Suppose we have a directed graph $G=(V, E)$ with $N=|V|$ nodes. Define normalized graph laplacian as $L=I-AD^{-1}$ where $A$ is a adjacency matrix of $G$ and $D$ is a degree matrix of $G$.
I'm interested in solving $Lx=b$ with nonsymmetric LAMG method, called NS-LAMG. NS-LAMG requires right null-space vector $v$ by solving $Lv=0$. However, there exist a inverse of $L$ if $L$ is a full-ranked, which means $v=0$. Therefore, it's impossible to solve the full-rank graph laplacian with NS-LAMG. So I was wondering, what does it mean that a laplacian of directed graph has full rank?
[EDIT]: suppose we have graph $G$ like below,
so we have $$ A =\left [ \begin{matrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ \end{matrix} \right ], D =\left [ \begin{matrix} 3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{matrix} \right ] $$ then $$ L = I - AD^{-1} = \left [ \begin{matrix} 1 & -\frac{1}{3} & -\frac{1}{2} & 0 \\ 0 & 1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & 0 \\ -\frac{1}{3} & -\frac{1}{3} & 0 & 1 \\ \end{matrix} \right ] $$ and $rank(L)=4$
The rank of the (normalized) Laplacian is never full, because the vector $(1,...,1)$ is always an eigenvector associated with eigenvalue $0$ (which however solves your initial problem). See e.g. wikipedia.
For an interpretation of the multiplicity of eigenvalue $0$ you can look at this question