Identify \(\psi \) with its coefficient matrix \(C\in M_{d_A,d_B}(\mathbb {C})\). The spectral theorem applied to the positive semidefinite matrix \(C\, C^\dagger \) supplies the eigenbasis \(\{ e_j\} \) of \(\mathbb {C}^{d_A}\) and the eigenvalues \(\lambda _j\); the normalized images of the eigenvectors under \(C^\dagger \), conjugated and completed to an orthonormal basis, supply \(\{ f_j\} \).

The case \(d_A\le d_B\) is Lemma 1.1.2. The case \(d_B{\lt}d_A\) reduces to it by applying the lemma to \(\psi \) with its two factors exchanged, then exchanging the two resulting bases back. This matches Wolf’s proof sketch: a change of local bases in the coefficient matrix corresponds to left and right multiplication by two unitaries, so the Schmidt coefficients are its singular values.

The Schmidt decomposition factors the coefficient matrix of \(\psi \) through the isometries with columns \(\{ e_j\} \) and \(\{ f_j\} \), so its rank equals the rank of the diagonal matrix of Schmidt coefficients, which is the number of nonzero entries.

Tracing out the second factor of \((\operatorname{id}\otimes T)(|\psi \rangle \! \langle \psi |)\) applies \(T\) to each rank-one block \(|a\rangle \! \langle b|\mapsto C_{i,a}\, \overline{C_{j,b}}\) of the coefficient matrix and takes its trace; the trace-pairing identity \(\operatorname{tr}[T(M)]=\operatorname{tr}[M\, T^*(\mathbb {1})]\) turns this into the displayed bilinear form in the rows of \(C\). Wolf derives this identity for the canonical purification \(\psi =(\sqrt\rho \otimes \mathbb {1})|\Omega \rangle \); the argument here uses only the coefficient matrix of \(\psi \) and linearity, so it holds for an arbitrary purification.

Write \(C\) for the coefficient matrix of \(\psi \), so \(C\, C^\dagger =\rho \), and let \(C^+:=C^\dagger \rho ^+\) using the generalized inverse of \(\rho \) on its support. Then \(C\, C^+\) is the support projection of \(\rho \), and \(C^+C\) is an orthogonal projection on \(\mathbb {C}^{d_B}\) (the subspace of \(\mathbb {C}^{d_B}\) needed to purify \(\rho \)). For each \(i\), the support projection of \(\rho \) absorbs \(\lambda _i\rho _i\) on both sides (its range lies in the range of \(\rho \)), so \(C\, (C^+(\lambda _i\rho _i)(C^+)^\dagger )\, C^\dagger =\lambda _i\rho _i\). Transposing \(C^+(\lambda _i\rho _i)(C^+)^\dagger \) gives an effect operator \(E_i\) with \(C\, E_i^{\mathsf T}\, C^\dagger =\lambda _i\rho _i\); adding the transpose of the orthogonal complement of \(C^+C\) to one fixed \(E_{i_0}\) keeps this identity (the complement contributes zero under \(C(\cdot )C^\dagger \)) while making \(\sum _i E_i=\mathbb {1}\). Taking \(T_i\) to be the Lüders instrument \(\sigma \mapsto \sqrt{E_i}\, \sigma \, \sqrt{E_i}\) gives \(T_i^*(\mathbb {1})=E_i\), so \(\{ T_i\} \) is an instrument, and Theorem 1.2.3 recovers the displayed identity.

This route proves the proposition directly for an arbitrary purification and an arbitrary convex decomposition, in place of Wolf’s route through the canonical purification \(\psi =(\sqrt\rho \otimes \mathbb {1})|\Omega \rangle \) and the transposition operator \(\theta \) with respect to its Schmidt basis.

Write \(P\) for the support projection of \(A\). The kernel inclusion compresses \(B\) onto the range of \(A\), so \(A^{-1/2}BA^{-1/2}\) lies entirely inside \(\mathrm{ran}(P)\), and conjugating by \(A^{-1/2}\) and by \(A^{1/2}\) give mutually inverse congruences

so \(A-cB\ge 0\) iff \(P-c\, A^{-1/2}BA^{-1/2}\ge 0\). Since \(A^{-1/2}BA^{-1/2}\) is supported on \(\mathrm{ran}(P)\), this last inequality is in turn equivalent to \(1-c\, A^{-1/2}BA^{-1/2}\ge 0\), which by the \(C^*\)-identity \(\| X\| _\infty \le 1\iff X\le 1\) (for \(X\ge 0\)) holds iff \(c\| A^{-1/2}BA^{-1/2}\| _\infty \le 1\).

If \(\rho v=0\) for some vector \(v\), every term of \(\rho =\sum _i\lambda _i\rho _i\) is positive semidefinite, so each summand’s quadratic form at \(v\) vanishes; in particular \(c\langle v,\rho _1v\rangle =0\), and since \(c{\gt}0\) and \(\rho _1\) is positive semidefinite, \(\rho _1v=0\).

Necessity of the kernel inclusion is Lemma 1.3.3.

The existence of a decomposition with weight \(c\) on \(\rho _1\) is equivalent to \(\rho -c\rho _1\ge 0\): given the kernel inclusion, the remaining terms of the decomposition sum to \(\rho -c\rho _1\), itself a positive-semidefinite combination; conversely, when \(\rho -c\rho _1\ge 0\) its trace is \(1-c\ge 0\), so either \(c=1\) and \(\rho =\rho _1\) (a zero-trace positive-semidefinite matrix vanishes), or \(c{\lt}1\) and \(\sigma :=(1-c)^{-1}(\rho -c\rho _1)\) is a density operator with \(\rho =c\rho _1+(1-c)\sigma \).

Finally \(\rho -c\rho _1\ge 0\) is equivalent to \(c\| \rho ^{-1/2}\rho _1\rho ^{-1/2}\| _\infty \le 1\) by Lemma 1.3.2, applied with \(A=\rho \), \(B=\rho _1\) under the kernel inclusion.

As the \(T(a_{ij})\) mutually commute, choose a basis in which they are simultaneously diagonal, and write \(\psi _{j,\beta }\) for the components of \(|\psi _j\rangle \) in this basis. Then

by positivity of \(T\) and of \([a_{ij}]\): the bracketed matrix is \(B_\beta ^\dagger [a_{ij}]B_\beta \) for the column vector \(B_\beta =(\psi _{j,\beta })_j\), hence positive semidefinite.

By Theorem 3.1.10 it suffices to show that the Choi matrix \(\tau \) of \(T\) satisfies \(\langle \phi |\tau |\phi \rangle \ge 0\) for every \(\phi \in \mathbb {C}^D\otimes \mathbb {C}^D\). Let \(a_{ij}\in M_D(\mathbb {C})\), for \(i,j\in \{ 1,\dots ,D\} \), be the blocks of \(|\Omega \rangle \! \langle \Omega |\), the Choi matrix of the identity map (Theorem 3.1.6), and let \(\psi _k:=\phi (\cdot ,k)\in \mathbb {C}^D\) be the reshaping of \(\phi \) into a \(D\)-indexed family of vectors. Then

the block quadratic form of Lemma 1.4.2 at \(T\), \([a_{ij}]\), and \(\{ \psi _i\} \). Since \([a_{ij}]\ge 0\) (it is the block family of the rank-one projector \(|\Omega \rangle \! \langle \Omega |\)) and, by commutative range, the images \(T(a_{ij})\) pairwise commute, Lemma 1.4.2 gives \(\langle \phi |\tau |\phi \rangle \ge 0\).

For \(X\ge 0\), \(T^*(X)\) is Hermitian since \(T\) preserves the adjoint. For every \(B\ge 0\), the trace-pairing identity \(\operatorname{tr}[T^*(X)B]=\operatorname{tr}[X\, T(B)]\) and positivity of \(X\) and of \(T(B)\) (as \(T\) is positive) give \(\operatorname{tr}[T^*(X)B]\ge 0\); a Hermitian matrix with nonnegative trace pairing against every positive semidefinite matrix is itself positive semidefinite, so \(T^*(X)\ge 0\).

Complete positivity in Kraus form is preserved by the trace-pairing adjoint, and the adjoint is an involution, so the two directions are the same statement applied to \(T\) and to \(T^*\).

By Lemma 1.4.4, \(T^*\) is positive, and by hypothesis it has commutative range, so Lemma 1.4.3 gives that \(T^*\) is completely positive. Lemma 1.4.5 then gives that \(T\) is completely positive.

If \(T\) has commutative range, apply Lemma 1.4.3. If instead \(T^*\) has commutative range, apply Lemma 1.4.6.

Expand both traces in product bases and sum first over the identity entry on \(H_R\).

If \(\rho ^{(k)}\) is the principal submatrix indexed by \(\{ k\} \times H_R\), then

Each \(\rho ^{(k)}\) is positive semidefinite, and hence so is their finite sum.

Expanding in product bases gives

Set \(Q_L=\mathbb {1}_L-P_L\). Since \(Q_L\operatorname{tr}_R\rho =0\), the adjoint identity gives

Positivity of \(\rho \) then implies \((Q_L\otimes \mathbb {1}_R)\rho =0\). Finally, \(\mathbb {1}_{L\otimes R}-(Q_L\otimes \mathbb {1}_R)=P_L\otimes \mathbb {1}_R\) gives the claim.

Apply \((\cdot )^\dagger \) to the left absorption identity. Since \(\rho ^\dagger =\rho \) and \((P_L\otimes \mathbb {1}_R)^\dagger =P_L\otimes \mathbb {1}_R\),

Set \(Q=\mathbb {1}_R-P_R\). The preceding trace identity gives

since \(P_R\) is the support projector of \(\operatorname{tr}_L\rho \). Positivity of \(\rho \) then gives \((\mathbb {1}_L\otimes Q)\rho =0\), and hence \((\mathbb {1}_L\otimes P_R)\rho =\rho \).

Take adjoints in the left absorption identity.

For all \(i,j\), \([\operatorname{tr}_{A}(X\otimes Y)]_{ij} =\sum _t X_{tt}Y_{ij}=\operatorname{tr}(X)Y_{ij}\).

The identity matrix is positive semidefinite, and \(m^{-1}\) is nonnegative.

Since \(\operatorname{tr}(\mathbb 1_m)=m\), scalar linearity of the trace gives the result.

Interchange the two tensor factors, trace out the new right factor, prepare the maximally mixed state \(m^{-1}\mathbb 1_m\), and interchange the factors back. Theorem 1.6.2 and Theorem 1.6.3 show that the prepared matrix is positive semidefinite and has trace one. Theorem 1.8.8, Lemma 1.8.22, Lemma 1.8.27, and Lemma 1.8.9 then show that the resulting map is trace-preserving and completely positive. Expanding the composition gives

The Kronecker-product partial-trace identity gives

so \(E(\mathbb 1_m\otimes X)=\mathbb 1_m\otimes X\). Pointwise fixation gives both idempotence and unitality, while the evaluation identity gives the range inclusion.

Compress \(A\) to the family \((A_{kk})_{k\in K}\) and apply the normalized partial-trace expectation from Theorem 1.6.4 in each coordinate. The coordinatewise Kraus theorem makes the resulting map on the finite sum completely positive, and its canonical full-matrix extension discards the off-diagonal blocks. Each coordinate preserves trace, so preservation of the total block trace implies preservation of the trace on \(\operatorname{End}(\mathcal H)\). The displayed formula shows that the range lies in the coordinate subalgebra. The one-block fixation identity holds in every summand, hence \(E\) fixes that subalgebra pointwise; idempotence and unitality follow.

Unitary conjugation preserves positive semidefiniteness, as does reindexing a matrix along an equivalence. Apply these facts in the two possible orders.

The forward map is the composition of conjugation by \(U^\dagger \) with reindexing by \(e^{-1}\), while the inverse is reindexing by \(e\) followed by conjugation by \(U\). Unitarity makes both single-Kraus conjugations trace preserving. Equivalence reindexing is trace preserving and completely positive, and composition preserves these properties.

By the finite-dimensional block form, choose unitary coordinates in which

Let \(E_0\) be the coordinate direct-sum expectation and set

Theorem 1.6.7 and closure under composition show that \(E\) is trace-preserving and completely positive. The identities \(\Phi \Phi ^{-1}=\operatorname{id}\) and \(E_0^2=E_0\) give \(E^2=E\), and all three maps are unital. Finally, the block-membership equivalence identifies \(\Phi (\mathcal B)\) with the coordinate subalgebra. Thus the range of \(E\) lies in \(\mathcal B\), and the pointwise fixation of that coordinate subalgebra by \(E_0\) implies \(E(B)=B\) for every \(B\in \mathcal B\).

The matrix \(\| A\| _\infty \mathbb 1-A\) is positive semidefinite and lies in \(S\otimes M_{n}(\mathbb {C})\), so complete positivity of \(T\) at level \(n\) gives that its image under \(T\otimes \mathrm{id}_n\) is positive semidefinite. Testing against \(\Omega \) and using linearity of \(\tau \),

both sides are real since \(\| A\| _\infty \mathbb 1-A\) is Hermitian, and rearranging gives the bound.

The functional \(f(A):=\| A\| _\infty \, \tau (\mathbb 1)\) is sublinear, so by Lemma 1.6.11 the Hahn–Banach theorem extends \(\tau \), dominated by \(f\), from the Hermitian matrices of \(S\otimes M_{n}(\mathbb {C})\) to those of \(M_{m}(\mathbb {C})\otimes M_{n}(\mathbb {C})\) as a real-linear functional. Decomposing each matrix of \(M_{m}(\mathbb {C})\otimes M_{n}(\mathbb {C})\) into its Hermitian and skew-Hermitian parts, both of which stay inside \(S\otimes M_{n}(\mathbb {C})\) whenever the original matrix does (since \(S\) is closed under the adjoint), extends this further to a \(\mathbb {C}\)-linear functional \(\tau '\) agreeing with \(\tau \) on all of \(S\otimes M_{n}(\mathbb {C})\) and obeying the same bound throughout.

The matrix \(\| A\| _\infty \mathbb 1-A\) is positive semidefinite (hence Hermitian) of norm at most \(\| A\| _\infty \) whenever \(A\) is positive semidefinite, so the domination hypothesis applied to it gives

since \(\operatorname{Re}\varphi (\mathbb 1)\ge 0\), rearranging gives \(\operatorname{Re}\varphi (A)\ge 0\).

Because \(\tau '\) is built by the Hermitian-decomposition construction, it sends the adjoint of a matrix to the conjugate of its value, so \(Y\) is Hermitian. For a vector \(v\), the trace-pairing identity gives \(v^\dagger Yv=\tau '(vv^\dagger )\), and \(vv^\dagger \) is positive semidefinite, so the hypothesis gives \(\operatorname{Re}(v^\dagger Yv)\ge 0\); since \(Y\) is Hermitian, \(v^\dagger Yv\) is already real, hence nonnegative.

Both sides expand, entry by entry, to the same double sum \(n\sum _k\sum _{p,q}\overline{v_k(p,i)}\, B_{pq}\, v_k(q,j)\).

Direct computation from the definition of \(\tau \), expanding the trace against \(\Omega \) and collapsing the resulting sum using the structure of \(B\otimes |i\rangle \langle j|\).

By Lemma 1.6.12, \(\tau \) extends to a \(\mathbb {C}\)-linear functional \(\tau '\) on \(M_{m}(\mathbb {C})\otimes M_{n}(\mathbb {C})\) agreeing with \(\tau \) on \(S\otimes M_{n}(\mathbb {C})\) and obeying the same norm-domination bound throughout. By Lemma 1.6.13, \(\tau '\) is nonnegative on positive semidefinite matrices, so by Lemma 1.6.14 the matrix \(Y\) representing \(\tau '\) through the trace pairing is positive semidefinite, hence a sum of rank-one terms \(Y=\sum _kv_kv_k^\dagger \). Lemma 1.6.15 reshapes these into a rectangular Kraus family for \(T'(B)_{ij}:=n\cdot \tau '(B\otimes |i\rangle \langle j|)\), so \(T'\) is completely positive. Lemma 1.6.16 gives \(\tau (B\otimes |i\rangle \langle j|)=\tfrac 1n\, T(B)_{ij}\) for \(B\in S\), and \(\tau '\) agrees with \(\tau \) on \(S\otimes M_{n}(\mathbb {C})\), so \(T'\) agrees with \(T\) on \(S\).

The first assertion follows from the corresponding block-diagonal positivity equivalence and invariance of positivity under simultaneous row and column reindexing. The second follows by evaluating both sides of (4) at a pair of ambient matrix indices.

Expanding matrix entries gives

Each summand preserves positive semidefiniteness, and so does their sum.

The image of an operator system under an injective unital \(*\)-homomorphism is an operator system. For complete positivity, a level-\(j\) matrix over the image \(\iota (S)\) compresses blockwise to a family over \(S\): the block-diagonal compression left-inverts the embedding on every slice,

and each block is positive semidefinite by invariance of positivity under simultaneous row and column reindexing. Complete positivity of \(T\) on \(S\) applies to this family, and the linear equivalence of \(S\) and \(\iota (S)\) converts the conclusion back to \(T''\).

Let \(\iota =\iota _1\) be the block-diagonal embedding. By Theorem 1.7.6, \(\iota (S)\) is an operator system in \(M_M(\mathbb {C})\) and the induced map \(T''\) is completely positive at every matrix level. Theorem 1.6.17 gives a rectangular Kraus extension \(\widetilde T:M_M(\mathbb {C})\to M_{p}(\mathbb {C})\) of \(T''\). Define \(T'=\widetilde T\circ \iota \).

To verify (2), fix \(j\) and an entrywise positive family \(X\). The level embedding \(\iota _j(X)\) is positive by Theorem 1.7.4. The Kraus ampliation theorem (Theorem 1.7.5) therefore gives \((\widetilde T\otimes \operatorname{id}_j)(\iota _j(X))\ge 0\). Using (4) identifies this matrix with \((T'\otimes \operatorname{id}_j)(X)\), proving complete positivity of \(T'\). The defining agreement of \(\widetilde T\) with \(T''\) shows that \(T'\) agrees with \(T\) on \(S\).

At each matrix level \(j\), the ampliations satisfy

If every component of \(X\) is positive semidefinite, complete positivity of \(F\) makes the inner matrix positive semidefinite. The Kraus representation of \(E\otimes \operatorname{id}_j\) then preserves its positivity.

Theorem 1.7.7 gives a completely positive extension \(F\) with values in \(M_{p}(\mathbb {C})\). Choose the conditional expectation \(E:M_{p}(\mathbb {C})\to M_{p}(\mathbb {C})\) from Theorem 1.6.8; its range lies in \(\mathcal B\). Set \(G=E\circ F\). Theorem 1.7.8 shows that \(G\) is completely positive, while \(E(M_{p}(\mathbb {C}))\subseteq \mathcal B\). For \(x\in S\), one has \(F(x)=T(x)\in \mathcal B\), and therefore \(G(x)=E(T(x))=T(x)\).

Apply Theorem 1.7.9 to the composite of \(T\) with the inclusion \(\mathcal B\hookrightarrow M_{p}(\mathbb {C})\). Since the resulting ambient map \(G\) takes every value in \(\mathcal B\), it defines a map \(\widetilde T\) with codomain \(\mathcal B\). Its composite with the inclusion is \(G\), hence is completely positive, and equality in \(M_{p}(\mathbb {C})\) gives \(\widetilde T|_S=T\).

Take \(r=1\) and \(A_0=I\); then \(\operatorname{id}(X)=IXI^\dagger =X\) and \(A_0^\dagger A_0=I\).

Take the sole Kraus operator to be \(V\). Its resolution of the identity is precisely \(V^\dagger V=I_H\).

Let \(P_e:\mathbb {C}^I\to \mathbb {C}^J\) be the permutation matrix of \(e\). Then \(R_e(X)=P_eXP_e^\dagger \) and \(P_e^\dagger P_e=I_{\mathbb {C}^I}\). The claim follows from the preceding isometry lemma.

The Kraus form of the composite and its resolution of the identity are

The displayed family is already a Kraus representation, and the assumed identity is precisely its trace-preserving normalization.

Reindexing the finite sum gives \(\sum _{b\in I'}A_{e^{-1}(b)}XA_{e^{-1}(b)}^\dagger =\sum _{a\in I}A_aXA_a^\dagger \).

If \(j\ne k\), then every zero-extended Kraus operator satisfies \([\widetilde A_{j,a}X\widetilde A_{j,a}^\dagger ]_{(k,b),(k,c)}=0\). Hence only the \(j=k\) terms survive, and \([\mathcal C(X)]_{kk}=\sum _a A_{k,a}X_{kk}A_{k,a}^\dagger \).

For every \((j,a)\) and \(k\ne l\), the zero-extended operator satisfies \([\widetilde A_{j,a}X\widetilde A_{j,a}^\dagger ]_{(k,b),(l,c)}=0\). Therefore \([\mathcal C(X)]_{kl}=\sum _{j,a}0=0\).

Each embedded operator has support in one summand, and therefore

Apply Theorem 1.8.12 to the combined rectangular Kraus family \((\widetilde A_{k,a})_{k,a}\).

Choose Kraus operators \((A_{s,a})_a\) for \(\Phi _s\), and set \(B_{s,a}=A_{s,a}P_s\). These operators give the displayed map, and

Apply Theorem 1.8.12.

For all \(i,j\), \([\operatorname{tr}_B(X\otimes Y)]_{ij} =\sum _t X_{ij}Y_{tt}=\operatorname{tr}(Y)X_{ij}\).

For each basis vector \(e_k\) of \(B\), define \(V_k:A\otimes B\to A\) by \(V_k(e_i\otimes e_\ell )=\delta _{k\ell }e_i\). Then

Theorem 1.8.12 applies.

Put \(R=\sqrt\rho \). Since \(R\) is Hermitian and \(R^2=\rho \), the \(((a,s),(b,t))\) entry of the left-hand side is

Since \(R\) is Hermitian and \(R^2=\rho \), the diagonal entries of the sum are \(\sum _{j,t}\overline{R_{tj}}R_{tj} =\operatorname{tr}(R^2)=\operatorname{tr}(\rho )=1\), while its off-diagonal entries vanish.

Theorem 1.8.24 identifies the map with the rectangular Kraus family \((A_j)_j\), and Theorem 1.8.25 gives \(\sum _j A_j^\dagger A_j=I_A\). Apply Theorem 1.8.12.

Diagonalize \(A=UDU^\dagger \), where \(D=\operatorname {diag}(\lambda _0,\ldots ,\lambda _{d-1})\). Unitary conjugation commutes with powers and leaves the trace unchanged, while \(D^k=\operatorname {diag}(\lambda _0^k,\ldots ,\lambda _{d-1}^k)\).

For \(i{\lt}j\), monotonicity of the ordered eigenvalues and positivity of \(\varepsilon \) give

Thus the diagonal entries in (11) are distinct. The same formula is continuous in \(\varepsilon \), and at \(\varepsilon =0\) it is the spectral decomposition of \(A\).

The common spectrum is the common finite set underlying the two root multisets. The roots of \(\chi _A\) are distinct, so this set has \(d\) elements. Both characteristic polynomials split over \(\mathbb {C}\) and have degree \(d\), so the root multiset of \(\chi _B\) also has cardinality \(d\). Hence it cannot repeat any root, and the two root multisets agree.

Theorem 1.9.4 gives equality of the root multisets. Formula (9) then gives equality of trace powers.

Finite matrix multiplication is continuous, hence so is \(A\mapsto A^k\); the trace is a finite sum of matrix entries.

Let \(e_m\) denote the \(m\)th elementary symmetric polynomial in the eigenvalues, with \(e_0 = 1\). The Newton–Girard recursion

expresses the coefficients of the characteristic polynomial in terms of the power-sum traces \(\operatorname{tr}(A^i)\). Equal power sums for \(A\) and \(B\) therefore give equal coefficients, hence equal characteristic polynomials.

Fix Hermitian \(A\) and use the positive simple-spectrum perturbations \(A_{\varepsilon _n}\) from Theorem 1.9.3. The hypothesis (14) and Theorem 1.9.5 give

for every \(n\) and \(k\). Linearity makes \(T\) continuous in finite dimension. Since \(A_{\varepsilon _n}\to A\), continuity from Theorem 1.9.6 yields \(\operatorname{tr}(T(A)^k)=\operatorname{tr}(A^k)\). For \(1\le k\le d\), Newton–Girard identities determine all coefficients of the degree-\(d\) characteristic polynomial, proving \(\chi _{T(A)}=\chi _A\). When \(d=0\), the same argument consists of empty sums and degree-zero characteristic polynomials.

Equality of the characteristic polynomials gives equality of the ordered eigenvalue lists. Write the two spectral decompositions as

Likewise,

Set

This matrix is unitary, and cancellation of \(U_A^\dagger U_A\) gives

The rank-one identity \(|v\rangle \! \langle v|^2=\langle v,v\rangle |v\rangle \! \langle v|\) gives idempotence, while taking the adjoint gives Hermiticity. The trace of \(|v\rangle \! \langle v|\) is \(\langle v,v\rangle \), and

Equality \(P_p=P_q\) implies that a representative of one ray is a nonzero scalar multiple of a representative of the other. Finally, matrix multiplication sends \(|v\rangle \! \langle v|\) to \(|Uv\rangle \! \langle Uv|\), and coordinatewise conjugation exchanges the two factors of the outer product.

The matrix \(P_p\) is Hermitian. Equality of the characteristic polynomials therefore gives a unitary \(U\) such that

Taking \(q=U\cdot p\) and using \(P_{U\cdot p}=UP_pU^\dagger \) proves the claim.

Substitute the definition \(P_{[v]}=\langle v,v\rangle ^{-1}|v\rangle \! \langle v|\) and use \(\langle v,v\rangle \ne 0\).

For \(v\ne 0\), multiply the equality at \(p=[v]\) by \(\langle v,v\rangle \) to obtain \(T(|v\rangle \! \langle v|)=S(|v\rangle \! \langle v|)\). The same equality is immediate for \(v=0\). Rank-one extensionality now gives \(T=S\).

By the trace-product identity above, \(\operatorname {tp}([v],[w])=\operatorname{tr}(P_{[v]}P_{[w]})\). Moreover,

which gives the displayed quotient.

For \(d\geq 2\), choose nonzero representatives \(v,w\) of \(p,q\), and define \(A\in M_{d,2}(\mathbb {C})\) and \(B\in M_{2,d}(\mathbb {C})\) by

Define the second factor by

Then \(AB=P_p+P_q\). The characteristic-polynomial identity for rectangular products gives

Direct calculation yields

It also gives

Substitution into the quadratic characteristic polynomial of \(BA\) proves (36). In dimension one, trace one forces each pure-state matrix to equal the identity. The zero-dimensional claim follows because a projective ray would require a nonzero vector in \(\mathbb {C}^0\).

For \(d\geq 2\), apply (36) to both sums and cancel the nonzero polynomial \(X^{d-2}\). Equality of the remaining constant terms gives the result. In dimension one both traces equal one, while in dimension zero there are no rays.

The canonical linear isometry between \(\mathbb {C}^d\) and its Euclidean-space realization induces inverse maps on rays. This identification preserves transition probabilities and commutes with both the unitary action and coordinatewise conjugation. Apply projective Wigner rigidity on the Euclidean-space realization and return the two alternatives through the inverse identification.

Use the bijection between rays and the range of \(p\mapsto P_p\) to regard \(F\) as a map of projective rays. The trace identity \(\operatorname{tr}(P_pP_q)=\operatorname {tp}(p,q)\) shows that this map preserves transition probabilities. Projective Wigner rigidity gives a unitary \(U\) and either \(p\mapsto U\cdot p\) or \(p\mapsto U\cdot \overline p\). The identities \(P_{U\cdot p}=UP_pU^\dagger \) and \(P_{\overline p}=P_p^{\mathsf T}\) give the two stated formulas.

The adjoint identity shows that \(T(A)\) is Hermitian whenever \(A\) is Hermitian. Hence Theorem 1.9.8 gives

for every Hermitian \(A\). For each projective ray \(p\), recognize \(T(P_p)\) by its characteristic polynomial and choose a ray \(f(p)\) such that

For rays \(p,q\), linearity and characteristic-polynomial preservation give

Theorem 1.10.9 therefore shows that \(f\) preserves transition probabilities. Projective Wigner rigidity gives a unitary \(U\) and either

for every \(p\), or

for every \(p\). Thus \(T\) agrees on every \(P_p\) with, respectively, \(X\mapsto UXU^\dagger \) or \(X\mapsto UX^{\mathsf T}U^\dagger \). Pure-state extensionality gives the corresponding equality on all matrices.

Write \(N=M+1\). Induction on \(M\) gives \((LQ)^{M+1}=L(QL)^M Q\). Moving the first factor cyclically through the trace and reassociating the product gives \(\operatorname{tr}((QL)^{M+1})\).