If \(v\) is a nonzero vector in the first block of coordinates, extend it by zero on the second block. Positive definiteness of \(Q\) gives \(v^TQ_{++}v>0\), so \(Q_{++}\) is positive definite. If \(v\) is a nonzero vector supported on the minus coordinates, extend it by zero on the plus coordinates. Positive definiteness of \(C\) gives \(v^TC_{{-}{-}}v>0\), so \(C_{{-}{-}}\) is positive definite. The off-diagonal block of \(QC=I\) is \[Q_{++}C_{+-}+Q_{+-}C_{{-}{-}}=0.\] Hence \[C_{+-}=-Q_{++}^{-1}Q_{+-}C_{{-}{-}}.\] Left multiplication by the invertible matrix \(Q_{++}^{-1}\) and right multiplication by the invertible matrix \(C_{{-}{-}}\) preserve rank, so the two blocks have the same rank.

For the cycle, the positive half is \(\{1,\ldots ,M\}\) and its image under \(\theta _\ell (t)=1-t\) is the complementary half, with \(M\ge 2\) and length \(2M\). The quadratic form of the massive scalar law has off-diagonal entry \(-1\) on each nearest-neighbor edge and no other off-diagonal entry. The only edges with one end in each half are \(\{0,1\}\) and \(\{M,M+1\}\). Their rows in the positive half are the distinct sites \(1\) and \(M\), and their columns in the negative half are the distinct sites \(0\) and \(M+1\). The crossing block therefore has rank \(2\). The covariance crossing block has the same rank. For real coordinate functions \(x_a\) on the positive half, \(\Theta x_a=x_{\theta (a)}\), so the reflection Gram matrix is that crossing block of \(C\) with columns ordered by \(\theta \). Its rank is \(2\). Every positive-half polynomial lies in the reflection-positive domain of the massive scalar law, so the Gram matrix is positive semidefinite.

Let \(\mu \) bound the absolute values of the entries of \(H\). By induction, the entries of \(H^k\) are at most \(n^{k-1}\mu ^k\). The exponential series is therefore dominated by a scalar exponential and converges entrywise. Since \(H\) commutes with \(-H\), the Cauchy product of the series for \(e^{\beta H}\) and \(e^{-\beta H}\) is the series for \(e^{0}\), which is the identity. The coefficients are real and \(H^*=H\), so \(e^{-\beta H}\) is Hermitian. The same product of series gives \(e^{-\beta H}=e^{-\beta H/2}e^{-\beta H/2}\). For a column \(v\), the squared length \(v^*e^{-\beta H}v\) vanishes only if \(e^{-\beta H/2}v=0\), hence only if \(v=0\). The exponential is positive definite, so its trace \(Z\) is positive. For finite matrices \(\operatorname {Tr}(XY)=\operatorname {Tr}(YX)\). Cycling the first factor gives \[\operatorname {Tr}(e^{-\beta H}Be^{-\beta H}Ae^{\beta H}) =\operatorname {Tr}(Be^{-\beta H}A),\] and cycling \(B\) gives \(\operatorname {Tr}(e^{-\beta H}AB)\). Dividing by \(Z\) is the identity for \(\omega \).