\mathbf{Z}
}_1 = (1, \ldots , 1)^T/\sqrt{\sum_k\tau_k}$ of $L^T$. All other eigenvalues are strictly positive since $A = S^T\mathcal{L}S$ with $S$ being invertible and $\mathcal{L}$ being (symmetric) positive-semidefinite with exactly one vanishing eigenvalue. The eigenvalues can be ordered such that $$\begin{aligned}
  0 = \lambda_1 < \lambda_2 \leq \lambda_3 \ldots.\end{aligned}$$

Participation ratio {#sec:Y2Eig}
===================

We want to express the temporal evolution of $Y_2 = {
  \mathbf{p}
}^2$ in terms of eigenvectors and eigenvalues of $L$ and $L^T$. It is easily verified that $$\begin{aligned}
  {
  \mathbf{p}
} = \sum_i \left( {
  \mathbf{p}
}_0^T {
  \mathbf{Z}
}_i \right) {
  \mathbf{X}
}_i e^{-\lambda_i t},\end{aligned}$$ and thus $$\begin{aligned}
   {
  \mathbf{p}
}^T {
  \mathbf{p}
}  &= \sum_{ij} \left( {
  \mathbf{Z}
}_i^T {
  \mathbf{p}
}_0 {
  \mathbf{p}
}_0^T {
  \mathbf{Z}
}_j \right) \left( {
  \mathbf{X}
}_i^T {
  \mathbf{X}
}_j\right) e^{-(\lambda_i + \lambda_j)t}.
   \label{equ:pTpInitial}\end{aligned}$$ If we average over all initial conditions ${
  \mathbf{p}
}_0$ fulfilling ${
  \mathbf{p}
}_0^2 = 1$, then $$\begin{aligned}
   \langle {
  \mathbf{p}
}_0 {
  \mathbf{p}
}_0^T\rangle_{p_0} &= \frac{\mathbb{I}_N}{N},
   \label{equ:p0average}\end{aligned}$$ with $\mathbb{I}_N$ the $N \times N$ identity matrix. Note, that we have not yet applied any disorder average. Averaging Eq. (\[equ:pTpInitial\]) over the initial conditions and applying Eq. (\[equ:p0average\]) we obtain $$\begin{aligned}
   \langle Y_2\rangle_{p_0} &= \frac{1}{N} 