A symmetric/anti-symmetric reformulation

Now set

$${\cal S} = \sum_{j=1}^{n} s_{j} \rho_{j} + \sum_{l=1}^{n-1} \sum_{j=1}^{n-l} (s_{j+l} \rho_{j} + \rho_{j+l} s_{j})$$

that is, ${\cal S}$ is the sum of products of s and $\rho$ over all instances and lags, and

$${\cal A} = \sum_{l=1}^{n-1} \sum_{j=1}^{n-l} (s_{j+l} \rho_{j} - \rho_{j+l} s_{j})$$

so that ${\cal S}$ and ${\cal A}$ are, respectively, the symmetric and anti-symmetric components of the principal diagonal of the second-order matrix. The n-layer transformation of Equation 10 is now  

$$\pmatrix{ I & 0 \cr 0 & I } - i \omega z \pmatrix{ 0 &... ...{2} z^{2}}{2} \pmatrix{ {\cal S+A} & 0 \cr 0 & {\cal S-A} }$$ (11)