The normalized Laplacian of a graph is defined as $\mathcal{L}=D^{-1/2}(D-A)D^{-1/2}$, where $A$ and $D$ are the adjacency and degree matrices of the graph. If the graph is simple (no multiple edges between two nodes) and symmetric the relation between the $\mathcal{L}$ and $A$ is unique. Am I correct? If the answer is afirmative, there should be a way of recovering the adjacency matrix of the graph given its normalized laplacian.
When the laplacian is writen explicitely for each component we obtain a set of $n(n-1)/2$ coupled equations with the same number of variables (i.e. the weights of the adjacency variables).
$$\omega_{ij}=\mathcal{L}_{ij} \sqrt{d_i} \sqrt{d_j}$$
where $d_i$ is the degree of the $i-th$ node and $i \neq j$. The diagonal part of the $A$ is full of zeros (no self-loops). The degree of each node is related to the weights by $d_i \sum_{k=1}^{N}\omega_{ik}$ therefore the non-linear coupling between equations.
For the unnormalized laplacian $L=D-A$ the recovery of $A$ is straightforward, but for the normalized version $\mathcal{L}$ it is not so simple. There should exist methods for solving non-linear systems of equations, but I can't seem to find any of them. I would appreciate it if somebody pointed me towards an available and "friendly" software that does this? Or even better, if there is an exact analytical solution to this issue.
It goes without saying that if I made a mistake at some point, corrections are more than welcomed! Graph theory is not my specialty so I am still getting used to all these problems.
Thanks in advance!