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The Laplacian matrix of an undirected graph can be written as $M^T M$ with $M$ being the incidence matrix of the graph. This makes the (otherwise tedious) proof of Kirchhoff's theorem into a beautiful application of the Cauchy-Binet formula (and indeed, this is one of the proofs in "Proofs from The Book").

If the graph is directed, $M^T M$ does not work anymore; the diagonal of the resulting matrix contains the total degree of vertices, whereas for Kirchhoff's theorem to work, only the in-degree should appear. Can this approach still be salvaged by a slightly different definition of $M$ that eludes me, or is the "tedious" proof necessary and Cauchy-Binet simply can't be used here?

Rodrigo de Azevedo
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Gadi A
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1 Answers1

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I believe I have found a way to write the Laplacian as a product. First recall the definition in the directed-graph case: The Laplacian of $G=(V,E)$ is a matrix of size $|V|\times |V|$ such that $L_{ii}$ is the in-degree of $v_i$, and $L_{ij}$ for $i\ne j$ is minus the number of edges from $v_i$ to $v_j$.

Since the Laplacian needs not be symmetric in the directed case, it can't be written as $M^{T}M$. However, it can be written as $AB^T$ for two very similar matrices:

Define both A and B to be $n\times m$ matrices, where each row represents a vertex and each column represents an edge of $G$. Now, for each edge $e_k = v_i\to v_j$:

  1. $A_{ik} = 1, A_{jk} = -1$
  2. $B_{ik} = 0, B_{jk} = -1$

The rest of the entries are 0.

Unless I have some computational error, we have $L=AB^T$.

Gadi A
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  • $n$ vertices, $m$ edges. Matrix $M$ is $n\times m$. Matrix $M^T$ is $m\times n$. So, $M^TM$ is which a matrix $m\times n$ times $n\times m$. $M^TM$ should be $m\times m$. When you say incidence $M$, you should describe the $M_{ij}$ clearly. In wikipedia Laplacian matrix, $M$ is $m\times n$. each row represents an edge and each column represents an vertex of $G$. But you are wrong here. – Nick Dong Jun 17 '19 at 07:51
  • By the way the incidence matrix $M$($m\times n$) in wikipedia Laplacian matrix is oriented incidence matrix. I read resources now (Jun 2019), $L$ can be symmetric. – Nick Dong Jun 17 '19 at 07:58
  • The stuff about what I confused is for directed network $D-A\neq M_{ev}^TM_{ev}$, since $A$ is not symmetric, while $M_{ev}^TM_{ev}$ is symmetric. For undirected network, $M_{ev}^TM_{ev}=2(D-A)$. In my test, $D-A$ is never equal to $M_{ev}^TM_{ev}$. Here $M$ is $m\times n$ oriented incidence matrix. each row is edge, each column is vertex. – Nick Dong Jun 17 '19 at 08:02
  • What do you mean $AB^T$ for two very similar matrices. They mathematically similar or just looks similar? – Nick Dong Jun 17 '19 at 08:18