As long as \(C_r(L,V)\) is nonzero, nilpotency prevents two consecutive terms from being equal: equality would make every later term equal, in contradiction with eventual vanishing. Thus each nonzero step strictly lowers dimension, and at most \(\dim _K(V)\) such steps are possible.

Regard \(A\) as a Lie algebra under the commutator bracket. Engel’s theorem makes \(V\) a nilpotent Lie module, so the dimension bound gives

For every nonnegative integer \(r\), every ordered family \(a_1,\ldots ,a_r\) in \(A\), and every vector \(v\) in \(V\), induction on \(r\) gives

Taking \(r=\dim _K(V)\) proves that the product annihilates every vector and is therefore zero.

Transport the matrix algebra to the endomorphisms of the coordinate space through the standard algebra equivalence. Nilpotency, membership, and the ordered product are preserved, so the finite-dimensional nil-matrix theorem applies; injectivity of the equivalence returns the zero matrix.

Simultaneous reindexing preserves the identity and matrix multiplication, and it commutes with the conjugate transpose. Reindex the equation \(AA^\dagger =\mathbb {1}\).

Expand the unitary-group condition for \(U^{(N)}\).

Use the equivalent left-multiplied characterization of a unitary matrix.

For every \(N{\gt}1\), both \(U^{(N)}\) and \(V^{(N)}\) are unitary. The defining contraction of the product tensor gives the periodic operator \(U^{(N)}V^{(N)}\), which is unitary because the unitary matrices form a group under multiplication.

At system size \(N\), closing the blocked tensor gives \(U^{(NL)}\), up to the canonical bijection between \(N\) blocked configurations and \(NL\) unblocked configurations. If \(N{\gt}1\) and \(L{\gt}0\), then \(NL{\gt}1\), so \(U^{(NL)}\) is unitary. Simultaneous reindexing preserves unitarity.

The operator family is multiplicative under the local product tensor, and the physical adjoint generates the conjugate transpose without reflecting the chain.

Applying the physical adjoint twice conjugates every entry twice and exchanges the physical indices twice. Closing a physical-adjoint tensor exchanges the two physical configurations and conjugates the resulting scalar. Hence its periodic operator is the conjugate transpose. The first MPU unitarity equation for \(U^{(N)}\) is therefore the second unitarity equation for \((U^\sharp )^{(N)}\).

Expand the double layer and then apply the physical adjoint to its upper factor. The two exchanges of physical indices leave the contracted index \(j\) in the second physical position of both factors. The product-index convention keeps \((\alpha ,\gamma )\) and \((\beta ,\delta )\) in the displayed order.

Apply (12) sequentially between neighboring letters. The inserted rank-one operators split the periodic contraction into the product of the one-letter contractions \((a|W^{\sigma _x\tau _x}|b)\). Identity (11) turns this product into \(\prod _x\delta _{\sigma _x,\tau _x}\), which is the matrix element of the physical identity.

For every \(N{\gt}1\), the closed-double-layer identity and the sequential contraction give

Hence \(U^{(N)}\) is unitary.

Identity (15) holds entrywise since physical adjunction conjugates each matrix entry without reversing the virtual word; (16) follows because physical blocking commutes with the local operator product.

On one site, expanding the diagonal double-layer contraction gives the transfer matrix after the stated product reindexing. The factor \(d^{-1}\) is the square of the local factor \(d^{-1/2}\). Summing blocked diagonal letters over words and distributing the ordered product shows that normalized diagonals turn blocking into powers. Simultaneous reindexing preserves multiplication, giving the displayed formula. Transfer stabilization gives the final rank-one specialization.

Expand the \(K\)th power of the finite sum defining \(R\). Distributivity indexes the resulting terms by physical words \(\tau \), and each term is the ordered diagonal word displayed above.

The trace functional is nonzero on the identity. Extend the identity by a basis of the kernel of the trace.

The trace functional is nonzero on the identity. Extend the identity to a basis adapted to the kernel of the trace; all remaining basis vectors are therefore traceless. The bilinear pairing \((X,Y)\mapsto \operatorname{tr}(XY)\) is nondegenerate. Its dual basis gives the virtual coefficients by physical contraction. The dual vector paired with the identity is \(d^{-1}\mathbb {1}\), so its virtual coefficient is precisely \(E\). Expanding matrix units in the dual basis yields the decomposition. Subtracting the trace component gives the equivalent residual formula.

Expand each contraction and distribute the ordered product. The closed tensor coefficient is a Kronecker delta between the two physical words, so only equal words survive. The remaining independent diagonal sums factor into the product of the physical traces.

At lengths greater than one, represent identity factors by \(d^{-1}\mathbb {1}\) and residual factors by traceless physical arguments. The multilinear contraction identity then vanishes. For a single residual coefficient \(A\), repeat the same coefficient at every length \(q{\gt}1\) to obtain \(\operatorname{tr}(A^q)=0\). Thus the one-letter case is recovered after the longer words: the shifted zero-trace-power theorem makes \(A\) nilpotent, and therefore \(\operatorname{tr}(A)=0\).

Represent identity factors by \(d^{-1}\mathbb {1}\) and residual factors by their traceless physical arguments. Since \(N{\gt}1\), the multilinear contraction identity applies, and one trace factor vanishes. For a nonempty residual word, adjoin \(E\) as one extra factor; its closed trace has length at least two and therefore vanishes. Cyclicity of trace and

give \(EPE=0\). The blocked statement follows from the transfer-power formula with the same product-index reindexing in both matrix indices.

A nonempty product in the multiplicative semigroup generated by \(\mathcal S\) is a nonempty residual word, so its trace vanishes by the positive-length mixed residual identity. The generated nonunital algebra is the linear span of this semigroup. Linearity of the trace therefore gives \(\operatorname{tr}(A)=0\) for every \(A\in \mathfrak S\). Since every positive power of \(A\) remains in \(\mathfrak S\), the same argument gives \(\operatorname{tr}(A^N)=0\) for \(N{\gt}1\). The shifted zero-trace-power theorem now implies that \(A\) is nilpotent.

The algebra \(\mathfrak S\) acts on a space of dimension \(D^2\), and every element is nilpotent. The finite-dimensional nil-matrix theorem therefore annihilates every ordered product of \(D^2\) elements. Each residual generator belongs to \(\mathfrak S\), giving the stated specialization.

Expand the product as a sum over binary choices of \(E\) or \(S_k\), retaining the original order. The all-\(E\) choice is the subtracted term because \(E^q=E\) for \(q{\gt}0\). The all-residual choice vanishes by the length-\(N\) hypothesis. In every remaining word, contiguous copies of \(E\) collapse. If two resulting copies of \(E\) are separated by a nonempty residual word, the factor \(EAE\) makes the word zero. Thus each surviving word has one contiguous projector block and belongs to one of the three summands of \(\mathcal T\).

First extend \(EPE=0\) from nonempty residual words \(P\) to every element of \(\mathfrak S\) by writing the generated algebra as the linear span of the multiplicative semigroup of residual generators. For a blocked matrix unit, each local double-layer entry is

The ordered expansion theorem applies with \(N=D^2\). Its all-\(E\) coefficient is the blocked Kronecker delta and is exactly removed by the residual slice. Hence every blocked matrix-unit residual lies in \(\mathcal T\). Expanding an arbitrary physical matrix in matrix units and using linearity proves the result.

It is enough by bilinearity and linearity of trace to consider the generators \(EA\), \(AE\), and \(AEB\) of the three summands. Representative products reduce as

The remaining generator pairs follow by the same use of associativity, \(E^2=E\), and \(EAE=0\) for \(A\in \mathfrak S\).

The pairing normalization makes \(E\) idempotent. Write every blocked double-layer coefficient as \(\delta _{ij}E+R_{ij}\), where \(R_{ij}\in \mathcal T\). The trace identity for \(R_{ij}E\) gives the diagonal contraction, and \(R_{ij}R_{k\ell }=R_{ij}ER_{k\ell }\), together with \(E^2=E\), gives the rank-one insertion identity after expanding both coefficients.

Induct simultaneously along the two words. The empty pair gives the identity matrix, mismatched lengths give zero, and the nonempty case applies the defining equality to the first letters and the induction hypothesis to the tails.

Apply the word identity to every pair of configurations and take the virtual trace. This is simultaneous row and column reindexing by \(e_*\).

This follows entrywise from the definitions of physical reindexing and the double layer.

Expand the two blocked words. The canonical regrouping concatenates their \(n\) words of length \(m\) into the direct word of length \(mn\).

Let \(E^J=|\rho )(\Phi |=p\) with \(J{\gt}0\), \((\Phi |\rho )=1\) and \(E|\rho )=|\rho )\). Then \(Ep=pE=p^2=p\), so \((E-p)^N=E^N-p\) for \(N{\gt}0\) and \(E-p\) is nilpotent. Hence \(\operatorname{tr}E=\operatorname{tr}p=1\). Writing \(W^{ij}=\delta _{ij}E+R_{ij}\) with a traceless residual \(R_{ij}\) gives the one-letter case. For \(N{\gt}1\), the trace of the word is a matrix entry of \((\mathcal U^{(N)})^\dagger \mathcal U^{(N)}=\mathbb {1}\). The algebra \(\mathcal A\) is the linear span of nonempty words, and the trace of a concatenated word is the product of the traces of its two parts; bilinearity gives multiplicativity on \(\mathcal A\).

Let \(\mathcal A\) be the double-layer algebra, \(p=E^J=|\rho )(\Phi |\) the stabilized transfer power, and

Since the trace is a character on \(\mathcal A\), the set \(\mathcal N\) is a subalgebra, and \(\operatorname{tr}(n^m)=\operatorname{tr}(n)^m=0\) for every \(n\in \mathcal N\) and \(m{\gt}0\). Every element of \(\mathcal N\) is therefore nilpotent, and by the nil-matrix theorem every product of \(D^2\) elements of \(\mathcal N\) vanishes. Moreover \(p\in \mathcal A\) and \(\operatorname{tr}p=1\), so

The ordered expansion theorem, applied with \(p\) in place of the projector and \(\mathcal N\) in place of the residual algebra, writes every blocked letter of length \(D^2\) as \(\delta _{IJ}p+t_{IJ}\) with \(t_{IJ}\) in the three-form subspace of \(p\) and \(\mathcal N\). The three-form contraction identity \(t\, t'=t\, p\, t'\) and \(p^2=p\) give

MPU unitarity then gives the one-letter contraction with the same witnesses, and simplicity persists at every later blocking.

First block \(J\) sites. The normalized diagonal of this block is the reindexed rank-one matrix \(|\rho )(\mathbb {1}|\), and the trace-one condition is exactly \((\mathbb {1}|\rho )=1\). The exact second-block contraction theorem then supplies both identities after a further \(D^2\) sites. The canonical equivalence between iterated blocking and direct length-\(JD^2\) blocking transports each double-layer coefficient without changing either virtual witness.

Trace normalization gives \((\mathbb {1}|\rho )=1\), so \(|\rho )(\mathbb {1}|\) is idempotent and every positive power of it equals itself. The blocked normalized diagonal is the corresponding power of \(E\), and \(E^{JD^2}\) is the \(D^2\)-th power of \(E^J\).

Apply input-first unitarity for each fixed tail. The inner output sum is \(\delta _{p,q}\). Summing over the \(d^K\) tails and multiplying by \(d^{-K}\) leaves the same Kronecker delta.

First suppose vectors \(a,b\) satisfy the rank-one insertion identity

Set \(x_{ij}=(a|W^{ij}|b)\). Applying this identity around the constant two-site and three-site words, and then using MPU unitarity, gives

Hence \(x_{ii}=1\), while \(x_{ij}=0\) for \(i\ne j\), so the same vectors obey the one-letter contraction \((a|W^{ij}|b)=\delta _{ij}\).

For the successor step, take four words \(I,J,K,L\) of length \(k+1\) and write \(W^{I,J}=W^{i_0j_0}\cdots W^{i_kj_k}\), with the analogous notation for \(W^{K,L}\). Apply the length-\(k\) rank-one insertion identity to the suffixes \(I_+=(i_1,\ldots ,i_k)\), \(J_+=(j_1,\ldots ,j_k)\) and the prefixes \(K_-=(k_0,\ldots ,k_{k-1})\), \(L_-=(\ell _0,\ldots ,\ell _{k-1})\). Then

Thus the length-\((k+1)\) block has the same insertion witnesses, and the two- and three-site argument gives their one-letter contraction. Induction on \(k'-k\) proves the claim.

A tensor of bond dimension \(D\) is simple after blocking any \(k\geq D^2\) sites. The sum of the squares of the bond dimensions is positive and bounds every square from above, so it is a common simple blocking length for two or for three tensors.

Every tensor is simple after blocking any \(k\geq D^2\) sites, in particular after blocking \(D^4\geq D^2\) sites. For fixed blocked ket and bra words, every matrix entry of the blocked tensor is a finite sum of products of entries of the original tensor. These entries are continuous, and the finite product and sum operations preserve continuity. Applying this fixed blocking map to the original path proves the path statement.

Choose the witnesses \(a,b\) from simplicity and put \(R=|b)(a|\). Expanding the normalized diagonal as the average of the diagonal letters, the two defining simple contractions give

Indeed, insert \(R\) on both sides of each diagonal letter; its middle sandwich is \(R W^{qq}R=R\) by the first contraction, and the normalized sum contains \(d\) equal terms. The same identity makes \(W^{ij}E^N=W^{ij}E\) for every \(N{\gt}0\). Hence

Substituting the supplied equality \(E^J=|\rho )(\Phi |\) proves the claim. No uniqueness of rank-one factorizations is used.

Substitute \(R^J=\lvert \rho )(\Phi \rvert \) in each expression and use \((\Phi \rvert \rho )=1\).

Replace the rank-one insertion by \(R^J\), apply the diagonal-word expansion, and distribute the two retained letters across the normalized finite sum.

Apply the supplied two-letter contraction and then replace its rank-one insertion by the normalized diagonal-word tail.

Let \(E^J=|\rho )(\Phi |\) be a stabilized transfer power. Decompose \(W^{ij}=\delta _{ij}E+R_{ij}\). Fixedness and normalization give

The mixed residual trace identity applied to one residual factor followed by \(J\) diagonal factors gives

Since \(E^J=|\rho )(\Phi |\), these two trace identities are exactly the displayed contraction.

The first two items follow from \(1^k=1\) and \(0^k=0\) for \(k\ge 1\). The characteristic polynomial is computed from \(\chi _{\operatorname{diag}(d)}(X) = \prod _i (X - d_i)\).

Choose any index \(i_0\) (the hypothesis \(n\ge 1\) makes this possible) and let \(B\) be the rank-one diagonal idempotent from Lemma 18.4.2. For \(1\le k\le n\) we have \(\operatorname{tr}(A^k)=1=\operatorname{tr}(B^k)\). The finite-range half of the Newton–Girard the corresponding theorem in the companion quantum-channel volume  [ 216 , “Newton–Girard trace recursion” ] then yields \(\chi _A = \chi _B = X^{\, n-1}(X-1)\).

If \(n=0\) the hypothesis \(\operatorname{tr}(A)=1\) contradicts the fact that the trace of an empty matrix is \(0\); the conclusion follows vacuously. When \(n\ge 1\), the hypothesis on all positive powers implies the finite-range hypothesis of Theorem 18.4.3.

For \(r\ge 1\), one has \((A^m)^r=A^{mr}\) and \(mr{\gt}1\), so the hypothesis at exponent \(mr\) gives the result.

For every positive integer \(r\), the shifted moments give

Hence

If \(\mu \in \sigma (A)\) is nonzero, spectral mapping places \(\mu ^2\) in \(\sigma (A^2)\) and \(\mu ^3\) in \(\sigma (A^3)\), so

Conversely, spectral mapping from \(1\in \sigma (A^2)\) supplies a nonzero spectral value of \(A\), which must be \(1\).

The commuting factorization

gives, for every nonnegative integer \(k\),

Therefore every vector annihilated by a power of \(A-\mathbb {1}\) is annihilated by the corresponding power of \(A^2-\mathbb {1}\).

With \(G_1\) as above, algebraic multiplicity equals the dimension of the maximal generalized eigenspace. Hence

Applying the all-positive trace-power theorem to \(A^2\) gives

The multiplicity inequality therefore bounds \(\operatorname {mult}_{1}(\chi _A)\) by one. The set-spectrum theorem shows that \(1\) occurs, so this multiplicity is positive and hence equals one. The same set-spectrum theorem excludes every other nonzero root. Splitting the monic characteristic polynomial now yields

Choose an arbitrary basis of \(V\) and represent \(cS\) by a matrix \(A\). The shifted trace-power theorem gives

Since \(cS\) is semisimple, its generalized zero-eigenspace is its ordinary zero-eigenspace, namely its kernel. The dimension of the generalized zero-eigenspace is the root multiplicity of zero in the characteristic polynomial, so

Rank–nullity and \(c\ne 0\) therefore give \(\dim \operatorname {range}(S)=1\).

For \(k{\gt}0\), commutation gives the binomial expansion

If \(m{\gt}0\), write \(N^m=N N^{m-1}\). The first factor commutes with the remaining factors, so nilpotence of \(N\) makes the entire summand nilpotent and its trace vanishes. Only the term \(m=0\) remains, giving \(\operatorname{tr}((N+S)^k)=\operatorname{tr}(S^k)\). This is the trace reduction used in [ 221 , Proposition 20, line 3865 ] .

Commutation makes \(\operatorname {range}(S)\) invariant under \(N\). The restriction of \(N\) to this one-dimensional space is both nilpotent and a scalar endomorphism, hence zero. Therefore \(NS=0\), and commutation also gives \(SN=0\). For every positive \(k\), induction using these two identities yields

Taking \(J\) at least the nilpotency index of \(N\) removes the first summand.

The zero matrix has the same positive trace powers as \(A\) and characteristic polynomial \(X^n\). Newton–Girard therefore identifies the two characteristic polynomials. This is the characteristic-polynomial consequence of the all-positive trace condition obtained in [ 16 , proof of Proposition III.3 ] .

For \(r\geq 1\), one has \((A^m)^r=A^{mr}\) and \(mr{\gt}1\), so the hypothesis at exponent \(mr\) gives the result.

Every positive power of \(A^2\) has vanishing trace. Hence

If \(\mu \) is a spectral value of \(A\), spectral mapping shows that \(\mu ^2\) is a root of this characteristic polynomial. Therefore \(\mu ^2=0\), and hence \(\mu =0\). Thus every root of the split monic polynomial \(\chi _A\) is zero, so \(\chi _A(X)=X^n\). Cayley–Hamilton now proves that \(A\) is nilpotent.

Apply the spectral theorem to \(MM^\dagger \). Its nonzero eigenvalues are positive, and their number is \(\operatorname {rank}(MM^\dagger )=\operatorname {rank}(M)=r\). For an orthonormal eigenvector \(e_j\) with eigenvalue \(\lambda _j{\gt}0\), put \(w_j=M^\dagger e_j\). Then \(\langle w_j,w_k\rangle =\lambda _j\delta _{jk}\). Hence the vectors \(\lambda _j^{-1/2}w_j\) are orthonormal. Expanding each column of \(M\) in the eigenbasis and discarding the zero-eigenvalue terms gives

which has the form in (99).

Physical adjunction exchanges the upper and lower physical indices and complex conjugates every entry. Under the source-cut regroupings, it also exchanges rows and columns. Hence the first cut becomes the conjugate transpose of the second, and conversely.

For every pair of physical indices \((i,j)\),

where the intermediate steps follow by definition of normalized flattening, physical adjunction, and the reality of \(d^{-1/2}\).

Theorem 7.6.4 independently preserves normality and left-canonical normalization under entrywise conjugation, and Theorem 7.6.8 preserves these properties under the physical-pair swap. Conjugating the coisometry equation gives

Conjugating the reconstruction at the swapped physical index gives

Finally, each diagonal positive matrix \(\Lambda _k\) is real, and therefore

Physical relabeling leaves this transfer map unchanged, which proves the transformed fixed-point equation.

Complex conjugation of the weights leaves the block indices and their bond dimensions unchanged, so the full-support sum is unchanged.

Apply Theorem 18.4.19 and use invariance of matrix rank under conjugate transpose.

Evaluate each cut after relabeling. In both cuts the upper physical index lies in the row and the lower physical index lies in the column, with the virtual index occupying the other product coordinate. Applying \(e\) to both physical coordinates gives the two displayed row-and-column equivalences.

Matrix rank is unchanged by bijective row and column relabelings. Apply this fact to the two source-cut identities.

Induct on the length of \(w\). For the empty word, use \(\mathbb {1}_D\otimes \mathbb {1}_E=\mathbb {1}_{DE}\) and invariance of the identity under the coordinate bijection. For a leading letter \(\pi _{d,e}(i,k)\), equation (125) and the induction hypothesis reduce the claim to

after transporting multiplication through the same coordinate bijection.

For every product word \(w\), let \((u,v)\) be its pointwise component words. The bijection between \(w\) and \((u,v)\) and Theorem 18.4.29 give

which proves (132) after taking linear spans. Pure tensors span \(M_{D}(\mathbb {C})\otimes M_{E}(\mathbb {C})\). The Kronecker linear equivalence, followed by simultaneous reindexing along \(\pi _{D,E}\), is an isomorphism onto \(M_{DE}(\mathbb {C})\); this proves the fixed-length consequence. If \(S_{N_A}(A)=M_{D}(\mathbb {C})\) and \(S_{N_B}(B)=M_{E}(\mathbb {C})\) for positive \(N_A,N_B\), then persistence at positive multiples gives

Applying the fixed-length consequence at \(N_A N_B\) proves algebraic normality of \(A\boxtimes B\).

The adjoint and multiplication laws for Kronecker products, together with the fact that simultaneous reindexing preserves adjoints, products, and finite sums, give

proving (136). Spectral normality gives positive lengths \(N_A,N_B\) at which the two homogeneous word spans are full. Theorem 18.4.30 therefore makes \(A\boxtimes B\) algebraically normal at the common length \(N_A N_B\). The algebraically normal, left-canonical tensor is CPSV normal by Theorem 7.6.3.

Each equality follows by composing the inverse product encoding at every site with the corresponding projection.

Grouping the two auxiliary indices makes every closed-word matrix a Kronecker product. The identity

then gives the stated finite-size operator after the sitewise configuration bijection.

The multiplication rule \((A\otimes B)(C\otimes D)=AC\otimes BD\) gives

Simultaneous reindexing preserves this unitarity equation.

Evaluate both sides on blocked ket and bra letters. Word multiplication of local Kronecker products gives

The two decoding identities identify \(I_1,I_2\) and \(J_1,J_2\) with the corresponding letters of \(q(I)\) and \(q(J)\).

Evaluate both source cuts entry by entry after the displayed row and column shuffles. Each entry is the product of the corresponding entries of the two source cuts.

Matrix rank is unchanged by row and column bijections and satisfies \(\operatorname {rank}(A\otimes B)=\operatorname {rank}(A) \operatorname {rank}(B)\).

The contracted physical index of the product runs over pairs \((m,n)\), and each summand factors as a product of an entry of \(\sum _m(\mathcal U^{mi})^{*}\otimes \mathcal U^{mj}\) and an entry of \(\sum _n(\mathcal V^{nk})^{*}\otimes \mathcal V^{n\ell }\).

By Lemma 18.4.41, and since \(|b\otimes b')(a\otimes a'|=|b)(a|\otimes |b')(a'|\),

The rule \((A\otimes B)(C\otimes D)=AC\otimes BD\) reduces (12) for the product to the same identity for each factor.

For the first cut, choose a compact factorization of the source matrix of \(\mathcal U^{[L]}\) through a space of dimension \(r_L\). Split each physical word of length \(L+1\) into its first letter and its remaining word of length \(L\). For suitable rectangular matrices \(A_{1,L}\) and \(B_{1,L}\), the successor source matrix factors as

The intermediate space is \(\mathbb C^d\otimes \mathbb C^{r_L}\), and hence

Interchanging the upper and lower physical indices gives matrices \(A_{2,L}\) and \(B_{2,L}\) with

This factorization passes through \(\mathbb C^d\otimes \mathbb C^{\ell _L}\), so

Induct on \(L-L_0\). The case \(L=L_0\) is equality. For the right rank, the successor bound and the induction hypothesis give

The same multiplication by \(d\) gives

The blocking bounds give

Their upper bounds have product

Since \(d{\gt}0\), the latter product is positive, so both \(r_L\) and \(\ell _L\) are positive. Hence neither upper bound can be strict, and both equalities follow.

The Kronecker product of the positive-definite matrices \(\mathbb {1}_d\) and \(\rho \) is positive definite.

The map \(V_1^\dagger \) is injective because \(V_1V_1^\dagger =\mathbb {1}_r\). Positive definiteness of \(W_\rho \) is therefore preserved by this congruence.

Since \(A\) is positive definite, its support projection is the identity and

Therefore

For the second cut, \(V_2V_2^\dagger =U_2U_2^\dagger =\mathbb {1}_\ell \) and \(D_2D_2^{-1}=\mathbb {1}_\ell \) give

From \(UU^\dagger =\mathbb {1}\) and the invertibility of \(D\), \(V^\dagger =MU^\dagger D^{-1}\), and \(V'^\dagger V'M=M\); hence \(V'^\dagger V'\, V^\dagger V=V^\dagger V\), and by symmetry the other product; both projections are Hermitian, so they are equal. Then \(x=V'V^\dagger \) is unitary with \(xV=V'\) and \(D'U'=V'M=xVM=xDU\). The compressed metric transforms as \(A'=x_1Ax_1^\dagger \), and the continuous functional calculus commutes with a unitary conjugation, so \(A'^{\pm 1/2}=x_1A^{\pm 1/2}x_1^\dagger \); the factor and gate identities follow by substitution.

The gates of any choice are those of the fixed choice multiplied by the unitaries \(x_2\otimes x_1\) and \(x_1^\dagger \otimes x_2^\dagger \); isometry and unitarity are preserved, the standard-form data transport along the gauge, and the gate relation of the fundamental theorem composes with the change-of-decomposition unitaries of each tensor.

The source-factor construction supplies \(M_1=X_1Y_1\). The second compact singular-value decomposition gives \(M_2=X_2Y_2\) directly.

Evaluate the two matrix factorizations at \(((i,\beta ),(\alpha ,j))\) and \(((\alpha ,i),(j,\beta ))\), respectively.

The source-factor construction supplies \(X_1^\dagger W_\rho X_1=\mathbb {1}_r\); evaluating at \((r,r')\) gives the displayed weighted sum. Since \(X_2=V_2^\dagger \), the coisometry identity \(V_2V_2^\dagger =\mathbb {1}_\ell \) gives \(X_2^\dagger X_2=\mathbb {1}_\ell \); evaluating at \((l,l')\) gives the second displayed entry sum.

Substitute the entry factorization \(M_2=X_2Y_2\) into the left-hand side and reorder the finite sums. The contraction of the two \(X_2\) factors is \(X_2^\dagger X_2=\mathbb {1}_\ell \), so the two source indices coincide and the remaining sum is the displayed Gram matrix of \(Y_2\).

Multiply \(M_1=X_1Y_1\) on the left by \(X_1^\dagger W_\rho \) and use \(X_1^\dagger W_\rho X_1=I_r\). Similarly, multiply \(M_2=X_2Y_2\) on the left by \(X_2^\dagger \) and use \(X_2^\dagger X_2=I_\ell \).

The source-factor construction supplies \(Y_1Z_1=\mathbb {1}_r\). For the second cut,

Expand the periodic trace through \(M_1=X_1Y_1\) and \(M_2=X_2Y_2\). Reordering the four scalar finite sums gives

Reindexing both physical configuration spaces and swapping \((l,r)\) to \((r,l)\) gives the matrix identity.

Expand \(v\) as the contraction of \(X_1\) with \(X_2\). Contracting \(X_1\) with \(Y_1\) gives the first tensor letter, while contracting \(X_2\) with \(Y_2\) gives the second. Reordering the finite sums proves the open product identity. The concrete blocked formula follows by encoding both physical pairs through \(\{ 0,\ldots ,d{-}1\} \times \{ 0,\ldots ,d{-}1\} \simeq \{ 0,\ldots ,({-}1\} d^2)\).

Expand the first tensor letter through \(M_2=X_2Y_2\) and the second through \(M_1=X_1Y_1\). The common auxiliary index contracts \(Y_2\) with \(Y_1\) to give \(u\). Summing the two remaining source indices gives the formula. The concrete blocked identity follows by decoding its physical indices into ordered pairs.

Closing the common output word gives the double-layer identity

Set \(\sigma =(p,\tau )\) and \(\sigma '=(q,\tau )\). Splitting off the two retained sites rewrites the right-hand side as

Finally,

Substitute this expansion and use linearity of the trace.

Expand the ordinary and rank-one-inserted matrix products entrywise. For the tensorized Gram entry, use \(X_1^\dagger (\mathbb {1}_d\otimes \rho )X_1=\mathbb {1}_r\) on the first cut and \(X_2^\dagger X_2=\mathbb {1}_\ell \) on the second cut, then reorder the finite sums. No ambient identity \(X_2X_2^\dagger =\mathbb {1}\) is used.

Expand the matrix product and tensor product, reorder the finite sums, and contract \(X_1Y_1\) and \(X_2Y_2\) by the two source-cut factorizations.

Multiplicativity of the Kronecker product gives

Evaluate both sides of \(Y^\dagger (v^\dagger v)Y=Y^\dagger Y\) at \((x,y)\) and expand the three matrix products. Conversely, apply the displayed equality after regrouping each pair of open source-cut indices by \(R\).

Multiplying \(Y^\dagger GY=Y^\dagger Y\) by \(Z^\dagger \) on the left and \(Z\) on the right, and using \(YZ=\mathbb {1}\), gives

The converse is immediate.

Apply right-inverse cancellation with \(G=v^\dagger v\) and \(Z=Z_1\otimes Z_2\).

The matrix \(M_1\) has \(dD\) columns. Its rank is at most its number of columns, hence at most \(dD=Dd\).

The matrix \(M_2\) has \(dD\) columns. Its rank is at most its number of columns, hence at most \(dD=Dd\).

Remove the zero irreducible blocks of \(\widetilde U\) and write

for every \(N{\gt}0\) and every word \(\sigma \) of length \(N\). At \(N=2\) the common self-overlap is one, so the retained sum is nonempty and \(D_0=\sum _kD_k{\gt}0\). For every integer \(N{\gt}1\), overlap invariance gives

Let \(J_k:\mathbb C^{D_k}\to \mathbb C^{D_0}\) be the inclusion of the \(k\)th block. Then

Thus every nonzero transfer eigenvalue of \(B_k\) is a nonzero eigenvalue of \(E_{A_0}\) and hence equals one by (231). Perron–Frobenius theory supplies \(\rho _k{\gt}0\) and \(s_k{\gt}0\) with \(\mathcal{E}_{B_k}(\rho _k)=s_k\rho _k\). Therefore \(s_k=1\), and every peripheral eigenvalue of \(B_k\) is also one. Each \(B_k\) is consequently normal.

The direct sum now has literal canonical-form data with full support. Apply the nonsingular blockwise canonical-form-II gauge. It preserves the block dimensions and all periodic vectors, so the resulting tensor \(A_1\) still satisfies

for every integer \(N{\gt}1\). The shifted traces force \(r=1\) and \(\left|\mu _1\right|=1\), and full support identifies the unique retained block with the ambient bond space, so its inclusion \(J_1\) is a unitary of \(\mathbb C^{D_{\mathrm{red}}}\):

The retained block carries a diagonal positive definite fixed matrix \(\Lambda _0\) of its own transfer map; rescale it to \(\Lambda =\operatorname{tr}(\Lambda _0)^{-1}\Lambda _0\), so that

the last equality because \(\left|\mu _1\right|=1\). Now conjugate the ambient tensor by \(J_1\) and read the block matrix in the resulting coordinates:

Since \(J_1\) is unitary this is a change of ambient coordinates: \(A_1^i=J_1A_{\mathrm{red}}^iJ_1^\dagger \), so \(A_1\) and \(A_{\mathrm{red}}\) have the same matrix product vectors at every positive length. The canonical-form-II data travel along the same factor: if \(\Phi \) is the ambient coisometry reconstructing \(A_1\) from the blocks \(B_k\) and the weights \(\mu _k\), then \(\Phi J_1\) is again a coisometry and reconstructs \(A_{\mathrm{red}}\) from those same blocks and weights,

The fixed matrix transports along the same conjugation:

and \(\rho \) is positive definite, diagonal, and of trace one because \(\Lambda \) is. This is (228).

It remains to see that the conjugated tensor is normal. Transport the trace identity to it: for every integer \(N{\gt}1\),

These shifted traces force the spectral radius of \(E_{A_{\mathrm{red}}}\) to be one and \(E_{A_{\mathrm{red}}}\) to be primitive. Together with \(r=1\) and full support, both preserved by the conjugation, they make \(A_{\mathrm{red}}\) normal.

Apply Theorem 18.5.2 and write its representative as \(A_{\mathrm{red}}\). Define

Since \(d{\gt}0\), normalization recovers \(A_{\mathrm{red}}\) exactly. For every \(N{\gt}0\) and every ket–bra pair \((\sigma ,\tau )\), the doubled-index coefficient identity gives

Hence the periodic operators agree at every positive length. For \(N{\gt}1\) the right side is unitary, so \(U_{\mathrm{red}}\) is an MPU. Its normalized doubled-index tensor is \(A_{\mathrm{red}}\), so the canonical-form-II data, the full-support identity, and the diagonal positive trace-one ambient right fixed matrix supplied by Theorem 18.5.2 assemble into a canonical-form-II presentation of \(U_{\mathrm{red}}\).

Positive blocking supplies canonical-form data with the same block dimensions, so the full-support sum is unchanged.

Physical relabeling leaves the blocks and their bond dimensions unchanged, so the full-support sum is unchanged.

If \(k\) is the block with \(D_k=D\), the total-dimension bound gives

Since every retained block dimension is positive, these equalities force \(k\) to be the only retained block. After reindexing its bond coordinates, the ambient coisometry \(U\) is square, hence unitary, and reconstruction reads

where \(B\) is the reindexed normal block and \(\mu _k\ne 0\). Thus Theorem D.3.4 transports irreducibility of \(B\) through the nonzero rescaling and unitary conjugation.

Let \(k\) be the unique block. The singleton sum and full-support identity give

Thus \(k\) has ambient bond dimension, so Theorem 18.5.6 applies.

Combine Theorem 18.5.7 with the supplied spectral-radius and primitivity properties.

Column stacking is a linear bijection and intertwines \(T\) with multiplication by \(\widehat T\). It therefore carries nonzero eigenmatrices to nonzero eigenvectors and conversely.

Replace the matrix power by the transfer matrix of the corresponding operator power, identify matrix trace with operator trace, and expand the latter over periodic words. For the ordinary self-transfer statement, specialize to \(B=A\) and use the identity \(F_{A,A}=\mathcal E_A\) from Theorem B.1.1.

Apply the two product-coordinate entry formulas after the regrouping in (256). The scalar normalizations agree because

Multiplying this identity by \(\mathcal U^{i,j}_{\alpha ,\beta } \mathcal V^{k,\ell }_{\gamma ,\delta }\) gives (257).

For physical labels \(i\) and \(k\), set

and

Regrouping retained coordinates by \(\chi \) gives the weighted-block-diagonal identity

Substitute \(A^i=V_A^\dagger \mathsf B_A^iV_A\) and \(B^k=V_B^\dagger \mathsf B_B^kV_B\). The mixed-product law for Kronecker products and the definition of \(W\) then give (265). The relation \(W W^\dagger =\mathbb {1}\) follows from \(V_AV_A^\dagger =\mathbb {1}\) and \(V_BV_B^\dagger =\mathbb {1}\). Left-canonicality and normality follow from Theorem 18.4.31.

For arbitrary matrices \(X\) and \(Y\), direct expansion of the finite Kraus sums gives

Taking \(X=\Lambda _a\) and \(Y=\Gamma _b\) proves the fixed-point equation; Kronecker products and bijective reindexing preserve positive definiteness and diagonality. The finite double-sum identity proves the supplied full-support conclusion. Finally apply the normalized-flattening identity and physical relabeling to obtain the MPO statement.

Tensor the chosen canonical blocks and their diagonal positive fixed matrices, then apply the doubled physical-coordinate relabeling. The retained dimensions multiply and their sum is the product bond dimension. Independent tensor products preserve the MPU property. The product of the two ambient fixed matrices is positive definite, diagonal, and of trace one. In the product bond coordinates, the transfer identity gives

Substitute \(U^{ij}=\sqrt d\, \widetilde U^{ij}\) and \(V^{jk}=\sqrt d\, \widetilde V^{jk}\) into the defining contraction for \(W^{ik}\), and then multiply the result by \(d^{-1/2}\).

A blocked letter is a product of \(L\) original letters. The identity

identifies its normalization with the product of the \(L\) one-site normalizations. Theorem 20.15.2 identifies the two physical indexings for every \(L\geq 0\); at \(L=0\) both blocked words are empty and both local matrix products are the identity.

Expand (282) as a sum over length-\(L\) words and sum one letter at a time, repeatedly applying the one-site left-canonical identity. For \(L=0\), the unique empty word evaluates to the identity. The equivalence (283) follows by reindexing the finite sum along \(e\).