Complete positivity follows from the Kraus representation (??) with the single Kraus operator \(X\). Positivity follows immediately. If \(X^\dagger X=\mathbb {1}\), cyclicity of the trace gives \(\operatorname{tr}(X\rho X^\dagger )=\operatorname{tr}(X^\dagger X\rho )=\operatorname{tr}(\rho )\).

Expanding with the trace-pairing adjoint gives

By Theorem I.1.1, \(X\rho X^\dagger \) is positive whenever \(\rho \) is positive; positivity of \(T\) therefore makes the filtered map positive. For trace preservation, the trace-pairing adjoint identity and cyclicity of the trace give

Since \(A\) is positive definite, \(S\) is invertible and self-adjoint, and \(S^2=A\). Therefore \((S^{-1})^\dagger A S^{-1}=S^{-1}S^2S^{-1}=\mathbb {1}\).

A positive definite matrix is invertible, hence so is its square root \(A^{1/2}\), and the inverse of an invertible matrix is invertible.

Let \(S=(T^*(\mathbb {1}))^{1/2}\). Since \(T^*(\mathbb {1})\) is positive definite, \(S\) is invertible and self-adjoint, and \(S^2=T^*(\mathbb {1})\). For \(X=S^{-1}\),

Theorem I.1.3 applies.

The hypothesis gives \(\operatorname{SR}(\psi )\le k\), and transitivity with \(k\le l\) gives \(\operatorname{SR}(\psi )\le l\).

The range of the linear map represented by \(A\) has dimension at most \(k\). Embed this range into \(\mathbb {C}^k\), compose with the range map, and then map back to the ambient codomain. The corresponding matrices give the desired factorization.

The partial trace of \(|\phi _i\rangle \! \langle \phi _i|\) over the second factor has entries \(\sum _b(C_i)_{a,b}\overline{(C_i)_{a',b}}\), which is the \((a,a')\) entry of \(C_iC_i^\dagger \).

By the coefficient-matrix form, \(\rho _i=C_iC_i^\dagger \), which is positive semidefinite for any matrix \(C_i\).

Diagonalizing \(\tau =U\operatorname{diag}(\nu _i)U^\dagger \) with \(U\) the unitary of eigenvectors, the change of variable \(y=U^\dagger \psi \) gives \(\langle \psi |\tau |\psi \rangle =\sum _i\nu _i|y_i|^2\), and \(y_i=\langle \phi _i | \psi \rangle \) is the \(i\)-th eigenvector overlap.

The eigenvectors form an orthonormal basis, so with \(y=U^\dagger \psi \) one has \(\sum _i|\langle \phi _i | \psi \rangle |^2=\sum _i|y_i|^2=\| y\| ^2=\| \psi \| ^2\), where the last equality follows because \(U\) is unitary.

Diagonalize \(A=U\operatorname{diag}(\lambda _i)U^\dagger \) and take \(P\) to be the projection onto the span of the eigenvectors with the \(\min (k,D)\) largest eigenvalues. Then \(\operatorname{Re}\operatorname{tr}(PA)=\sum _{i\le k}\lambda _i=S_k(A)\).

Conjugating \(P\) by \(U\) gives an orthogonal projection \(Q\) of the same rank, and \(\operatorname{Re}\operatorname{tr}(PA)=\sum _iw_i\lambda _i\), where \(w_i\) are the diagonal entries of \(Q\). Each \(w_i\) lies in \([0,1]\) and the weights sum to \(\operatorname{tr}(P)=k\). With the threshold \(c=\lambda _{k+1}\),

since for \(i\le k\) the eigenvalues dominate \(c\) while \(1-w_i\ge 0\), and for \(i{\gt}k\) they are dominated by \(c\) while \(w_i\ge 0\).

By (??), the \((i,p),(j,q)\) entry of the left-hand side of (??) is

which is (??).

The coefficient matrix of \(\psi _X\) is a scalar multiple of \(X\). Therefore its rank is bounded by the number of columns, which is \(k\).

The coefficient matrix of \(\psi _X\) is \(D^{-1/2}X\). Since \(D{\gt}0\), the scalar \(D^{-1/2}\) is nonzero, and multiplication by this scalar preserves rank.

Take \(X_{i,p}=D^{1/2}\psi _{(i,p)}\). Since \(D{\gt}0\), this gives \(D^{-1/2}X_{i,p}=\psi _{(i,p)}\). The rank identity follows from Theorem I.4.3. The bounded-rank statement follows by comparison with the assumed Schmidt-rank bound.

The forward direction is the bounded-rank part of Theorem I.4.4. Conversely, if \(\psi _{(i,j)}=D^{-1/2}X_{i,j}\), then \(C_\psi =D^{-1/2}X\), so \(\operatorname{rank}(C_\psi )=\operatorname{rank}(X)\) because multiplication by a nonzero scalar preserves rank.

If \(\eta \) is read as a \(D\times k\) coefficient matrix \(E\), then the coefficient matrix of \(R_X^\dagger \eta \) is \(EX^\dagger \). Therefore \(\operatorname{rank}(EX^\dagger )\le \operatorname{rank}(X^\dagger )=\operatorname{rank}(X)\le k\).

Let \(C_\psi \) be the coefficient matrix of \(\psi \). The rank assumption gives a factorization \(C_\psi =BC\) through \(\mathbb {C}^k\). Taking \(X=C^\dagger \) and reading \(B\) as the coefficient matrix of \(\eta \) gives the required identity, because the coefficient matrix of \(R_X^\dagger \eta \) is \(BX^\dagger =BC\).

Substituting (??) gives

In entries, both sides are

If \(\psi _{(a,p)}=D^{-1/2}X_{a,p}\), then for fixed \(p,q\), the corresponding block of \(|\psi \rangle \! \langle \psi |\) is

Hence

where the second equality uses \(\tau _{(i,a),(j,b)}=D^{-1}(T(E_{a,b}))_{i,j}\).

This is an entrywise calculation: the Kronecker delta in the physical index forces the two right-tensor factors to have the same physical coordinate, and the remaining sum is the matrix product \(PX\).

Since \(PX=X\), Lemma I.4.10 gives \(R_XR_P=R_X\). Therefore

which is positive because positive semidefiniteness is preserved under compression. The left-hand side of (??) is the rectangular compression by (??).

The column space of \(X\) has dimension at most \(k\). Since \(k\le D\), extend this column space to a \(k\)-dimensional subspace \(U\subseteq \mathbb {C}^D\). Let \(P\) be the orthogonal projection onto \(U\). Then \(P\) is Hermitian and idempotent, its rank is \(\dim U=k\), and it fixes each column of \(X\).

This is the entrywise calculation obtained by expanding the definition of \(R_V\) and summing over the middle physical index.

Replace \(R_{VV^\dagger }\) by \(R_V^\dagger R_V\) using Lemma I.4.13 and reassociate the matrix products.

By (??), \(R_{VV^\dagger }\tau R_{VV^\dagger }^\dagger =R_V^\dagger (R_V\tau R_V^\dagger )R_V\). The factor \(R_V\tau R_V^\dagger \) is positive by Theorem 22.2.3.7, and positive semidefiniteness is preserved under compression by \(R_V\).

Regard \(P\) as a linear operator on \(\mathbb {C}^D\). Hermiticity and idempotence say that this operator is the orthogonal projection onto its range. Choose an orthonormal basis of the range, indexed by \(\{ 0,\ldots ,k-1\} \) using the rank hypothesis, and let \(V\) be the matrix whose columns are these basis vectors. The standard rank-one expansion of an orthogonal projection gives \(P=VV^\dagger \).

Embed \(M_{k}(\mathbb {C})\) as the upper-left corner of \(M_{k+1}(\mathbb {C})\). For \(X\succeq 0\) in \(M_{D}(\mathbb {C})\otimes M_{k}(\mathbb {C})\), extend \(X\) by a zero row and column, apply the positive \((k+1)\)-fold ampliation, and compress back to the same corner. Compression preserves positive semidefiniteness, and the compressed image is the \(k\)-fold ampliation of \(X\).

Repeat the corner-extension and compression argument from dimension \(k\) to dimension \(m\). Equivalently, iterate the preceding theorem \(k-m\) times.

Every ampliation of the zero map is zero, and the zero matrix is positive semidefinite.

For \(X\succeq 0\), both \(E^{(k)}(X)\) and \(F^{(k)}(X)\) are positive semidefinite. Since \((E+F)^{(k)}=E^{(k)}+F^{(k)}\), their sum is positive semidefinite.

Ampliation commutes with scalar multiplication, and a non-negative scalar multiple of a positive semidefinite matrix is positive semidefinite.

Let \(E_j\to E\) with every \(E_j\) \(k\)-positive. For each fixed \(X\succeq 0\), continuity of ampliation and evaluation gives \(E_j^{(k)}(X)\to E^{(k)}(X)\). The positive semidefinite cone is closed, so \(E^{(k)}(X)\succeq 0\). Since this holds for every positive semidefinite \(X\), the limit map \(E\) is \(k\)-positive.