zero) and $S = \text{diag}\left(\frac{1}{\sqrt{\tau_1}}, \ldots, \frac{1}{\sqrt{\tau_N}}\right)$, the Laplacian $L$ is given by $$\begin{aligned}
  L = \mathcal{L}S^2.\end{aligned}$$ The transpose of $L$ is $$\begin{aligned}
  L^T &= S^2 \mathcal{L} = S^2 L S^{-2},\end{aligned}$$ and the matrix $$\begin{aligned}
  A &= SLS^{-1}
  \label{equ:A}\end{aligned}$$ is symmetric and hence diagonalizable over $\mathbb{R}$. This diagonalizability is inherited to $L$ and $L^T$. We will denote the eigenvectors of $L$ by ${
  \mathbf{X}
}_i$, of $A$ by ${
  \mathbf{Q}
}_i$ and of $L^T$ by ${
  \mathbf{Z}
}_i$. Since $A$ is symmetric, we can choose the ${
  \mathbf{Q}
}_i$ such that they form an orthonormal basis of $\mathbb{R}^N$ and define the eigenvectors of $L$ and $L^T$ such that for eigenvectors to the same eigenvalue $\lambda_i$ the following relation holds: $$\begin{aligned}
  {
  \mathbf{Z}
}_i &= S {
  \mathbf{Q}
}_i = S^2 {
  \mathbf{X}
}_i.\end{aligned}$$ From this follows: $$\begin{aligned}
  \delta_{ij} &= {
  \mathbf{Q}
}_i^T {
  \mathbf{Q}
}_j = (S {
  \mathbf{X}
}_i)^T (S^{-1} {
  \mathbf{Z}
}_j) = {
  \mathbf{X}
}_i^T {
  \mathbf{Z}
}_j,
  \label{equ:Orthogonality}\end{aligned}$$ where we have used the fact that $S$ is diagonal. Furthermore, there is an equilibrium state of $L$, namely the eigenvector ${
  \mathbf{X}
}_1 = (\tau_1, \tau_2, \ldots , \tau_N)^T/\sqrt{\sum_k\tau_k}$ to eigenvalue zero, corresponding to the eigenvector ${
  