(\rho )\ge 0\). For the second summand, ampliating \(\Phi _{\mathrm{ccp}}\) after transposition equals ampliating \(\Phi _{\mathrm{ccp}}\) after the first-factor partial transpose of the input,

because both sides evaluate \(\Phi _{\mathrm{ccp}}\) on the transposed \((i_{2},j_{2})\)-slice of \(\rho \). Since \(\Phi _{\mathrm{ccp}}\circ \theta \) is completely positive and \(\rho ^{T_{1}}\ge 0\) is the PPT hypothesis, the right-hand side is positive semidefinite. The sum of the two positive images is positive.

By Theorem 4.13.2, entanglement means that \(\rho \) has Schmidt number larger than one. The rectangular detector in Theorem 4.13.10, at \(n=1\), gives a positive map \(T\colon M_{d}(\mathbb {C})\to M_{d'}(\mathbb {C})\) for which \((T\otimes \operatorname{id})(\rho )\) is not positive semidefinite. If \(T\) were decomposable, Theorem 4.10.10 would make this ampliation positive because \(\rho \) is PPT, a contradiction.

Let \(\mathcal D_1\) be the compact convex set in Theorem 4.10.7. Since \(W\notin \mathcal D_1\), strict separation gives a real-linear functional \(f\) and a constant \(c\) such that

Represent \(f\) on Hermitian operators as \(f(X)=\operatorname{Re}\operatorname{tr}(XH)\), with \(H\) Hermitian, and put \(R=H-c\mathbb {1}\). Since \(W\) and every \(D\in \mathcal D_1\) have trace one,

At the endpoint \(a=1\), this gives \(\operatorname{Re}\operatorname{tr}(PR)\geq 0\) for every positive trace-one \(P\); normalization extends the inequality to every \(P\geq 0\), so self-duality of the positive cone gives \(R\geq 0\). At \(a=0\), it gives \(\operatorname{Re}\operatorname{tr}(Q^{T_1}R)\geq 0\) for every positive trace-one \(Q\). By Lemma 4.10.8, this is \(\operatorname{Re}\operatorname{tr}(QR^{T_1})\geq 0\), and the same argument gives \(R^{T_1}\geq 0\).

Take \(R\) from Theorem 4.10.12. The strict inequality \(\operatorname{Re}\operatorname{tr}(WR){\lt}0\) implies \(R\neq 0\). Hence \(\operatorname{tr}(R){\gt}0\), and

is a density operator with \(\rho ^{T_1}\geq 0\) and \(\operatorname{Re}\operatorname{tr}(W\rho ){\lt}0\). The assumed nonnegativity of \(W\) on Schmidt-rank-one vectors and the witness criterion show that \(\rho \) has Schmidt number larger than one. It is therefore not separable.

Let \(T^*\colon M_{d'}(\mathbb {C})\to M_{d}(\mathbb {C})\) be the trace-pairing adjoint and set

This is the output-first rectangular Choi orientation: the first tensor factor has dimension \(d\), and every partial transpose below is \(T_1\) on that factor. Positivity of \(T^*\) and the Schmidt-rank Choi criterion give

Indecomposability implies \(T\neq 0\). For a positive map, \(T(\mathbb {1}_d)=0\) would imply \(T=0\), so \(T(\mathbb {1}_d)\) is a nonzero positive operator. The rectangular Choi trace formula therefore gives

Put \(W=W_0/\operatorname{tr}(W_0)\). Then \(W\) is Hermitian, has trace one, and retains the preceding nonnegativity. If \(W\) were decomposable, positive rescaling and Theorem 4.10.9 would make \(T\) decomposable. Thus \(W\) is not decomposable, and Theorem 4.10.13 gives the required state.

The forward implication is Theorem 4.10.14; the converse is Theorem 4.10.11.

When \(d,d'\geq 1\), apply Theorem 4.10.15. If either dimension is zero, every relevant linear map is the zero map and hence decomposable, while every matrix on the empty tensor product has trace zero and cannot be a density operator.

This is the contrapositive of Theorem 4.10.10. Together with the assumed positivity of \(\Phi \), it is exactly indecomposability.

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.

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.

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 rectangular matrix \(X\in M_{d\times k}(\mathbb {C})\) 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.

Apply Theorem 4.13.5 with output dimension \(r=d\) and bystander dimension \(k=d'\).

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 forward implication is the separating-hyperplane construction (4.21.5). 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 entanglement witness \(W\) for \(\rho \) (4.21.5) is the Choi matrix of an \(n\)-positive map \(P\colon M_{d'}(\mathbb {C})\to M_{d}(\mathbb {C})\) (4.22.2). Its trace-pairing adjoint \(T=P^{*}\) is again \(n\)-positive (4.3.2). By the Choi trace pairing through the trace-pairing adjoint (4.22.3), \(\langle \Omega _{d'}|(T\otimes \mathbb {1}_{d'})(\rho )|\Omega _{d'}\rangle =\operatorname{tr}(W\rho )\), whose real part is negative. A positive semidefinite matrix has non-negative quadratic forms, so \((T\otimes \mathbb {1})(\rho )\) is not positive semidefinite.