Which are the sufficient and necessary conditions for an undirected graph with no self edges (i.e. no loop of length $1$) to have an invertible adjacency matrix?
In this case, the adjacency matrix is symmetric (i.e. $A = A^\top$). Moreover, all the diagonal elements are $0$ and there is a $1$ in both the entries $(i,j)$ and $(j,i)$, with $i\neq j$, if and only if vertices $i$ and $j$ are connected.
A first necessary condition is the following: all vertices must have at least one connection, otherwise the relative row of $A$ is null. But what else?
As far as I know, there is no "simple" necessary and sufficient condition of $G$ for an invertible adjacency matrix. A few comments:
This necessary condition is not sufficient. For example, a $4$ cycle has the expansion property, but a singular adjacency matrix.
A special case of the above: If $G$ is bipartite with non-singular adjacency matrix, both halves of the bipartition have equal size.
The property is not in general what is termed "monotone" -- adding edges to an invertible graph can both turn a non-singular matrix into a singular one and vice versa (e.g. closing a path on $3$ vertices to a triangle vs. closing a path on $4$ vertices to a $4$ cycle).
It's been conjectured that the vast majority of singular adjacency matrices are caused by pairs of vertices having identical neighborhoods (these correspond to pairs of equal rows in the adjacency matrix). In other words, if we look at all graphs on $n$ vertices, then it is believed that $$\frac{|\{G \textrm{ with at least one pair of identical neighborhoods }\}|}{|\{G \textrm{ with singular adjacency matrix }\}|} \rightarrow 1$$ as $n$ goes to infinity. however, this is still an open problem. The numerator is an $O\left(n^2 2^{-n}\right)$ fraction of all $n$-vertex graphs, and the best known result is that the denominator is a $\exp(-n^c)$ fraction of graphs for some small positive $c$ (Vershynin was the first to prove a bound of this form. The current best value of $c$ is due to Ferber and Jain).