If \(S_i(A)=M_{D}(\mathbb {C})\) and \(\varphi \neq 0\), then \(S_i(A)\varphi =\mathbb {C}^D\), proving (2)\(\Rightarrow \)(1). For (1)\(\Rightarrow \)(3), first obtain a nonzero positive fixed point \(\rho \). The spreading condition makes \(\rho \) positive definite and makes the tensor irreducible. The blocking-periodicity theorem then gives \(p{\gt}0\) such that \(\mathcal{E}_A^p\) is primitive. If \(\mathcal{E}_A(X)=\lambda X\) with \(|\lambda |=1\), then \(\mathcal{E}_A^p(X)=\lambda ^pX\); uniqueness of the peripheral eigenvalue of the primitive power gives \(\lambda ^p=1\). If \(\lambda \neq 1\), the eigenvector yields a nonzero trace-zero Hermitian matrix \(H\) fixed by \(\mathcal{E}_A^p\). Decompose \(H=Q_1-Q_2\) into positive semidefinite \(\mathcal{E}_A^p\)-fixed parts. Paper primitivity makes every nonzero positive fixed point of the power proportional to \(\rho \), while \(\operatorname{tr}(H)=0\) makes the two proportionality constants equal. Hence \(H=0\), a contradiction. Thus every peripheral eigenvalue equals \(1\), and \(\mathcal{E}_A\) is strongly irreducible.

For (3)\(\Rightarrow \)(2), write \(\mathcal{E}_A^m=P_\rho +(\mathcal{E}_A-P_\rho )^m\). The complementary powers tend to zero. If \(S_m(A)\neq M_{D}(\mathbb {C})\) for every \(m\), choose a nonzero matrix \(B\) orthogonal to \(S_m(A)\). The associated trace pairing vanishes on the left, while positive definiteness gives a lower bound \(c\lVert B\rVert ^2\) for the \(P_\rho \) term and the complementary term is \(o(\lVert B\rVert ^2)\). For large \(m\) this is a contradiction.

By Theorem 8.1.1.7, \(\iota (A)\) is a positive length with \(S_{\iota (A)}(A)=M_{D}(\mathbb {C})\). Therefore \(H_{\iota (A)}(A,\varphi )=\mathbb {C}^D\) for every \(\varphi \neq 0\), so the defining minimum for \(q(\mathcal{E}_A)\) is at most \(\iota (A)\).

The span \(T_{n+1}\) includes all words of length at most \(n+1\) and therefore contains all words of length at most \(n\).

Since \(T_n(A)\subseteq M_{D}(\mathbb {C})\) and \(\dim M_{D}(\mathbb {C})=D^2\), the result follows.

If \(T_n=T_{n+1}\), then \(S_{n+1}\subseteq T_n\). Left-multiplying any element of \(S_{n+1}\) by \(A^i\) gives an element of \(S_{n+2}\), so \(S_{n+2}\subseteq T_{n+1}=T_n\). By induction, \(S_m\subseteq T_n\) for every \(m{\gt}n\), hence \(T_m=T_n\).

The inclusion \(T_n\le T_{n+1}\) from Lemma 8.1.1.9, together with \(T_n\neq T_{n+1}\), gives strict subspace inclusion and hence strict dimension growth.

Both conditions assert that the span of all products of length \(L\) equals \(M_{D}(\mathbb {C})\).

Every word of length \(n+1\) factors into its first letter and a word of length \(n\), which proves the displayed equality by bilinearity. Write \(L=n+1\). Since \(S_L(A)=M_{D}(\mathbb {C})\), one has \(S_n(A)\le S_L(A)\). Hence

Iterating this implication gives \(S_m(A)=M_{D}(\mathbb {C})\) for every \(m\ge L\).

This is Theorem D.2.7.

If \(T_n(A)\neq T_{n+1}(A)\), then the dimension strictly increases. Since \(S_1(A)\subseteq T_1(A)\) and \(\dim S_1(A)=d'\), one has \(\dim T_1(A)\ge d'\). After at most \(D^2-d'\) additional strict-growth steps, the dimension reaches \(D^2=\dim M_{D}(\mathbb {C})\). Stabilisation before reaching \(M_{D}(\mathbb {C})\) contradicts normality. Hence \(T_{D^2-d'+1}(A)=M_{D}(\mathbb {C})\).

Theorem 8.1.1.7 gives eventual full Kraus rank, hence normality. Put \(b=D^2-\operatorname{kr}(A)+1\) and let \(V_n\) be the span of \(A^w\) with \(1\le |w|\le n\). Then \(S_1(A)\subseteq V_1\), so \(\dim V_1\ge \operatorname{kr}(A)\). If \(V_n=V_{n+1}\) for some \(n\ge 1\), then \(V_n\) is closed under left multiplication by every \(A^i\). It therefore contains every positive-length word product; normality forces \(V_n=M_{D}(\mathbb {C})\). Thus, until the full matrix algebra is reached, the dimensions increase strictly, and the bound on \(\dim M_{D}(\mathbb {C})\) gives \(V_b=M_{D}(\mathbb {C})\). Express \(\mathbb {1}\) as a linear combination of the positive-length words defining \(V_b\). Since \(\operatorname{tr}(\mathbb {1})=D\neq 0\), at least one has nonzero trace.

By Theorem 8.2.2, the identity \(\mathbb {1}\) lies in \(T_{D^2-d'+1}(A)\) and is therefore a linear combination of word products of length at most \(D^2-d'+1\). At least one such product has nonzero trace.

Define the positive-level cumulative span \(V_n=\operatorname{span}_{\mathbb {C}}\{ A^w:1\le |w|\le n\} \). The same stabilization-or-strict-growth dichotomy shows \(V_{D^2-d'+1}=M_{D}(\mathbb {C})\): if \(V_n\) stabilises and is not \(M_{D}(\mathbb {C})\), then \(V_n\) is closed under left multiplication by each \(A^i\), so \(S_{N}(A)\subseteq V_n\neq M_{D}(\mathbb {C})\) for every \(N\ge 1\), contradicting normality. Since \(V_{D^2-d'+1}=M_{D}(\mathbb {C})\), the identity \(\mathbb {1}\) lies in \(V_{D^2-d'+1}\) and is therefore a linear combination of word products of positive length. At least one such product has nonzero trace because \(\operatorname{tr}(\mathbb {1})=D\neq 0\) for \(D\ge 1\).

The vector span \(K_n\) is monotone and \(\dim K_n\le D\). If \(K_n=K_{n+1}\) for some \(n{\lt}D-1\), the span stabilises by Theorem D.2.11. But \(T_{D^2}(A)=M_{D}(\mathbb {C})\) by Theorem 8.2.1, so Lemma D.2.12 gives \(K_{D^2}(A,\varphi )=\mathbb {C}^D\), contradicting early stabilisation. Hence \(\dim K_n\) grows by at least \(1\) at each step, reaching \(D\) by step \(D-1\).

Proposition 3, in the form of Theorem 8.1.1.7, gives eventual full Kraus rank and hence normality. The cumulative spaces \(K_n(A,\varphi )\) cannot stabilise below \(\mathbb {C}^D\), because every matrix eventually lies in an exact word span. Their dimensions therefore reach \(D\) by \(n=D-1\), giving \(K_{D-1}(A,\varphi )=\mathbb {C}^D\). Finally, the eigenvector relation pads every shorter word to length \(D-1\) by a nonzero power of \(\mu \), so \(K_{D-1}(A,\varphi )=H_{D-1}(A,\varphi )\).

Adjoin \(X\) as an extra first Kraus operator, obtaining \(\widetilde A\). Theorem 8.1.1.7 gives normality of \(A\), hence normality of \(\widetilde A\); the Kraus rank and all exact word spans are unchanged. The first Kraus operator of \(\widetilde A\) is invertible, so Theorem D.3.5 gives \(S_{D^2-\operatorname{kr}(\widetilde A)+1}(\widetilde A)=M_{D}(\mathbb {C})\). Transfer this equality back to \(A\).

The word-span equality (??) gives an admissible value in the defining minimum for \(\iota (A)\).

Let \(\mathcal L(M)=A^{i_0}M\), and choose \(r\) so that \(\mathcal L^r\) annihilates the generalized zero-eigenspace of \(\mathcal L\); equivalently, \(\operatorname{range}(\mathcal L^r)\) is its stabilized nonzero-spectral part. The dimension-growth argument gives a level \(n_0\) for which

Since \(A^{i_0}\varphi =\mu \varphi \) and \(\mu \neq 0\), iterating gives \((A^{i_0})^r\varphi =\mu ^r\varphi \) and hence \(\varphi =\mu ^{-r}(A^{i_0})^r\varphi \in \operatorname{range}((A^{i_0})^r)\). Therefore, for every \(\psi \in \mathbb {C}^D\), \(|\varphi \rangle \! \langle \psi |\in S_{r+n_0}(A)\). The corresponding Fitting-index estimate gives \(r+n_0\le D^2-D+1\). Since

one has \(|\varphi \rangle \! \langle \psi | =\mu ^{-1}A^{i_0}|\varphi \rangle \! \langle \psi |\). Thus \(|\varphi \rangle \! \langle \psi |\in S_k(A)\) implies \(|\varphi \rangle \! \langle \psi |\in S_{k+1}(A)\), because left multiplication by \(A^{i_0}\) maps \(S_k(A)\) into \(S_{k+1}(A)\). Iterating from \(k=r+n_0\) and using \(r+n_0\le D^2-D+1\) gives \(|\varphi \rangle \! \langle \psi |\in S_{D^2-D+1}(A)\).

Theorem 8.1.1.7 first gives normality. Apply the Fitting decomposition to left multiplication \(\mathcal L(M)=A^{i_0}M\). Choose \(r\) so that \(\mathcal L^r\) kills its generalized zero-eigenspace. Dimension growth then gives \(n_0\) with

Since \((A^{i_0})^r\varphi =\mu ^r\varphi \) and \(\mu \neq 0\), the vector \(\varphi \) lies in the range of \((A^{i_0})^r\). Hence \(|\varphi \rangle \! \langle \psi |\in S_{r+n_0}(A)\) for every \(\psi \). The identity \(|\varphi \rangle \! \langle \psi |=\mu ^{-1}A^{i_0} |\varphi \rangle \! \langle \psi |\) pads this element to the exact length \(D^2-D+1\).

Theorem 8.1.1.7 gives normality. Adjoin \(X\) as an extra first Kraus operator. The augmented tensor remains normal and has the same exact word spans as \(A\). The eigenvector relation for \(X\) is now a one-generator eigenvector relation. The eigenvector-spreading argument gives full fixed-length vector span by length \(D-1\), and Lemma 8.4.4 gives the rank-one operators \(|\varphi \rangle \! \langle \psi |\) at length \(D^2-D+1\). Multiplying these two fixed-length spans gives \(S_{D^2}\) for the augmented tensor, hence for \(A\).

The equality \(S_{D^2}(A)=M_{D}(\mathbb {C})\) gives an admissible value in the defining minimum for \(\iota (A)\).

Theorem 8.1.1.7 gives normality. By Theorem 8.2.3, there exists a positive-length word \(w\) with \(|w|\le D^2-d'+1\) and \(\operatorname{tr}(A^w)\neq 0\), where \(d'=\operatorname{kr}(A)\). The matrix \(A^w\) has nonzero trace, hence a nonzero eigenvalue \(\mu \) and eigenvector \(\varphi \) by Theorem D.2.9. Set \(n=|w|\) and let \(B=A^{[n]}\) be the blocked tensor. Then \(B\) is normal by Theorem D.4.1, and the encoding of \(w\) as a single block index gives \(B^{i_0}=A^w\).

1. If \(A^w\) is invertible, then \(S_{D^2}(B)=M_{D}(\mathbb {C})\), whence \(S_{D^2n}(A)=M_{D}(\mathbb {C})\).

2. If \(A^w\) is non-invertible, then the eigenvector \(\varphi \) spans a proper subspace, and the rank-one extraction argument gives \(S_{D^2}(B)=M_{D}(\mathbb {C})\), whence \(S_{D^2n}(A)=M_{D}(\mathbb {C})\).

In the invertible case, the invertible Kraus operator \(B^{i_0}\) allows direct dimension growth, giving \(S_{D^2-k_B+1}(B)=M_{D}(\mathbb {C})\), where \(k_B=\operatorname{kr}(B)\). In the non-invertible case, the bound \(S_{D^2}(B)=M_{D}(\mathbb {C})\) follows from tracking indices through the rank-one extraction argument of [ SPGWC10 , Lemma 2(b) ] ; the separate blocked support construction in Theorem D.4.6 does not supply this explicit bound. In both cases, \(\iota (A)\le D^2n\le D^2(D^2-d'+1)\).

Any word \(w\) with \(|w|\le n\) can be padded to exact length \(n\) by appending \(k=n-|w|\) copies of the index \(i_0\). The eigenvector relation gives \(A^{w\cdot i_0^k}\varphi =\mu ^kA^w\varphi \), so the padded vector is a nonzero scalar multiple of the original. Hence every generator of \(K_n\) lies in \(H_n\). The reverse inclusion \(H_n\le K_n\) is immediate.

Fix a matrix unit \(E_{ij}\). By hypothesis (1), there exists \(M_i\in S_n(A)\) with \(M_i\varphi =e_i\). By hypothesis (2), \(|\varphi \rangle \! \langle e_j|\in S_m(A)\). Their product satisfies

Lemma D.1.1 therefore gives \(E_{ij}\in S_{n+m}(A)\). Since the matrix units span \(M_{D}(\mathbb {C})\), \(S_{n+m}(A)=M_{D}(\mathbb {C})\).

By the blocked-tensor definition (??), each length-\(n\) word in the blocked alphabet \(\{ 0,\ldots ,d^L-1\} \) decodes to a length-\(nL\) word in the original alphabet \(\{ 0,\ldots ,d-1\} \), so \(S_n(A^{[L]})\subseteq S_{nL}(A)\). For \(n=1\), the \(d^L\) blocked letters are indexed by the length-\(L\) words of \(A\), giving \(S_1(A^{[L]})=S_L(A)\), which proves the stated equivalence.

Normality is eventual full Kraus rank, so Theorem 8.1.1.7 gives paper primitivity. Theorem 8.4.8 then yields

Trace preservation gives a nonzero Kraus operator, hence \(\operatorname{kr}(A)\ge 1\) and the right-hand side is at most \(D^4\). Set \(L=\iota (A)\). By definition, \(L{\gt}0\) and \(S_L(A)=M_{D}(\mathbb {C})\). Lemma 8.5.3 identifies the latter equality with one-site injectivity of \(A^{[L]}\).

Choose \(L\) with \(0{\lt}L\le D^4\) and \(S_L(A)=M_{D}(\mathbb {C})\), and write \(D^4=L+k\). Induction on \(k\), using positive-length propagation at each step, gives \(S_{L+k}(A)=M_{D}(\mathbb {C})\).

Apply Theorem 8.5.4 to each block. The product of the finitely many positive lengths is a common positive multiple, and exact block injectivity persists under positive multiples.

Apply Theorem 8.5.4 to each block. Taking the product of the finitely many resulting positive blocking lengths gives a common positive multiple, and injectivity persists under positive further blocking.

By contrapositive. Assume \(A\) has an invariant projection \(P\) (\(P\neq 0\), \(P\neq \mathbb {1}\), \((\mathbb {1}-P)A^iP=0\) for all \(i\)). Set \(\sigma =P\rho P\).

1. Invariance of \(\sigma \) under \(\mathcal{E}_A^n\): the projection relation \((\mathbb {1}-P)A^iP=0\) implies \((\mathbb {1}-P)\mathcal{E}_A^n(\sigma )=0\) for all \(n\) (by induction on \(n\)).

2. Convergence: By Lemma D.5.6, \(\mathcal{E}_A^n(\sigma )\to \frac{\operatorname{tr}(\sigma )}{\operatorname{tr}(\rho )}\rho \). Hence \((\mathbb {1}-P)\frac{\operatorname{tr}(\sigma )}{\operatorname{tr}(\rho )}\rho =0\).

3. Nonzero scalar: Since \(P\neq 0\) and \(\rho \) is positive definite, \(\operatorname{tr}(P\rho P){\gt}0\), so \(\frac{\operatorname{tr}(\sigma )}{\operatorname{tr}(\rho )}\neq 0\). Therefore \((\mathbb {1}-P)\rho =0\).

4. Contradiction: Since \(\rho \) is positive definite, it is a unit in \(M_{D}(\mathbb {C})\). From \((\mathbb {1}-P)\rho =0\) we get \(\mathbb {1}-P=0\), i.e. \(P=\mathbb {1}\), contradicting \(P\neq \mathbb {1}\).

Let \(W\le \mathbb {C}^D\) be an \(A\)-invariant subspace. In finite dimension the orthogonal projection \(P\) onto \(W\) exists. Invariance of \(W\) implies \((\mathbb {1}-P)A^iP=0\) for all \(i\). Thus \(P\) is a nontrivial invariant projection unless \(W=0\) or \(W=\mathbb {C}^D\), contradicting irreducibility of the tensor.

Identify \(\operatorname{alg}(A)\) with its image under the algebra equivalence \(M_{D}(\mathbb {C})\simeq _{\mathbb {C}}\operatorname{End}_{\mathbb {C}}(\mathbb {C}^D)\). The irreducible-action hypothesis makes \(\mathbb {C}^D\) a simple module for this algebra. Over the algebraically closed field \(\mathbb {C}\), Schur’s lemma identifies the endomorphisms commuting with the action with scalars, and Jacobson density then forces the action map to be surjective onto all of \(\operatorname{End}_{\mathbb {C}}(\mathbb {C}^D)\).

Lemma D.5.7 gives irreducibility of \(\mathcal{E}_A\). The complementary-gap hypothesis gives peripheral primitivity directly: the only eigenvalue of modulus one is \(1\). Together with the positive-definite fixed point \(\rho \), these are exactly the three conditions in Definition 8.1.1.6. The complementary-gap consequences are proved in Section D.5.

By Lemma 8.7.4.1, \(A\) is strongly irreducible. Theorem 8.1.1.7 identifies strong irreducibility with eventual full Kraus rank, hence with normality.

If \(S_n(A)=M_{D}(\mathbb {C})\), then for every \(X\in M_{D}(\mathbb {C})\), substituting (??) gives

Indeed, \((A^i)^\dagger X\in S_n(A)\) because \(S_n(A)=M_{D}(\mathbb {C})\). Thus \(S_{n+1}(A)=M_{D}(\mathbb {C})\). Iterating this implication gives the result for every \(m\ge L\).

Since \(\operatorname{alg}(A)=M_{D}(\mathbb {C})\), the ascending chain condition for finite-dimensional subspaces gives an \(N\) such that \(T_N(A)=M_{D}(\mathbb {C})\). Apply Theorem D.6.5.

If \(D=1\), trace preservation gives a nonzero one-site Kraus matrix, and every nonzero scalar matrix spans \(M_{1}(\mathbb {C})\). Hence \(A\) is one-site injective, so Lemma 2.4.3 gives the positive witness \(L=1\). Suppose now that \(D\ge 2\). Theorem 9.11.1.1 gives a block-injectivity witness \(L\). If \(L=0\), Corollary D.1.3 would force \(D=1\), a contradiction. Thus \(L{\gt}0\), and the blocked-chain equivalence identifies this \(L\)-block injectivity with one-site injectivity of the blocked tensor \(A^{[L]}\).