Every product \(A^{w_1}A^{w_2}\) with \(|w_1|=m\) and \(|w_2|=n\) equals \(A^{w_1w_2}\), where \(|w_1w_2|=m+n\).

The zero-length products consist only of the identity, so \(S_0(A)=\operatorname{span}_{\mathbb {C}}\{ \mathbb {1}\} \) has dimension \(1\). Since \(\dim M_{D}(\mathbb {C})=D^2\ge 4\), this span cannot equal \(M_{D}(\mathbb {C})\).

By Lemma 8.1.1.13, \(0\)-block injectivity gives \(S_0(A)=M_{D}(\mathbb {C})\). Lemma D.1.2 excludes \(D\ge 2\), so the nonzero bond dimension must be \(1\).

Choose a nonzero entry \(C_{pq}\). Then every matrix unit \(E_{ij}\) is a scalar multiple of \(E_{ip}CE_{qj}\), so the span in (??) contains the standard matrix basis.

By \(L\)-block injectivity, the length-\(L\) word products span \(M_{D}(\mathbb {C})\). Bilinearity of \((R,S)\mapsto RXS\) reduces the claim to Lemma D.1.4, with \(R\) and \(S\) chosen among those word products.

If \(P\) were a nontrivial invariant orthogonal projection, then its range would be invariant under every letter of \(A\). Irreducibility of the action forces this range to be either \(0\) or all of \(\mathbb {C}^D\), so \(P\) must be \(0\) or \(\mathbb {1}\), a contradiction.

Every element of \(\operatorname{alg}(A)\) lies in some cumulative span \(T_n(A)\): the generators lie in \(T_1(A)\), scalar multiples of the identity lie in \(T_0(A)\), and the family \(\{ T_n(A)\} _{n\ge 0}\) is closed under addition and under multiplication with lengths adding. Since \(M_{D}(\mathbb {C})\) is finite-dimensional, the ascending chain \(T_0(A)\subseteq T_1(A)\subseteq \cdots \) stabilizes, so one can choose a single \(N\) with \(T_N(A)=M_{D}(\mathbb {C})\).

Nilpotency on the zero generalized eigenspace follows from the definition of generalized eigenspaces. Invertibility on nonzero generalized eigenspaces follows because \(f-\mu \) is nilpotent there, so \(f=\mu (1-(1-f/\mu ))\) is invertible. Spanning and independence are standard results for generalized eigenspaces over algebraically closed fields. In  [ SPGWC10 ] and [ Wol12 , Lemma 6.3(b) ] , the Jordan normal form is used directly. The argument is phrased in terms of generalized eigenspaces instead.

A nilpotent endomorphism on an \(n\)-dimensional space has nilpotency index at most \(n\).

By Theorem 8.1.1.12, if \(T_n\neq T_{n+1}\), then \(\dim T_{n+1}{\gt}\dim T_n\). Since \(\dim T_n\le D^2\) by Lemma 8.1.1.10, after at most \(D^2\) steps either \(T_n=M_{D}(\mathbb {C})\) or \(T_n\) has stabilised. Stabilisation before reaching \(M_{D}(\mathbb {C})\) contradicts normality, which requires \(S_{L_0}=M_{D}(\mathbb {C})\) for some \(L_0\) and hence \(T_{L_0}=M_{D}(\mathbb {C})\).

By Theorem 8.2.1, \(\mathbb {1}\in T_{D^2}(A)\), so \(\mathbb {1}\) is a linear combination of word products of length at most \(D^2\). At least one has nonzero trace because \(\operatorname{tr}(\mathbb {1})=D\neq 0\).

The trace is the sum of the eigenvalues of \(M\), counted with multiplicity, so a nonzero trace gives a nonzero eigenvalue \(\mu \). Choosing a nonzero vector \(\varphi \in \ker (M-\mu \mathbb {1})\) gives \(M\varphi =\mu \varphi \).

By Theorem D.2.8, there is a word \(w_0\) with \(\operatorname{tr}(A^{w_0})\neq 0\) and \(|w_0|\le D^2\). By Theorem D.2.9, \(A^{w_0}\) has a nonzero eigenvector.

The argument is the same as for Theorem 8.1.1.11: left multiplication by \(A^i\) cannot escape the stabilised span.

Every matrix \(M\in M_{D}(\mathbb {C})\) lies in \(T_N(A)\), so in particular every rank-one matrix sending \(\varphi \) to any target vector \(v\) lies in \(T_N(A)\). Applying such matrices to \(\varphi \) recovers all of \(\mathbb {C}^D\).

This is the length-one word with letter \(i\).

The new generator \(X\) is already in \(S_1(A)\), and each original generator remains in the augmented family. Hence the one-step spans agree. Multiplying word spans and arguing by induction on the word length gives equality at every length.

Since all exact word spans are unchanged by Lemma D.3.3, eventual full word span for \(A\) gives eventual full word span for \(\widetilde A\). The equality of Kraus ranks is the equality of the dimensions of the common one-step span.

The dimensions of the exact word spans cannot decrease, because left multiplication by the invertible Kraus operator is injective. If this dimension growth stopped before the full matrix algebra was reached, the same argument at every later length would contradict normality. Since \(\dim S_1(A)=\operatorname{kr}(A)\) and \(\dim M_{D}(\mathbb {C})=D^2\), the full algebra is reached by length \(D^2-\operatorname{kr}(A)+1\). The index bound follows from the definition of \(\iota (A)\).

Apply the one-step case to \(X=A^{i_0}\), which belongs to \(S_1(A)\) by Lemma D.3.2.

Apply the one-step case to \(X=A^{i_0}\), which belongs to \(S_1(A)\) by Lemma D.3.2.

Choose \(N\) with \(S_N(A)=M_{D}(\mathbb {C})\). Then \(\mathbb {1}\in S_N(A)\). For any \(M\in S_N(A)\), write \(\mathbb {1}=\sum _r c_rA^{w_r}\) with each \(|w_r|=N\). Then

By Lemma D.1.1, iterating (??) gives \(S_N(A)\subseteq S_{kN}(A)\) for every \(k\ge 1\). Taking \(k=L\) yields \(M_{D}(\mathbb {C})=S_N(A)\subseteq S_{NL}(A)\). By Lemma D.4.2, \(S_{NL}(A)\subseteq S_N(A^{[L]})\), so \(S_N(A^{[L]})=M_{D}(\mathbb {C})\).

Every length-\(nL\) word in the original alphabet can be split into \(n\) consecutive chunks of length \(L\), each of which is a single blocked Kraus operator by (??). The claim follows by induction on \(n\), using Lemma D.1.1.

By (??), \((A^{[L]})^{j_0}=A^{w_0}\).

Set \(B=A^{[L]}\), which is normal by Theorem D.4.1. The word eigenvectors transfer to single-index eigenvectors of \(B\) and \(B^T\) by Theorem D.4.3. Eigenvector spreading and padding then give \(H_{D-1}(B,\varphi )=H_{D-1}(B^T,\psi )=\mathbb {C}^D\).

For each standard basis vector \(e_j\), choose \(R_j\in S_{D-1}(B^T)\) with \(R_j\psi =e_j\). Reversing each word after transposition shows that \(R_j^T\in S_{D-1}(B)\), and therefore \((\varphi \psi ^T)R_j^T=\varphi e_j^T\) belongs to \(S_{m+D-1}(B)\). Theorem 8.5.2, applied with \(n=D-1\) and \(m'=m+D-1\), now gives \(S_{(D-1)+m+(D-1)}(B)=M_{D}(\mathbb {C})\). Finally, Lemma 8.5.3 transfers this identity to \(S_{((D-1)+m+(D-1))L}(A)=M_{D}(\mathbb {C})\).

Set \(B=A^{[N_0]}\). By (??) and \(S_{N_0}(A)=M_{D}(\mathbb {C})\), one has \(S_1(B)=M_{D}(\mathbb {C})\). Hence \(\mathbb {1}\in S_1(B)\), so not all blocked generators can have trace zero. Choose \(j_0\) with \(\operatorname{tr}(B^{j_0})\neq 0\). By Theorem D.2.9, \(B^{j_0}\) has a nonzero eigenvector \(\varphi \) with eigenvalue \(\mu \neq 0\). Writing \(j_0\) as the blocked index corresponding to a word \(\sigma _0\) of length \(N_0\) gives \(A^{\sigma _0}\varphi =\mu \varphi \). Applying the same argument to the transpose blocked tensor gives a word \(\tau _0\) of length \(N_0\), a nonzero vector \(\psi \), and \(\nu \neq 0\) with \((A^{\tau _0})^T\psi =\nu \psi \). Since \(A\) is normal and \(N_0\ge 1\), the blocked tensor \(B\) is normal by Theorem D.4.1. Therefore there exists \(M\) with \(S_M(B)=M_{D}(\mathbb {C})\). In particular, \(\varphi \psi ^T\in S_M(B)\). Taking \(m=M\) proves (3).

By normality, some \(N_0{\gt}0\) has \(S_{N_0}(A)=M_{D}(\mathbb {C})\). Theorem D.4.5 produces eigenvectors and a rank-one element \(\varphi \psi ^T\in S_m(A^{[N_0]})\). Theorem D.4.4 then gives \(S_{((D-1)+m+(D-1))N_0}(A)=M_{D}(\mathbb {C})\).

The claim follows from the self-overlap convergence under a complementary transfer-map gap (Theorem 7.11.2).

Irreducible Perron–Frobenius theory supplies a nonzero positive fixed point. Peripheral primitivity gives a strict spectral-radius bound on the complementary transfer map. Thus \(A\) is a primitive MPS tensor, and Theorem D.5.1 applies.

The PSD fixed point \(\rho \) of \(\mathcal{E}_A\) satisfies \(\mathcal{E}_A^n(\rho )=\rho \neq 0\). Since \(\mathcal{E}_A^n(\rho )=\sum _{|w|=n}A^w\rho (A^w)^\dagger \), at least one summand is nonzero, hence some \(A^w\neq 0\).

By Lemma D.5.3, there exists a word \(w\) of length \(n\) with \(A^w\neq 0\). The expansion \(\mathcal{E}_A^n(\mathbb {1})=\sum _{|u|=n}A^u(A^u)^\dagger \) contains the nonzero positive semidefinite summand \(A^w(A^w)^\dagger \), so \(\mathcal{E}_A^n(\mathbb {1})\neq 0\).

Set \(\sigma '=\sigma -\frac{\operatorname{tr}(\sigma )}{\operatorname{tr}(\rho )}\rho \). Then \(\operatorname{tr}(\sigma ')=0\) and \((\mathcal{E}_A-P_\rho )(\sigma ')=\sigma '\). If \(\sigma '\neq 0\), then \(1\) is an eigenvalue of \(\mathcal{E}_A-P_\rho \), contradicting the complementary transfer-map gap \(\rho _{\operatorname{spec}}(\mathcal{E}_A-P_\rho ){\lt}1\).

The spectral radius of \(\mathcal{E}_A-P_\rho \) is strictly less than \(1\) by the complementary transfer-map gap hypothesis. Apply the general result that powers of an operator with spectral radius \({\lt}1\) tend to zero.

By Lemma D.5.5, if \(\sigma \ge 0\) and \(\mathcal{E}_A(\sigma )=\sigma \), then \(\sigma =\frac{\operatorname{tr}(\sigma )}{\operatorname{tr}(\rho )}\rho \). Since \(\rho {\gt}0\), Wolf’s uniqueness criterion for irreducibility applies.

For any \(M\in S_n(A)\), one has \(M=M\mathbb {1}\in S_n(A)S_1(A)\subseteq S_{n+1}(A)\) by Lemma D.1.1.

By Theorem D.6.3, the word spans form the chain \(S_0(A)\le S_1(A)\le \cdots \le S_n(A)\). Therefore their sum \(T_n(A)=S_0(A)+\cdots +S_n(A)\) equals the largest term \(S_n(A)\).

Theorem D.6.4 gives \(S_N(A)=T_N(A)=M_{D}(\mathbb {C})\).