Let \(S\) be the set of peripheral eigenvalues and let \(m=|S|\). Finiteness of \(S\) and \(1\in S\) give \(m{\gt}0\). Every element of \(S\) is a root of unity, while \(S\) is closed under products and powers. Hence \(\mu ^{-1}\) belongs to \(S\) whenever \(\mu \in S\). Multiplication by such a \(\mu \) therefore permutes \(S\). Since no element of \(S\) vanishes,

Cancelling the nonzero product gives \(\mu ^m=1\).

Let \(\gamma =\exp (2\pi i/m)\), a primitive \(m\)-th root of unity. The set \(S\) is contained in the set of all \(m\)-th roots of unity, and both sets have cardinality \(m\). Thus they are equal. Consequently every element of \(S\) is \(\gamma ^j\) for some \(0\leq j{\lt}m\), and power closure gives the reverse inclusion.

The peripheral eigenvalues form a finite subgroup of \(\mathbb {C}^\times \): they contain \(1\), are nonzero, and are closed under multiplication and inversion. Every finite subgroup of \(\mathbb {C}^\times \) is cyclic. Choosing a generator \(\gamma \) gives the asserted primitive root and enumeration.

Apply Theorem 9.2.1 for the cyclic structure. Divisibility \(m\mid D\) follows from the cyclic decomposition: \(m\) mutually orthogonal projections summing to \(\mathbb {1}\) each have equal trace by trace preservation, so \(m\, \operatorname{tr}(P_0)=D\) with \(\operatorname{tr}(P_0)\in \mathbb {N}\).

The cyclic decomposition gives orthogonal projections \(P_0,\ldots ,P_{m-1}\) with \(\sum _kP_k=\mathbb {1}\) and \(E(P_{k+1})=P_k\). Trace preservation gives \(\operatorname{tr}(P_{k+1})=\operatorname{tr}(P_k)\) for every \(k\), hence

Since the trace of an orthogonal projection is an integer, \(m\mid D\).

Normalize an eigenvector for \(\gamma \) to a unitary \(U\). Its powers remain eigenvectors, and the Fourier projections of \(U\) give the stated cyclic decomposition.

Given a peripheral unitary eigenvector \(U\) for \(\gamma \), the multiplicative-domain identity gives \(E(XU^\dagger )=XU^\dagger \) for any \(\gamma \)-eigenvector \(X\). By the scalar fixed-point lemma for irreducible unital maps, \(XU^\dagger =c\mathbb {1}\), hence \(X=cU\) and the eigenspace is one-dimensional.

The operator trace is \(\operatorname{Tr}(T)=\sum _{p,q}\operatorname{tr}(T(E_{pq})E_{qp})\), and \(\operatorname{tr}(ME_{qp})=M_{pq}\).

Write \(B=U\Lambda U^\dagger \) with \(U\) unitary and \(\Lambda \) diagonal. Then

because \(U^\dagger AU\geq 0\) and \(\Lambda _{ii}\geq 0\) for every \(i\).

Choose \(r\) with \(\rho _{\operatorname{spec}}(a){\lt}r{\lt}1\). The Gelfand formula \(\| a^n\| ^{1/n}\to \rho _{\operatorname{spec}}(a)\) gives \(\| a^n\| \leq r^n\) for all sufficiently large \(n\), and \(r^n\to 0\).

\(P\) has the rank-one form \(P(X)=f(X)\rho \) for the linear functional \(f(X)=\operatorname{tr}(X)/\operatorname{tr}(\rho )\). The operator trace of a rank-one endomorphism \(X\mapsto f(X)\rho \) equals \(f(\rho )\), and \(f(\rho )=\operatorname{tr}(\rho )/\operatorname{tr}(\rho )=1\).

Positive semidefiniteness and \(\rho \ne 0\) imply \(\operatorname{tr}(\rho )\ne 0\). Every eigenvalue of \(E-P\) has modulus strictly less than one by irreducibility and primitivity. Since \(M_{D}(\mathbb {C})\) is finite-dimensional, the maximum eigenvalue modulus equals the spectral radius.

By the Kadison–Schwarz inequality (3), the gap \(G=E(X^\dagger X)-E(X)^\dagger E(X)\) is positive semidefinite. Pairing \(G\) with the positive-definite adjoint fixed point and using \(E(X)=\mu X\) and \(|\mu |=1\) gives weighted trace zero, hence \(G=0\). Therefore \(E(X^\dagger X)=E(X)^\dagger E(X)=X^\dagger X\), so \(X^\dagger X\) is a nonzero positive semidefinite fixed point. Irreducibility upgrades it to a positive-definite, hence invertible, matrix, so \(\det (X)\ne 0\).

Take eigenvectors \(X\) and \(Y\) for \(\mu \) and \(\nu \), respectively. The Kadison–Schwarz equality at \(X\) places \(X\) in the multiplicative domain. Thus the right multiplicative identity (12) gives \(E(YX)=E(Y)E(X)=(\nu \mu )YX\). Since \(X\) and \(Y\) are invertible by Lemma 9.8.1, \(YX\ne 0\), and \(|\mu \nu |=1\).

If \(E(X)=X\), then the Hermitian matrices \(X+X^\dagger \) and \(\mathrm{i}(X-X^\dagger )\) are also fixed by \(E\). The positive-semidefinite fixed-point uniqueness theorem gives \(X+X^\dagger =a\mathbb {1}\) and \(\mathrm{i}(X-X^\dagger )=b\mathbb {1}\), hence \(X=\frac{1}{2}(a-\mathrm{i}b)\mathbb {1}\).

For an eigenvector \(X\) at a peripheral eigenvalue \(\gamma \), the Kadison–Schwarz equality gives \(E(X^\dagger X)=E(X)^\dagger E(X)=X^\dagger X\), so \(X^\dagger X\) is a nonzero positive semidefinite fixed point. Irreducible uniqueness of the positive semidefinite fixed point makes it a positive scalar \(c\mathbb {1}\); rescaling \(X\) by \(c^{-1/2}\) produces a unitary eigenvector.

The Kadison–Schwarz equality at the unitary eigenvector places \(U\) in the multiplicative domain, so the right multiplicative identity (12) gives \(E(U^n)=E(U)^n=\mu ^nU^n\) by induction on \(n\).

Given a unitary eigenvector \(U\) for a primitive \(m\)-th root \(\gamma \), Lemma 9.8.5 gives \(E(U^m)=U^m\), a fixed point that is scalar by Lemma 9.8.3: \(U^m=\alpha \mathbb {1}\). Unitarity of \(U^m\) forces \(|\alpha |=1\); multiplying \(U\) by an \(m\)-th root \(\beta \) of \(\alpha ^{-1}\) produces \((\beta U)^m=1\) while preserving the eigenvector equation.

Given a unitary \(U\) of order \(m\) with \(E(U)=\gamma U\) for a primitive \(m\)-th root \(\gamma \), the Fourier projections \(P_k:=m^{-1}\sum _{j=0}^{m-1}\overline\gamma ^{kj}U^j\) are mutually orthogonal and sum to \(\mathbb {1}\) by discrete Fourier orthogonality of the characters \(j\mapsto \gamma ^{kj}\); since \(E(U^j)=\gamma ^jU^j\) (Lemma 9.8.5), each Fourier mode is shifted by one step, giving \(E(P_{k+1})=P_k\).

The eigenvalues of an orthogonal (Hermitian idempotent) projection are \(0\) or \(1\): the eigenvalues \(\lambda \) satisfy \(\lambda ^2=\lambda \) from \(P^2=P\). The trace is the sum of the eigenvalues, which counts the eigenvalue-\(1\) multiplicities, i.e. the rank \(n\).

Induction on the exponent, using the left- and right-multiplicative domain identities on each sector projection \(P_k\), shows \(T^n(P_{k+n\bmod m}XP_{k+n\bmod m})=P_k\, T^n(X)\, P_k\); specializing to \(n=m\), where \(P_{k+m}=P_k\), gives that \(T^m\) preserves the corner \(P_kM_{D}(\mathbb {C})P_k\).

The corner projection \(P\) lies in its own corner and is fixed by the restriction, so \(1\) is a peripheral eigenvalue of the restriction. Conversely, every eigenvalue of the restriction of unit modulus arises from a nonzero corner element \(X\) with \(E(X)=\mu X\); viewing \(X\) as an element of \(M_{D}(\mathbb {C})\) exhibits \(\mu \) as a peripheral eigenvalue of \(E\). Primitivity of \(E\) forces \(\mu =1\), so the peripheral spectrum of the restriction is exactly \(\{ 1\} \) and the restriction is primitive.

Let \(Q\) be an invariant projection for the corner restriction of \(T^m\) on \(P_kM_{D}(\mathbb {C})P_k\). The orbit-sum hypothesis supplies an ambient invariant projection \(R\) for \(T\) with \(Q=0\iff R=0\) and \(Q=P_k\iff R=\mathbb {1}\). Irreducibility of \(T\) forces \(R=0\) or \(R=\mathbb {1}\), hence \(Q=0\) or \(Q=P_k\) by the two biconditionals.

Every peripheral eigenvalue of \(T\) is an \(m\)-th root of unity, so the peripheral spectrum of \(T^m\) is exactly \(\{ 1\} \); combined with \(T^m(P_k)=P_k\), Lemma 9.8.1.9 makes the corner restriction of \(T^m\) to \(P_kM_{D}(\mathbb {C})P_k\) primitive.

The same induction as in Lemma 9.8.1.8, with the cyclic shift \(k\mapsto k+1\) replaced by a permutation \(\sigma \) of the sector index set, shows that \(T^{\mathrm{ord}(\sigma )}\) preserves each corner \(P_kM_{D}(\mathbb {C})P_k\).

Peripheral eigenvalues admit unitary eigenvectors (Lemma 9.8.4). If \(U_\alpha ,U_\beta \) are unitary eigenvectors for \(\alpha ,\beta \), the multiplicative-domain identity gives \(E(U_\alpha U_\beta )=E(U_\alpha )E(U_\beta )=\alpha \beta \, U_\alpha U_\beta \), and \(U_\alpha U_\beta \neq 0\) since both factors are unitary.

For a unitary eigenvector \(U_\alpha \) with \(|\alpha |=1\), conjugating the eigenvector equation gives \(E(U_\alpha ^\dagger )=\overline\alpha \, U_\alpha ^\dagger =\alpha ^{-1}U_\alpha ^\dagger \), and \(U_\alpha ^\dagger \neq 0\).