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I know that for a simple undirected graph $\mathcal{G}(V,E) $ the Laplacian matrix $L$ is defined as: $$ L:=D-A$$ where $D$ is the degree diagonal matrix and $A$ is the adjacency matrix of $\mathcal{G}$.

My question is: how can I define the Laplacian for a directed weighted graph $\mathcal{G}(V,W)$ ?

Leonardo Massai
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1 Answers1

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First let me give another common way of computing the Laplacian of an undirected graph $G=(V, E)$ that generalizes more easily to the directed weighted graphs you are interested in.

Suppose $V = (v_1, \dots, v_n)$ and $E = \{1, \dots, m\}$ and fix an arbitrary orientation on the edges. Consider the vertex space $\mathbb{R}^V$ with standard basis $\{\mathbf{e}_1, \dots, \mathbf{e}_n\}$ and, for each edge $i \in \{1,\dots,m\}$, let $\mathbf{n}_i = \mathbf{e}_j - \mathbf{e}_k$. Then the $n \times m$ matrix $N$ whose columns are the $\mathbf{n}_i$ ($i\in[m]$) is called the signed vertex-edge incidence matrix of $G$ (with respect to the fixed orientation).

The key fact is that the Laplacian $L$ of the $G$ is the (transpose of the) Gram matrix of $N$, that is, $L = NN^{\top}$. In particular, the matrix product $NN^{\top}$ does not depend on the orientation we fixed.

Now let $G$ be a weighted directed graph and let $W$ be the $m \times m$ diagonal matrix with entries $W_{ij}$ equal to the weight of edge $i$ if $i=j$ and zero else. Then we can define the Laplacian of $G$ as the matrix product $NWN^{\top}$ where $N$ is the signed vertex-edge incidence matrix of the underlying unweighted graph of $G$. And this definition generalizes the case for unweighted undirected graphs. See Section 3 of this paper for more on the subject.

Nick Dong
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Aaron Dall
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    If $N$ is $n\times m$, then actually $L=NN^\top$ rather than the other way around. Despite this detail, very useful answer. – Jonas Greitemann Feb 08 '19 at 13:25
  • Thanks for the catch, @Jonas. I edited the answer accordingly. – Aaron Dall Feb 08 '19 at 23:23
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    This might be not actual any longer, but still... In your formulation, $G$ is a symmetric matrix, while this should not be the case for directed graphs. What would be, e.g., the difference between G for a graph 1->2->1 with weights w(1,2)=2 and w(2,1)=1 and the same graph with weights w(1,2)=3 and w(2,1)=2? – Dmitry Jan 26 '21 at 11:24
  • @Dmitry: $G$ is a graph and not a matrix. The Laplacian $L$ of $G$ is always symmetric in this set up and I don't see why this "should not be the case for the directed graphs". It's true that the adjacency matrix $A$ of a weighted directed graph need not be symmetric but that matrix doesn't appear in the given definition of $L$. – Aaron Dall May 21 '21 at 04:31
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    @AaronDall the last comment means then that the definition of the laplacian as $L = D-A$ and $L=NWN^\top$ do not coincide (i.e. it is not the same matrix?), right? With the first definition, the Laplacian is not symmetric, whereas with the second defintion, it is always symmetric. – Trb2 Jul 19 '23 at 11:39