The singular-value sequence satisfies \(s_i(A)=0\) for all \(i\geq \operatorname{rank}(A)\), so summing over the support and summing over \(\{ 0,\ldots ,\operatorname{rank}(A)-1\} \) give the same value.

The singular values satisfy \(s_i(A)\geq 0\), so \(\lVert A\rVert _{\operatorname{tr}}=0 \iff \forall i,\, s_i(A)=0 \iff A=0\).

The map \(T^\dagger T\), for \(T\) the linear map on \(\mathbb {C}^D\) represented by \(A\), is represented by \(A^\dagger A\), so the singular values of \(A\) are \(s_i(A)=\sqrt{\lambda _i}\) with \(\lambda _0,\ldots ,\lambda _{D-1}\) the eigenvalues of \(A^\dagger A\). Diagonalizing \(A^\dagger A=U\operatorname{diag}(\lambda _i)U^\dagger \) gives \(\lvert A\rvert =U\operatorname{diag}(\sqrt{\lambda _i})U^\dagger \), hence

where the last step uses that \(\lvert A\rvert \) is positive semidefinite, so \(\operatorname{tr}\lvert A\rvert \geq 0\) is real.

From \((cA)^\dagger (cA)=\lvert c\rvert ^2\, A^\dagger A\) and uniqueness of the positive-semidefinite square root, \(\lvert cA\rvert =\lvert c\rvert \, \lvert A\rvert \). Linearity of the trace gives \(\operatorname{tr}\lvert cA\rvert =\lvert c\rvert \, \operatorname{tr}\lvert A\rvert \), and the claim follows from Theorem 13.1.5.

From \((UA)^\dagger (UA)=A^\dagger U^\dagger UA=A^\dagger A\) the absolute values agree: \(\lvert UA\rvert =\lvert A\rvert \). For the right factor, \((AV)^\dagger (AV)=V^\dagger (A^\dagger A)V\), and since \(V^\dagger \lvert A\rvert V\) is positive semidefinite with

uniqueness of the positive-semidefinite square root gives \(\lvert AV\rvert =V^\dagger \lvert A\rvert V\). Cyclicity of the trace then yields \(\operatorname{tr}\lvert AV\rvert =\operatorname{tr}(\lvert A\rvert \, VV^\dagger ) =\operatorname{tr}\lvert A\rvert \). For the two-sided case, \(\lVert UAV\rVert _{\operatorname{tr}} =\lVert UA\rVert _{\operatorname{tr}} =\lVert A\rVert _{\operatorname{tr}}\) by applying the right and left cases in turn.

Since \(s_i(A)=\sqrt{\lambda _i(A^\dagger A)}\) by the definition of singular value, the finite singular-value expansion (3) gives

Diagonalizing \(A^\dagger A=V\operatorname{diag}(\lambda _i)V^\dagger \) gives \(\lvert A\rvert =V\operatorname{diag}(\sqrt{\lambda _i})V^\dagger \), so \(\operatorname{tr}\lvert A\rvert =\sum _{i=0}^{D-1}\sqrt{\lambda _i}\), and the claim follows from Theorem 13.1.5.

Write \(s_0,\ldots ,s_{D-1}\geq 0\) for the singular values of \(A\). Then \(\lVert A\rVert _2^2=\sum _i s_i^2\) and \(\lVert A\rVert _1=\sum _i s_i\). The first inequality follows by expanding \((\sum _i s_i)^2\) and using nonnegativity. The second is the Cauchy–Schwarz bound \((\sum _i s_i)^2\leq D\sum _i s_i^2\).

For the upper bound, let \(v_0,\ldots ,v_{D-1}\) be an orthonormal eigenbasis of \(A^\dagger A\) with eigenvalues \(\lambda _i\), and let \(U\) be unitary. Expanding the trace in this basis and applying the Cauchy–Schwarz inequality,

since \(\lVert Av_i\rVert ^2=\langle v_i,\, A^\dagger A\, v_i\rangle =\lambda _i\) and \(\lVert Uv_i\rVert =1\).

For attainment, the vectors \(w_i=Av_i/\sqrt{\lambda _i}\), taken over the indices with \(\lambda _i\neq 0\), satisfy

so they form an orthonormal family. Extend it to an orthonormal basis \((w_i)_i\) of \(\mathbb {C}^D\) and let \(U\) be the unitary with \(Uv_i=w_i\) for all \(i\). Then

where the indices with \(\lambda _i=0\) contribute \(0\) to both sides because \(\lVert Av_i\rVert ^2=\lambda _i=0\) forces \(Av_i=0\).

Choose by Theorem 13.1.11 a unitary \(U\) with \(\operatorname{tr}[(A+B)^\dagger U]=\lVert A+B\rVert _{\operatorname{tr}}\). Splitting the trace and applying the upper-bound half of the same theorem to \(A\) and to \(B\) separately,

The sum \(H^++H^-\) is positive semidefinite, and by orthogonality its square is

so \(H^++H^-=\sqrt{H^\dagger H}=\lvert H\rvert \). Hence \(\lVert H\rVert _{\operatorname{tr}} =\operatorname{tr}\lvert H\rvert =\operatorname{tr}[H^+]+\operatorname{tr}[H^-]\) by Theorem 13.1.5.

Diagonalize \(A=U\operatorname{diag}(\lambda _1,\ldots ,\lambda _D)\, U^\dagger \) and set

where \(\chi \) is the indicator function of \((0,\infty )\). Each claim is read off eigenvalue-wise: \(0\leq \chi \leq 1\) gives \(0\leq \Pi \leq \mathbb {1}\), \(\chi ^2=\chi \) gives \(\Pi ^2=\Pi \), and \(\chi (\lambda ) \lambda =\max (\lambda ,0)\) gives \(\Pi A=A^+\).

The first bound is Lemma 9.5.1. For the second, \(\operatorname{tr}[X]-\operatorname{tr}[CX]=\operatorname{tr}[(\mathbb {1}-C)X]\geq 0\) by the same lemma, since \(\mathbb {1}-C\geq 0\).

Write \(H=H^+-H^-\) with \(H^\pm \geq 0\). Then \(T(H)=T(H^+)-T(H^-)\) is a difference of positive semidefinite matrices, hence Hermitian.

Let \(\Pi _+\) be the support projection of \((TH)^+\) from Lemma 13.1.14, so that \(\Pi _+ T(H) = (T(H))^+\) and \(0\leq \Pi _+\leq \mathbb {1}\). The estimate follows from the chain

using Lemmas 13.1.15 and 13.1.16.

Lemma 13.1.17 gives \(\operatorname{tr}[(TH)^+]\leq \operatorname{tr}[T(H^+)]\). The result follows because trace preservation gives \(\operatorname{tr}[T(H^+)]=\operatorname{tr}[H^+]\).

Write \(H=P_+-P_-\) and \(T(H)=Q_+-Q_-\) for the Jordan decompositions. Applying Lemma 13.1.18 to \(H\) and to \(-H\) gives \(\operatorname{tr}[Q_+]\leq \operatorname{tr}[P_+]\) and \(\operatorname{tr}[Q_-]\leq \operatorname{tr}[P_-]\), so by the Jordan trace-norm formula (Lemma 13.1.13),

By linearity \(T(\rho _1)-T(\rho _2)=T(\rho _1-\rho _2)\), and \(\rho _1-\rho _2\) is Hermitian as a difference of positive semidefinite matrices, so Theorem 13.1.19 applies.

An orthogonal pair of pure states gives a pair of distinct density matrices at trace-norm distance \(2\), which proves the lower bound. For the reverse inequality, set \(H=\rho _1-\rho _2\) and write its Jordan decomposition as \(H=H^+-H^-\). Since \(H\) is nonzero and traceless, \(\operatorname{tr}[H^+]=\operatorname{tr}[H^-]=t{\gt}0\). Thus \(P=H^+/t\) and \(Q=H^-/t\) are density matrices with orthogonal supports, and homogeneity reduces the quotient for \(H\) to \(\frac12\lVert T(P-Q)\rVert _{\operatorname{tr}}\).

Take spectral decompositions \(P=\sum _i\lambda _i|\psi _i\rangle \! \langle \psi _i|\) and \(Q=\sum _j\mu _j|\phi _j\rangle \! \langle \phi _j|\). Orthogonality of the supports gives \(\psi _i\perp \phi _j\) whenever \(\lambda _i\mu _j\neq 0\), while \(\lambda _i,\mu _j\geq 0\) and \(\sum _i\lambda _i=\sum _j\mu _j=1\). The product-weight expansion

is therefore a convex combination of differences of orthogonal pure states. The triangle inequality and homogeneity bound its image under \(T\) by the right-hand supremum. Taking the supremum over \(\rho _1\neq \rho _2\) proves the equality.

The commutation \(T_\phi T=TT_\phi \) iterates to \(T_\phi T^n=T^nT_\phi \), which proves the first equality by linearity. Idempotence of \(T_\phi \) gives \(T_\varphi (\rho -T_\phi (\rho ))=0\); every positive power of \(T_\varphi \) therefore vanishes on the same remainder, proving the second equality.

The proof is the upper-bound half of Lemma 13.1.21: normalize the positive and negative parts of \(\rho _1-\rho _2\) to orthogonally supported density matrices and expand both in eigenprojectors.

The first inequality is finite Cauchy–Schwarz applied to the singular values. The second is the defining application bound for the operator norm after Frobenius vectorization. Orthogonality makes the two rank-one projectors multiply to zero, so the squared Hilbert–Schmidt norm of their difference is the sum of their traces, namely \(2\).

Apply the numerator identity to \(S=T^n-T_\varphi ^n\). The peripheral projection is again positive and trace preserving, so Lemma 8.3 applies to the distinct density matrices \(\rho \) and \(T_\phi (\rho )\). Then use \(\lVert \cdot \rVert _{\operatorname{tr}}\leq \sqrt D\lVert \cdot \rVert _2\) and the \(\sqrt2\) Hilbert–Schmidt distance of orthogonal pure states. The constants simplify as \(\frac12\sqrt D\sqrt2=\sqrt{D/2}\); if \(\rho =T_\phi (\rho )\), both sides vanish directly. Finally, the positive-power identity identifies the transfer-matrix difference with the transfer matrix of \(T^n-T_\varphi ^n\), whose largest-singular-value norm equals the Hilbert–Schmidt operator norm \(\lVert T^n-T_\varphi ^n\rVert _{2\to 2}\).

Lemma 13.1.17 gives \(\operatorname{tr}[(TH)^+]\leq \operatorname{tr}[T(H^+)]\). The trace-scaling hypothesis then yields \(\operatorname{tr}[T(H^+)] = c\, \operatorname{tr}[H^+]\), giving the claimed bound.

Same argument as Theorem 13.1.19: apply Lemma 13.1.27 to \(H\) and \(-H\), then sum using the Jordan trace-norm formula.

The map \(S:=T-\varepsilon T'\) is positive by hypothesis. Since \(T'(\rho _1-\rho _2)=\operatorname{tr}[\rho _1-\rho _2]\, Y=0\) (the densities have unit trace), \(T(\rho _1)-T(\rho _2)=S(\rho _1-\rho _2)\). The trace of \(S(\rho )\) is \((\operatorname{tr}[\rho ])-\varepsilon \operatorname{tr}[\rho ]=(1-\varepsilon ) \operatorname{tr}[\rho ]\), so \(S\) satisfies the scaled-trace hypothesis of Theorem 13.1.28 with \(c=1-\varepsilon \). Positivity of \(S\) on \(\rho _1\) forces \(0\leq 1-\varepsilon \) (by PSD trace nonnegativity), so the scaled-trace theorem applies and yields the claimed bound.

The normalized omega slice is \(D^{-1}E_{ij}\), where \(E_{ij}\) is the matrix unit. Since \(\operatorname{tr}(E_{ij})=\delta _{ij}\), the trace of the slice is \(D^{-1}\delta _{ij}\), and \(T'_Y\) multiplies this scalar by \(Y\). On the other hand, \((Y\otimes \mathbb {1})_{(a,i),(b,j)} = Y_{ab}\, \delta _{ij}\), so both sides equal \(D^{-1}Y_{ab}\, \delta _{ij}\).

Since \(\operatorname{tr}[Y]=1\), \(T'_Y\) is trace-preserving: \(\operatorname{tr}[T'_Y(X)]=\operatorname{tr}[X]\operatorname{tr}[Y]=\operatorname{tr}[X]\). Since \(Y\) is Hermitian and \(\operatorname{tr}[X]\) is real for Hermitian \(X\), \(T'_Y\) is Hermiticity-preserving. By Lemma 13.1.31, \(\tau (T'_Y)=(1/D)(Y\otimes \mathbb {1})\). Linearity of the Choi assignment gives \(\tau (T-\varepsilon T'_Y)=\tau (T)-\varepsilon \tau (T'_Y) \ge 0\), so \(T-\varepsilon T'_Y\) is completely positive (the rectangular Choi CP equivalence, Theorem 2.6.6) and hence positive. Theorem 13.1.29 then yields the asserted bound.

For \(A_{ij}=P_i-P_j\), positivity of \(\rho \) gives \(A_{ij}\rho A_{ij}^{\dagger }\ge 0\). Since the projections are Hermitian and sum to the identity, expansion and collection of the four double sums gives

Thus \(kT(\rho )-\rho \) is positive semidefinite.

Positivity gives

Taking square roots in (30) at \(x-y\) gives \(\| x-y\| \le \| (I+P)(x-y)\| \). Use \((I+P)(x-y)=(Px-y)-(Py-x)\) and the triangle inequality. Similarly, (30) applied to \((I-P)x\) gives the final estimate.

Set \(P=(D^\dagger D)^{1/2}\). The equal Gram matrices \(D^\dagger D=P^\dagger P\) give an isometry from the range of \(P\) to the range of \(D\), taking \(Px\) to \(Dx\). These ranges have equal dimension, as do their orthogonal complements in \(\mathcal H\). Extending between the complements therefore gives a unitary \(U\) on \(\mathcal H\) with \(D=UP\).

Unitary invariance and Theorem 13.3.1 give

For the second inequality put \(\xi =U^\dagger \eta \). Then

Only positivity of \(P\) and unitarity of \(U\) enter this argument; neither an inverse of \(D\) nor a larger Hilbert space is required.

Unitarity follows from \((U\otimes I)^\dagger (U\otimes I)=I\). Positive square roots satisfy

so the fixed factorization tensors to \(E=V(E^\dagger E)^{1/2}\). Apply the two norm arguments of Theorem 13.3.2 on \(\mathcal H\otimes \mathcal K\).

Induct on the right cut, adjoining its channel on the right. The Kraus expansion follows by summing over successive indices.

Choose a density \(\theta \) fixed by \(F_{0:N}\) and put \(\sigma _a=F_{a:N}(\theta )\). Positivity and trace preservation give the density conditions. Composition gives transport, and the fixed-point equation identifies the two closing references.

The largest nonnegative eigenvalue of \(H\) is at most their sum. Apply this to \(B^\dagger B\) and use \(\| B^\dagger B\| =\| B\| ^2\).

The input-first unnormalized Choi matrix \(C_Q\) is \(bJ(Q)\) after exchanging tensor factors. It is positive semidefinite and has trace \(bt\). Its Frobenius norm equals that of the coefficient matrix of \(Q\). The sum of squares of its nonnegative eigenvalues is at most \((bt)^2\), which bounds the squared operator norm of that coefficient matrix.

Choi linearity makes \(F-\eta R_\tau \) completely positive. Its trace on any input density gives \(1-\eta \ge 0\). Apply the traceless-input equality to \(X-\operatorname{tr}(X)\rho \) to obtain \(F-R_{F(\rho )}=Q-R_{Q(\rho )}\). The two maps on the right are completely positive with trace scale \(t\), so their two norm bounds add.

Peel complete \(s\)-windows from the right. Their residuals each scale trace by \(1-\eta \); the shorter interval left at the beginning scales trace by one. On traceless inputs every subtracted reset vanishes. Compose these residuals and apply the scaled-trace estimate, including \(t=0\).

The inequality \(n/s\le \lfloor n/s\rfloor +1\) loses at most the factor \(\alpha ^{-1}\le 2\) when exponentiating.

Substitute the parameter and coordinate identifications.

Let \(\mathcal{DS}_d \subset M_d(\mathbb {R})\) be the set of \(d \times d\) doubly stochastic matrices and let \(\mathcal{P}_d\) be the set of \(d \times d\) permutation matrices. The Birkhoff–von Neumann decomposition gives

Indeed, every doubly stochastic matrix is a convex combination of permutation matrices, while convexity of \(\mathcal{DS}_d\) gives the converse inclusion in the first identity. Suppose that a permutation matrix \(P_\sigma \) is a convex combination \(P_\sigma =tA+(1-t)B\), where \(0{\lt}t{\lt}1\) and \(A,B\) are doubly stochastic. Nonnegativity and the row sums give

Thus \(A=B=P_\sigma \), so every permutation matrix is extreme. Conversely, a non-permutation matrix has a Birkhoff decomposition involving at least two distinct permutation matrices and is therefore not extreme, which proves the second identity.

Substitute the equality of the matrices in the defining eigenvalue sum.

All eigenvalues of the zero matrix are zero, and \(0\log 0=0\).

Each eigenvalue \(\lambda _i\) of a density matrix satisfies \(0\le \lambda _i\le 1\), and \(-x\log x\ge 0\) on \([0,1]\).

Follows from \(\operatorname{tr}(\rho )=\sum _i\lambda _i=1\).

Non-negativity comes from positive semidefiniteness. The upper bound follows because the eigenvalues are non-negative and sum to \(1\).

By Jensen’s inequality applied to the concave function \(-x\log x\), the entropy is maximized when all eigenvalues equal \(1/D\).

The entropy is the \(-x\log x\) sum over all eigenvalues; the \(D-r\) zero eigenvalues contribute nothing. Jensen’s inequality applied to \(-x\log x\) over the \(r\) nonzero eigenvalues, with uniform weights \(1/r\), gives \(S(\rho )\le \log r\), the maximum attained when each nonzero eigenvalue equals \(1/r\). The rank equals the number of nonzero eigenvalues of the Hermitian matrix \(\rho \).

The eigenvalues of a Hermitian matrix are exactly the roots of its characteristic polynomial, counted with multiplicity, so the eigenvalue sum defining \(S(\rho )\) equals the displayed sum over roots.

A Hermitian matrix and its logarithm are simultaneously diagonalized by a unitary \(U\) with \(\rho =U\operatorname{diag}(\lambda _i)U^*\), so

Its negative is \(\sum _i({-}\lambda _i\log \lambda _i)=S(\rho )\). Zero eigenvalues contribute nothing under the convention \(0\log 0=0\), so no full-support assumption is required.

Expand the matrix product over the difference \(\log \rho -\log \sigma \) and use linearity of the trace and of the real part.

The logarithmic difference \(\log \rho -\log \rho \) vanishes.

The trace-log expression is multiplied on the left by the zero matrix.

Apply Lemma 13.6.13 to write

The trace-logarithm identity \(\operatorname{Re}\operatorname{tr}(\rho \log \rho )=-S(\rho )\) then gives the result.

Diagonalize \(\rho =\sum _i p_i|e_i\rangle \! \langle e_i|\) and \(\sigma =\sum _j q_j|f_j\rangle \! \langle f_j|\) in their eigenbases, with all \(q_j{\gt}0\) since \(\sigma \) has full rank. The overlap numbers \(P_{ij}=\lvert \langle e_i | f_j \rangle \rvert ^2\) are non-negative with row sums and column sums equal to \(1\), because the two eigenbases are orthonormal. A trace computation gives

The row and column sums, together with the trace-one normalizations, give \(\sum _{i,j}P_{ij}(p_i-q_j)=\sum _i p_i-\sum _jq_j=0\). Hence

Each bracket is non-negative. If \(p_i{\gt}0\), put \(x=q_j/p_i{\gt}0\); the bracket is \(p_i(x-1-\log x)\geq 0\) by \(\log x\leq x-1\). If \(p_i=0\), the bracket equals \(q_j{\gt}0\). Therefore \(D(\rho \Vert \sigma )\geq 0\).

Regularize \(\sigma \) by the trace-one perturbation \(\sigma _\varepsilon '=(1+\varepsilon D)^{-1}(\sigma +\varepsilon \mathbb {1})\), which is positive definite for every \(\varepsilon {\gt}0\), hence of full rank. The full-rank Klein inequality (Theorem 13.6.17) gives \(D(\rho \Vert \sigma _\varepsilon ')\ge 0\). The perturbation shares the eigenbasis \(\{ |f_j\rangle \} \) of \(\sigma \), so the cross term reduces to a scalar sum over the eigenvalues \(q_j\) of \(\sigma \),

For \(q_j{\gt}0\) the scalar logarithm converges to \(\log q_j\); for \(q_j=0\) the eigenvector \(|f_j\rangle \) lies in \(\ker \sigma \subseteq \ker \rho \), so its diagonal weight \(\langle f_j|\rho |f_j\rangle \) vanishes and the summand is identically zero. No eigenvalue-continuity input is needed, so \(D(\rho \Vert \sigma _\varepsilon ')\to D(\rho \Vert \sigma )\) as \(\varepsilon \to 0^+\) and the inequality passes to the limit.

If \(\rho =\sigma \) then \(\log \rho -\log \sigma =0\) and \(D(\rho \Vert \sigma )=0\). For the converse, diagonalize \(\rho =\sum _i p_i|e_i\rangle \! \langle e_i|\) and \(\sigma =\sum _j q_j|f_j\rangle \! \langle f_j|\), with all \(q_j{\gt}0\) by full rank, and set \(P_{ij}=\lvert \langle e_i | f_j \rangle \rvert ^2\). As in Theorem 13.6.17,

a sum of non-negative terms: each is bounded below by the tangent inequality \(\log x\le x-1\) at \(x=q_j/p_i\), while the linear remainder telescopes through the doubly stochastic row and column sums, \(\sum _{i,j}P_{ij}(p_i-q_j)=\sum _ip_i-\sum _jq_j=0\). If the total vanishes then every term vanishes. On a row with \(p_i=0\) the term reads \(P_{ij}q_j\), so \(P_{ij}=0\); on a row with \(p_i{\gt}0\) the term forces the tangent inequality to be tight, \(\log (q_j/p_i)=q_j/p_i-1\), and strict concavity of the logarithm (\(\log x{\lt}x-1\) for \(x\neq 1\)) gives \(q_j=p_i\). Hence \(q_j=p_i\) whenever \(\langle e_i | f_j \rangle \neq 0\). Writing \(W=U_\rho ^\dagger U_\sigma \) for the overlap of the eigenvector unitaries, this matching says \(W\operatorname{diag}(q)=\operatorname{diag}(p)W\). Thus, conjugating \(\sigma \) into the eigenbasis of \(\rho \) and using the unitarity \(WW^\dagger =1\),

the spectral diagonal of \(\rho \). Therefore \(\sigma =U_\rho \operatorname{diag}(p)U_\rho ^\dagger =\rho \).

For \(s\in [0,1)\) consider the approximant

The trace \(\operatorname{tr}\rho \) is real-affine in the pair, hence convex, while \((\rho ,\sigma )\mapsto \operatorname{Re}\operatorname{tr}(\rho ^s\sigma ^{1-s})\) is jointly concave by the \(K=\mathbb {1}\) case of the Lieb concavity theorem (Corollary 7.7.14). Their difference, scaled by the non-negative factor \((1-s)^{-1}\), is therefore jointly convex, so each \(g_s\) is convex.

Writing \(\rho \) and \(\sigma \) in their eigenbases with eigenvalues \(p_i,q_j{\gt}0\) and overlap weights \(P_{ij}=\lvert \langle e_i | f_j \rangle \rvert ^2\), the approximant becomes the double sum \(g_s(\rho ,\sigma )=\sum _{i,j}(1-s)^{-1}(p_i-p_i^sq_j^{1-s})P_{ij}\). As \(s\to 1^-\) each per-pair term converges, via the scalar limit \((c^u-1)/u\to \log c\), to \(p_i(\log p_i-\log q_j)P_{ij}\), whose sum equals \(D(\rho \Vert \sigma )\). Thus \(g_s\to D\) pointwise on positive definite pairs, and since the pointwise limit of convex functions is convex, \(D\) is jointly convex.

The domain is convex: for a strict convex combination of two such pairs the kernel of \(a\sigma _1+b\sigma _2\) is \(\ker \sigma _1\cap \ker \sigma _2\), since the quadratic forms of the positive semidefinite summands are non-negative and their positively weighted sum vanishes only when each does, and a positive semidefinite matrix annihilates exactly the vectors of zero quadratic form; the two pointwise support inclusions then give \(\ker \sigma _1\cap \ker \sigma _2\subseteq \ker \rho _1\cap \ker \rho _2\).

Regularize both arguments through the affine trace-one perturbation \(M_\varepsilon =(1+\varepsilon N)^{-1}(M+\varepsilon \mathbb {1})\), where \(N\) is the matrix dimension, which is positive definite for every \(\varepsilon {\gt}0\). Because the perturbation is affine, it commutes with the convex combination, so the four regularized endpoints and the regularized mixture all lie in the positive definite domain and the positive definite joint convexity (Theorem 13.6.20) gives the two-point inequality for the regularized pairs. As \(\varepsilon \to 0^+\), the regularized relative entropy converges on each pair of the support domain: \(D(\rho _\varepsilon \Vert \sigma _\varepsilon ) \to D(\rho \Vert \sigma )\). Indeed the perturbation shares the eigenbasis of its argument, so each trace-logarithm term is the diagonal sum \(\sum _jw_j(\varepsilon ) \log ((1+\varepsilon N)^{-1}(q_j+\varepsilon ))\), where \(q_j\) runs over the eigenvalues of \(\sigma \) and \(w_j(\varepsilon )\) is the corresponding diagonal weight. For \(q_j{\gt}0\), the scalar factor converges to \(\log q_j\), while at a zero eigenvalue the support condition makes the weight vanish, with \(w_j(\varepsilon )=(1+\varepsilon N)^{-1}\varepsilon \), so the summand is

by \(x\log x\to 0\) as \(x\to 0^+\). The two-point inequality therefore passes to the limit, giving joint convexity on the support domain.

The conjugation \(\varphi :x\mapsto UxU^\dagger \) is a star-algebra automorphism of the matrix algebra, and the continuous functional calculus commutes with such automorphisms, \(\varphi (f(A))=f(\varphi (A))\); since \(\varphi (A)=UAU^\dagger \), this is the claim.

This is the \(f=\log \) case of Lemma 13.6.22: \(f(UAU^\dagger )=Uf(A)U^\dagger \) specialized to the real logarithm gives \(\log (UAU^\dagger )=U(\log A)U^\dagger \).

Carry the two logarithms through the conjugation by Lemma 13.6.23, so that

where the middle equality is trace cyclicity together with \(U^\dagger U=1\).

The tensor product factors as \(\rho \otimes \tau =(\rho \otimes \mathbb {1})(\mathbb {1}\otimes \tau )\) into a pair of commuting positive definite matrices, so the logarithm of the product is the sum of the logarithms of the factors. Each unital embedding \(A\mapsto A\otimes \mathbb {1}\) and \(B\mapsto \mathbb {1}\otimes B\) is a continuous star-algebra homomorphism, hence commutes with the functional calculus, which carries each logarithm onto its factor: \(\log (\rho \otimes \mathbb {1})=\log \rho \otimes \mathbb {1}\) and \(\log (\mathbb {1}\otimes \tau )=\mathbb {1}\otimes \log \tau \).

Splitting each tensor logarithm by Lemma 13.6.25, the common \(\mathbb {1}\otimes \log \tau \) terms cancel in the difference, leaving \(\log (\rho \otimes \tau )-\log (\sigma \otimes \tau ) =(\log \rho -\log \sigma )\otimes \mathbb {1}\). Hence

where the trace of a tensor product factors as a product of traces. As \(\operatorname{tr}\tau =1\), the right-hand side is \(D(\rho \Vert \sigma )\).

Each summand equals \(\xi ^b\) with \(\xi =\zeta ^i\overline{\zeta }^{\, j}\), and \(\xi ^d=1\). When \(i=j\) the base \(\xi \) equals \(1\) and the sum is \(d\). When \(i\ne j\) the base is a root of unity different from \(1\), so the geometric sum \((\xi -1)\sum _b\xi ^b=\xi ^d-1=0\) forces the sum to vanish; the equivalence \(\xi =1\iff i=j\) uses that \(\overline{\zeta }=\zeta ^{-1}\) and the injectivity of \(b\mapsto \zeta ^b\) on residues.

The cyclic shift is the permutation matrix of a cyclic permutation. The clock operator is diagonal, and each diagonal entry is a power of \(\zeta \), hence has modulus one. Thus \(X\) and \(Z\) are unitary, and so is every product \(X^aZ^b\).

The double average factors into a clock average followed by a shift average. The clock average \(\sum _bZ^bM(Z^b)^\dagger \) multiplies the entry \(M_{ij}\) by \(\sum _b\zeta ^{bi}\overline{\zeta ^{bj}}\), which by Lemma 13.6.27 is \(d\) when \(i=j\) and \(0\) otherwise; the result is \(d\) times the diagonal part of \(M\). The shift average \(\sum _aX^a(\operatorname{diag}v)(X^a)^\dagger \) cyclically permutes the diagonal entries, so each diagonal position receives the full sum \(\sum _kv_k=\operatorname{tr}M\), giving \((\operatorname{tr}M)\mathbb {1}\). Combining the two factors of \(d\) with the prefactor \(d^{-2}\) leaves \((\operatorname{tr}M/d)\mathbb {1}\).

On each pair of blocks indexed by the first factor, the conjugation by \(\mathbb {1}_S\otimes W(a,b)\) acts as the Weyl conjugation \(W(a,b)(\cdot )W(a,b)^\dagger \) of the corresponding block of \(M\). Averaging over the \(d_C^2\) Weyl operators sends each block to the depolarizing channel by Theorem 13.6.29, replacing it by its trace times \(\mathbb {1}_C/d_C\). The block trace is exactly the corresponding entry of the partial trace \(\operatorname{tr}_C M\), so the average is \((\operatorname{tr}_C M)\otimes (\mathbb {1}_C/d_C)\).

The identity is positive definite and \(d_C^{-1}{\gt}0\), so \(\tau _C\) is positive definite. Moreover, \(\operatorname{tr}\tau _C=d_C^{-1}\operatorname{tr}\mathbb {1}_C=1\).

Reindexing is a star-algebra isomorphism and therefore commutes with the continuous functional calculus for the real logarithm.

Write \(D(\rho \Vert \sigma ) =\operatorname{Re}\operatorname{tr}(\rho (\log \rho -\log \sigma ))\). The matrix logarithm is covariant under reindexing, \(\log (\rho _{e^{-1},e^{-1}}) =(\log \rho )_{e^{-1},e^{-1}}\), because reindexing is a star-algebra isomorphism and so commutes with the continuous functional calculus. The same isomorphism preserves products and the trace,

Taking \(M=\log \rho -\log \sigma \) and applying these three identities gives \(D(\rho _{e^{-1},e^{-1}}\Vert \sigma _{e^{-1},e^{-1}}) =D(\rho \Vert \sigma )\).

By ancilla additivity (Theorem 13.6.26) the reduced-state relative entropy equals \(D\bigl((\operatorname{tr}_C\rho )\otimes (\mathbb {1}_C/d_C)\Vert (\operatorname{tr}_C\sigma )\otimes (\mathbb {1}_C/d_C)\bigr)\). By Lemma 13.6.30 each tensored reduced state is the uniform average of the conjugations by the \(d_C^2\) unitaries \(U_{ab}=\mathbb {1}_S\otimes W(a,b)\), so this is the relative entropy of a convex combination of the conjugated pairs \((U_{ab}\rho U_{ab}^\dagger ,U_{ab}\sigma U_{ab}^\dagger )\) with equal weights \(d_C^{-2}\). Joint convexity bounds it above by the same convex combination of the per-term relative entropies, each of which equals \(D(\rho \Vert \sigma )\) by unitary invariance (Theorem 13.6.24). As the weights sum to one, the bound is \(D(\rho \Vert \sigma )\).

Let \(N\) be the dimension of the joint space. Regularize both arguments through the affine trace-shrinking perturbation \(M_\varepsilon =(1+N\varepsilon )^{-1}(M+\varepsilon \mathbb {1})\), which is positive definite for every \(\varepsilon {\gt}0\), so the positive-definite data-processing inequality (Theorem 13.6.34) gives

The right-hand side converges to \(D(\rho \Vert \sigma )\) as \(\varepsilon \to 0^+\), by the same shared-eigenbasis scalar-limit argument as the joint convexity on the support domain (Theorem 13.6.21).

For the left-hand side, the partial trace of the regularization is a differently scaled regularization of the marginal: because the partial trace is linear and \(\operatorname{tr}_C\mathbb {1}=d_C\mathbb {1}\),

This is the affine regularization of \(\operatorname{tr}_C M\) with the same scaling rate \(N\) but shift rate \(d_C\) and the smaller identity on the system factor. The support condition transfers to the marginals, \(\ker (\operatorname{tr}_C\sigma )\subseteq \ker (\operatorname{tr}_C\rho )\): a vector annihilated by \(\operatorname{tr}_C\sigma \) has vanishing marginal quadratic form, which splits into the non-negative joint quadratic forms of its single-ancilla lifts, so each lift lies in \(\ker \sigma \), hence in \(\ker \rho \), and reassembling the ancilla sum shows the vector lies in \(\ker (\operatorname{tr}_C\rho )\). With this support condition the arbitrary-rate affine regularization has the same both-arguments limit:

Passing the inequality through the two limits gives the support-domain bound.

On the non-negative spectrum of \(\tau \), the defining function satisfies

The spectral functional calculus therefore gives \(f(\tau )\geq 0\).

Every eigenvalue of \(\tau \) is strictly positive. Hence the function defining the support inverse square root agrees on the spectrum with the reciprocal of the positive square-root function.

In an eigenbasis of \(\tau \), the left-hand side has eigenvalue zero when \(\lambda _i=0\) and eigenvalue \(\lambda _i^{-1/2}\lambda _i\lambda _i^{-1/2}=1\) otherwise.

The support inverse square root commutes with \(\tau \) by spectral functional calculus, so

where the last step is Lemma 13.6.40. Taking adjoints gives the identity with \(\tau \) on the left.

Diagonalize \(A\) and \(B\). In the resulting product eigenbasis, the eigenvalues of \(A\otimes B\) are \(a_i b_j\) with \(a_i,b_j\geq 0\), and the claim follows from \(f(a_i b_j)=f(a_i)f(b_j)\).

Diagonalize both factors. The eigenvalues of \(A\otimes B\) are the products \(a_i b_j\). Both the square-root function and the function that equals \(x^{-1/2}\) for \(x{\gt}0\) and zero at \(x=0\) are multiplicative on the non-negative reals. For the support projectors, apply the sandwich identity to \(A\otimes B\), factor the support inverse square root and matrix products, and apply the sandwich identity to each factor. The cancellation identities follow entrywise. If \(A\) is positive definite, then \(A^{-1/2}_{\mathrm{supp}}=(\sqrt A)^{-1}\) and \(\sqrt A\) is invertible. Hence the cancellation identity gives \(P_A=\sqrt A(\sqrt A)^{-1}=\mathbf1\).

The scalar identity follows from the functional calculus and \(\sqrt{cx}=\sqrt c\sqrt x\) for \(c{\gt}0\). For the support-inverse function \(f(x)=x^{-1/2}\) when \(x{\gt}0\) and \(f(0)=0\), the functional calculus gives

This proves (76); the same argument with the identity as the left factor proves (77). Applying (75) with \(c=d_R^{-1}\) to \(\tau \otimes \mathbf1_R\) then gives (78).

Expand the marginal in a product basis. The remaining identities follow from functional calculus for positive semidefinite tensor products. The support projection is obtained by multiplying the factorized support inverse square root on both sides of the marginal.

Substitute the tensor factorizations of the square root and marginal support inverse into the Petz sandwich. The first-factor terms reduce to \(P_AA_0P_A\). This proves the formula for product operators. A finite product-operator decomposition proves the linear-map identity.

The support condition makes \(\mathcal S_{P_A}\) act as the identity on \(X\). If \(\rho _A\) is positive definite, its support projection is the identity, so the condition holds on the full matrix algebra.

Expand the two partial traces in a product basis. The sum over the \(d_A\) diagonal entries cancels the factor \(d_A^{-1}\). Trace invariance under a partial trace then gives the last identity.

The marginal identity gives \(\sigma _{AB}=d_A^{-1}\mathbf1_A\otimes \rho _B\). Apply the support-inverse scaling identity and the tensor identity for an auxiliary left factor. Since \(d_A{\gt}0\), \((\sqrt{d_A^{-1}})^{-1}=\sqrt{d_A}\).

Functional calculus gives \(\sqrt{\sigma _{ABC}} =d_A^{-1/2}\mathbf1_A\otimes \sqrt{\rho _{BC}}\). Theorem 13.6.55 gives the corresponding factorization of the marginal support inverse. The scalar factors cancel in the Petz sandwich. The identity follows first for \(A_0\otimes X_B\) and then for every operator by a finite sum of product operators.

Insert the factorized support inverse into \(P_{AB}=\sigma _{AB,\mathrm{supp}}^{-1/2} \sigma _{AB}\sigma _{AB,\mathrm{supp}}^{-1/2}\). The scalar factors cancel, and the remaining sandwich is \(\mathbf1_A\otimes (\rho _{B,\mathrm{supp}}^{-1/2}\rho _B \rho _{B,\mathrm{supp}}^{-1/2})=\mathbf1_A\otimes P_B\).

For an orthonormal basis \((e_r)_r\) of \(H_R\), let \(J_r:H_L\to H_L\otimes H_R\) be given by \(J_r(v)=v\otimes e_r\). Then

Thus the support Petz map has a rectangular Kraus representation.

Cyclicity of the trace and the defining property of the partial trace give

By (70), this equals \(\operatorname{tr}(P_\tau X)\).

The map \(X\mapsto Q_\tau XQ_\tau \) has the single Kraus operator \(Q_\tau \). Adjoining the positive semidefinite matrix \(\omega _R\) is completely positive, and the composition of these two maps is completely positive.

Since \(\operatorname{tr}(\omega _R)=\operatorname{tr}(\sigma )=1\) and \(Q_\tau ^2=Q_\tau \), factorization and cyclicity of the trace give \(\operatorname{tr}(\mathcal C_\sigma (X)) =\operatorname{tr}(Q_\tau XQ_\tau ) =\operatorname{tr}(Q_\tau ^2X) =\operatorname{tr}(Q_\tau X)\).

The two summands are completely positive. Since \(\operatorname{tr}(\omega _R)=1\), the complementary term satisfies (93). Adding (93) to (91) gives trace preservation.

From (90), \(Q_{AB}=\mathbf1_A\otimes Q_B\), while \(\operatorname{tr}_{AB}\sigma _{ABC}=\rho _C\). Hence

A finite product-operator decomposition gives the result for every input.

Add (89) and (95), and use additivity of the tensor product of linear maps.

By (87), \(\operatorname{tr}(\sigma _{ABC})=\operatorname{tr}(\rho _{BC})=1\). Apply the channel property of the completed Petz map.

The assumption implies \(Q_\tau XQ_\tau =0\). Hence \(\mathcal C_\sigma (X)=0\), and adding the complementary term to \(\mathcal R_\sigma (X)\) does not change its value.

Diagonalize \(A\) and \(B\). On a product eigenvector with eigenvalues \(a,b\geq 0\), equation (98) becomes

If \(a,b{\gt}0\), this is the ordinary product identity for the logarithm. If either eigenvalue is zero, both sides vanish. Conjugating the resulting diagonal identity by the product eigenbasis proves the formula.

The complementary projection \(\mathbf1-P_A\) has range contained in \(\ker A\), and hence in \(\ker B\). Thus \(B(\mathbf1-P_A)=0\). Taking adjoints gives \((\mathbf1-P_A)B=0\), so \(B=P_A B\), and therefore \(P_A B P_A=P_A B=B\).

Since \((J_r w)_{(j,c)}=w_j\mathbf1_{c=r}\), expansion of the matrix-vector product gives

Expanding the matrix products and the partial trace gives \(\sum _{i,j,r}\overline{w_i}X_{(i,r),(j,r)}w_j\) on both sides.

If \(w\in \ker (\operatorname{tr}_R\sigma )\), then

Each summand is non-negative and therefore vanishes. Positive semidefiniteness shows that \(\sigma (w\otimes e_r)=0\) for every \(r\), so the joint kernel inclusion gives \(\rho (w\otimes e_r)=0\). Summing the diagonal components over \(r\) yields \((\operatorname{tr}_R\rho )w=0\).

Kernel inclusion descends through the partial trace, giving \(\ker (\operatorname{tr}_R\sigma )\subseteq \ker (\operatorname{tr}_R\rho )\). Theorem 13.6.71 therefore yields \(P_\tau (\operatorname{tr}_R\rho )P_\tau =\operatorname{tr}_R\rho \), where \(\tau =\operatorname{tr}_R\sigma \). By Theorem 13.6.68,

By (70), the middle factor reduces to \(P_\tau \otimes \mathbf1_R\). Marginal-support absorption gives \((\mathbf1_{LR}-P_\tau \otimes \mathbf1_R)\sigma =0\), so

because that matrix is positive semidefinite of trace zero. Therefore \(\sqrt\sigma (P_\tau \otimes \mathbf1_R)\sqrt\sigma =\sigma \).

The marginal satisfies \(P_\tau \tau P_\tau =\tau \). Therefore the completed channel agrees with the support Petz map at \(\tau \), and Theorem 13.6.77 gives \(\widehat{\mathcal R}_\sigma (\tau )=\sigma \).

By Lemma 13.6.9 the entropy depends only on the characteristic polynomial. Reindexing conjugates \(\rho \) by a permutation matrix, so \(\chi _{(\rho _{e(i)\, e(j)})}=\chi _\rho \), and the entropies coincide.

The rectangular characteristic-polynomial identity gives \(X^n\chi _{AB}=X^m\chi _{BA}\). Thus the two characteristic polynomials have the same nonzero roots, with multiplicity; the additional zero roots contribute nothing because \(0\log 0=0\).

By Lemma 13.6.9, the entropy of a Hermitian matrix is its charpoly-root entropy sum. Lemma 13.6.80 identifies these two sums for \(AB\) and \(BA\).

The tensor product is unitarily conjugate to the diagonal of eigenvalue products \(\lambda _i\mu _j\). Using

and the unit eigenvalue sums collapses the double sum to \(S(\omega )+S(\tau )\).

The eigenvalues of \(c\, \omega \) are \(c\lambda _i\), so the entropy is \(\sum _i-(c\lambda _i)\log (c\lambda _i)\). The splitting identity

together with the unit eigenvalue sum \(\sum _i\lambda _i=1\) gives

the stated form.

Each block diagonalizes by a unitary congruence, and the block-diagonal assembly of the block unitaries diagonalizes the direct sum, whose eigenvalue multiset is the disjoint union of the block eigenvalue multisets. Hence

Write \(A=XY\), where \(X\) maps the direct sum of the labelled ambient spaces to the original space by the matrices \(M_j\), and \(Y\) maps back by the operators \(P_j\). The support and annihilation identities give \(YX=\bigoplus _jM_j\). Reversing the two rectangular factors preserves the nonzero eigenvalues, while any additional eigenvalues are zero. Entropy is therefore unchanged, and additivity on the block diagonal gives the result.

Entropy is additive over the orthogonal blocks. Applying the scaled-state formula to each block gives \(S(p_j\omega _j)=-p_j\log p_j+p_jS(\omega _j)\), and summing over \(j\) gives the result.

Each number \(p_j(R_j-L_j)\) is non-negative, and their sum vanishes. Therefore every one vanishes. Since \(p_j{\gt}0\), it follows that \(R_j-L_j=0\).

Follows from \(\overline{\rho _{ji}} = \rho _{ij}\) applied entry-wise inside the summation defining each partial trace.

When \(\rho _R\) is singular the tensor logarithm of the reference does not split. Regularize the reduced state by the affine perturbation \(\rho _{R,\varepsilon } = (1 + d_R\varepsilon )^{-1}(\rho _R + \varepsilon \mathbb {1})\), positive definite for \(\varepsilon {\gt} 0\). The reference \((\mathbb {1}_A / d_A) \otimes \rho _{R,\varepsilon }\) is then a positive definite tensor product, whose logarithm splits, so the cross trace term evaluates to \(-\log d_A + \operatorname{Re}\operatorname{tr}(\rho _R\log \rho _{R,\varepsilon })\). As \(\varepsilon \to 0^+\) the support condition makes the zero eigenvalues of \(\rho _R\) contribute nothing, so

Since \(D(\rho \| \sigma ) = -S(\rho ) - \operatorname{Re}\operatorname{tr}(\rho \log \sigma )\), the evaluation follows.

A vector \(v\) annihilated by \((\mathbb {1}_A / d_A) \otimes \rho _R\) is annihilated by \(\mathbb {1}_A \otimes \rho _R\), since \(\mathbb {1}_A / d_A\) is invertible. Let \(P\) be the orthogonal projection onto the range of \(\rho _R\). The complementary lift \(\mathbb {1}_A \otimes (\mathbb {1}- P)\) then fixes \(v\), while it annihilates \(\rho \) on the left because the reduced state of \(\rho \) on \(R\) is supported on the range of \(P\). Hence \(\rho v = 0\).

For indices \((a_1,b_1)\) and \((a_2,b_2)\), the corresponding matrix entry is

This is the corresponding entry of \((\mathbb {1}_A/d_A)\otimes \rho _B\).

Read the inequality as one instance of data processing under the partial trace over \(C\), with the singular reference state \(\sigma _{ABC} = (\mathbb {1}_A / d_A) \otimes \rho _{BC}\). Being a density operator is not enough to place the pair \((\rho _{ABC}, \sigma _{ABC})\) in the relative-entropy domain, which is the kernel inclusion \(\ker \sigma _{ABC} \subseteq \ker \rho _{ABC}\); the marginal support lemma supplies it. Against this reference the relative entropy of each pair evaluates to an entropy difference:

For a singular reference the tensor logarithm does not split, so each evaluation regularizes the reduced state through the affine perturbation \(\rho _{R,\varepsilon } = (1 + d_R\varepsilon )^{-1}(\rho _R + \varepsilon \mathbb {1})\), which is positive definite, and passes to the limit

where the zero eigenvalues of \(\rho _R\) contribute nothing because the kernel inclusion makes the corresponding diagonal weights vanish. Data processing on the singular support domain under the partial trace over \(C\), which sends \(\rho _{ABC} \mapsto \rho _{AB}\) and \((\mathbb {1}_A / d_A) \otimes \rho _{BC} \mapsto (\mathbb {1}_A / d_A) \otimes \rho _B\), reads

Substituting (115) and (116) into (117) cancels the common \(\log d_A\) and rearranges to the claimed inequality.

Read the inequality as one instance of data processing under the partial trace over \(C\), with reference state \(\sigma _{ABC} = (\mathbb {1}_A / d_A) \otimes \rho _{BC}\), which is positive definite because \(\rho _{BC}\) is. Against this reference the relative entropy of each pair evaluates to an entropy difference:

Each equality follows from the tensor logarithm split and the adjoint of the partial trace. Data processing under the partial trace over \(C\), which sends \(\rho _{ABC} \mapsto \rho _{AB}\) and \((\mathbb {1}_A / d_A) \otimes \rho _{BC} \mapsto (\mathbb {1}_A / d_A) \otimes \rho _B\), reads

Substituting (118) and (119) into (120) cancels the common \(\log d_A\) and rearranges to the claimed inequality.

Write \(P_\rho \), \(P_\sigma \), and \(P_\tau \) for the support projections. The kernel inclusion gives \(P_\sigma \rho P_\sigma =\rho \) and \(\rho P_\sigma =\rho \), while \(\rho P_\rho =\rho \) and \(\tau P_\tau =\tau \). Substituting the singular tensor-logarithm formulas into the relative entropy and using these four identities cancels the two contributions containing \(\log \tau \). Factoring the trace of the remaining tensor product gives

Let \(\tau _C=d_C^{-1}\mathbb {1}_C\). The twirl identity gives \(\overline X=(\operatorname{tr}_C X)\otimes \tau _C\). The support inclusion passes to the partial traces:

Hence support-domain ancilla additivity and the saturation hypothesis give

Every \(U_{ab}\) is unitary, and unitary invariance gives

Together, (125) and (126) prove (124).

Unitary invariance gives, for every \(c,e\),

Theorem 13.8.8 gives \(D(\overline\rho \Vert \overline\sigma )=D(\rho \Vert \sigma )\). Substituting into the right-hand side gives

The functional-calculus identity \(\log (cA)=(\log c)\mathbb {1}+\log A\), and its analog for \(B\), show that the scalar logarithmic terms cancel in \(\log (cA)-\log (cB)\). Linearity of the trace then gives the result.

On the support of a positive semidefinite matrix,

The support projection \(P_A\) fixes \(A\) on the right. The kernel inclusion also makes \(P_B\) fix \(A\) on the right. Hence the two terms containing \(\log c\) cancel in the trace-log difference, and linearity of the trace gives the result.

Positive definiteness of \(\sigma \) makes the support condition in Theorem 13.8.9 automatic. That theorem gives the equality with \(d_C^{-2}\) multiplying each scalar relative entropy. Lemma 13.8.10 moves this coefficient inside both matrix arguments, which is precisely the stated weighted gap.

The integrand is the derivative of \(a(\log (1+t)-\log (a+tb))\). Its value at \(t=0\) is \(-a\log a\), whereas its limit as \(t\to \infty \) is \(-a\log b\). The derivative has constant sign, according as \(a-b\) is positive or negative, and is therefore integrable. The fundamental theorem of calculus gives the stated value.

Expand both trace terms in eigenbases. Unitarity of the overlap matrix gives \(\sum _j|w_{ij}|^2=1\), so

Subtraction gives (132).

If \(\beta _j=0\), the kernel inclusion gives \(\alpha _i|w_{ij}|^2=0\), and both \(r_{ij}\) and \(e_{ij}\) are defined to be zero. If \(\beta _j{\gt}0\), the scalar integral applies when \(\alpha _i{\gt}0\), while the term is identically zero when \(\alpha _i=0\). Since the double sum is finite, integration term by term gives

The kernel inclusion also shows that replacing each \(\beta _j=0\) summand in Theorem 13.8.14 by zero does not change its value. Hence that theorem identifies the sum in (138) with \(D(A\Vert B)\).

Diagonalize \(A\) and \(B\) and write \(W=U_A^\ast U_B\). In these coordinates, the two equations

have entries

For every positive semidefinite \(S\), the identities \(S^+S=SS^+=P_S\) imply that \(Sx=b\) gives \(\langle b,S^+b\rangle =\langle b,x\rangle \). Applying this identity to (142) and (143), with (146) and (149), gives (139) and (140). Their coefficientwise combination is the scalar function appearing in Lemma 13.8.13. Continuity follows term by term from the finite sums.

Put

The source equation for the support relative-modular operator shows that \(b_i\) lies in the support of \(S_i\). The same argument applied to \(\sum _i A_i\) and \(\sum _i B_i\) shows that \(\sum _i b_i\) lies in the support of \(\sum _i S_i\). The support-resolvent residual identity writes the displayed defect as a sum of nonnegative quadratic residuals.

The source vectors lie in the supports of their left–right operators, and their sum lies in the support of \(S_\Sigma \). The vanishing real defect and the support-resolvent residual identity force every quadratic residual to vanish. The common-solution conclusion follows after projection to the support of each \(S_i\).

Put

The inclusion \(\ker B_i\subseteq \ker A_i\) gives \(A_iP_{B_i}=A_i\). The source-\(A\) left–right equation therefore places \(a_i\) in the support of \(S_i\). Positivity shows that a vector annihilated by \(\sum _i B_i\) is annihilated by every \(B_i\), and hence by every \(A_i\). Thus the summed source also lies in the support of the summed operator. The support-resolvent residual identity writes the displayed difference as a sum of nonnegative quadratic residuals.

The kernel inclusions for the summands imply \(\ker B_\Sigma \subseteq \ker A_\Sigma \). Apply the one-pair support-domain integral representation to every \((A_i,B_i)\) and to \((A_\Sigma ,B_\Sigma )\). The pointwise left–right identities identify the difference of the spectral integrands with \(F\); linearity of trace cancels the terms involving \(\operatorname {tr}(B_i)\). Finite summation commutes with the integral. Continuity follows from the corresponding one-pair continuity theorem and finite summation.

Both defects are nonnegative. Since \(t{\gt}0\) and \(1+t{\gt}0\), multiplying the source-\(B\) defect by \(t\), adding the source-\(A\) defect, and dividing by \(1+t\) preserves nonnegativity.

The relative-entropy equality and (150) make the integral of the nonnegative function \(F\) vanish. Hence \(F=0\) almost everywhere. Its continuity on the open positive half-line upgrades this to \(F(t)=0\) for every \(t{\gt}0\). Both defects are nonnegative, so \(\operatorname {Def}_A(t)+t\operatorname {Def}_B(t)=0\), and \(t{\gt}0\) forces \(\operatorname {Def}_B(t)=0\).

Vectorization sends \(L_A+tR_B\) to \(A\otimes \mathbb {1}+t\mathbb {1}\otimes B^{\mathsf T}\). The vectors \(u_i\otimes \overline{v_j}\) diagonalize this matrix with eigenvalues \(\alpha _i+t\beta _j\), which proves (151) and (152). Insert these identities into (153), use \(\sum _i|w_{ij}|^2=1\), and apply Lemma 13.8.13 term by term. The same finite spectral sum proves continuity and integrability.

Since \(G_iS_i=S_iG_i=P_i\) and \(P_i b_i=b_i\), expansion of the \(i\)th summand gives

Sum over \(i\). The support assumption on \(b\) gives \(Sx=SGb=b\), so the last three terms combine to \(-\langle b,Gb\rangle \).

Lemma 13.8.24 writes the complex defect in (155) as a finite sum of non-negative quadratic forms. Its imaginary part therefore vanishes automatically, while the hypothesis makes its real part vanish. Hence \(G_i(b_i-S_ix)=0\) for every \(i\). Multiplication by \(S_i\) shows that \(P_i(b_i-S_ix)=0\). Both \(b_i\) and \(S_ix\) lie in the support of \(S_i\), so \(b_i=S_ix\). Multiplication by \(G_i\) now gives (156).

Write \(S_g\) for the positive definite matrix representing \(T_g\) under the vectorization \(X\mapsto \operatorname{vec}(X^{\mathsf T})\), and put \(b_g=\operatorname{vec}(B_g^{\mathsf T})\) and \(x=S^{-1}\sum _g b_g\). For the residual \(r_g=b_g-S_gx\), direct expansion gives

Since \(S_g^{1/2}\) is invertible,

Summing proves the identity.

Apply the residual identity of Theorem 13.8.26 with \(a_g=\operatorname{vec}(A_g^{\mathsf T})\) and \(\bar a=\operatorname{vec}(A^{\mathsf T})\). Thus

which is the asserted source-\(A\) identity.

Apply Theorem 13.8.23 to each pair \((A_g,B_g)\) and to \((A,B)\), and interchange the finite sum with the integral. The trace terms cancel because \(B=\sum _gB_g\). The two fixed-resolvent identities write the defects as sums of squared norms, proving nonnegativity. Continuity and integrability follow from the corresponding spectral assertions before taking the finite difference.