Multiplicity of 0 eigenvalue of directed graph Laplacian matrix

I am looking for a result (if it exists) for directed graphs relating the multiplicity of 0 eigenvalues of the directed Laplacian matrix.

Consider a directed graph $\mathcal{G}=(\mathcal{V},\mathcal{E})$ and define the in-degree Laplacian as $L_{in}(\mathcal{G}) = \Delta_{in}(\mathcal{G}) - A_{in}(\mathcal{G})$, where $\Delta_{in}(\mathcal{G})$ is the diagonal in-degree matrix, and $A_{in}(\mathcal{G})$ is the adjacency matrix.

A well-known result states that $L_{in}(\mathcal{G})$ has rank $|\mathcal{V}|-1$ if and only if $\mathcal{G}$ contains a rooted out-tree.

For undirected graphs, is is also known that the rank of $L(\mathcal{G})$ is $|\mathcal{V}|-c$ where $c$ is the number of connected components.

Is there a similar result for directed graphs? It is straight-forward to show, for example, that $L_{in}(\mathcal{G})$ loses rank for every node with in-degree 0 that is, if there are $p$ nodes with in-degree 0, then $\mbox{rk}[L_{in}(\mathcal{G})] \leq n-p$. However, I am not sure if this is in fact an equality.

I hope this is clear. Thanks!

I'm a few years late but I think tst conjecture is correct, according to https://arxiv.org/pdf/2002.02605.pdf (which also cites other references).

They use the notion of Reach :

Definition 2.3 :

i) Let $i\in V$. The reachable set $R(i)$ consists of all $j\in V$ such that there exists a path from $i$ to $j$

ii) A reach $R$ is a maximal reachable set

so basically the number of reach is the number of (oriented) trees needed to cover the graph (notice that the number of reaches might not be the number of strongly connected components).

The paper then claim

Theorem 4.6:

Given a digraph $G$. The algebraic and geometric multiplicity of the eigenvalue $0$ of $L$ equals the number of reaches.

The paper does contain a proof (which is actually quite nice) and detailled examples