Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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The Laplacian equals BBT for every oriented incidence matrix B

Statement

Let G be a finite simple graph, let B be any oriented incidence matrix of G, and let L(G) be its Laplacian matrix. Then

L(G)=BBT.

Facts & Assumptions

Given: A finite simple graph G with oriented incidence matrix B=(bij) and Laplacian matrix L=L(G).

[F1]

Each column of B has one entry 1, one entry 1, and all other entries 0 (An oriented incidence matrix of a finite simple graph).

[F2]

The Laplacian has diagonal entries equal to vertex degrees and off-diagonal entries 1 on adjacent pairs and 0 otherwise (The Laplacian matrix of a finite simple graph).

Proof

technique · direct
1.1

The (i,i) entry of BBT is jbij2. By [F1], the term bij2 is 1 exactly when the edge ej is incident with vi, and is 0 otherwise. Therefore (BBT)ii is the number of edges incident with vi, namely deg(vi)=Lii by [F2].

F1F2
1.2

For ir, the entry (BBT)ir=jbijbrj receives a nonzero contribution only from edges incident with both vi and vr. In a simple graph there is at most one such edge. If vi and vr are not adjacent, the sum is 0. If they are adjacent, the unique common edge contributes (1)(1) or (1)(1), hence 1. So (BBT)ir=Lir by [F2].

F1F2
2.1

Steps 1.1 and 1.2 show that every entry of BBT matches the corresponding entry of L(G), so the matrices are equal.

step 1.1step 1.2

Depends on

Used by

Dependency tree · two levels

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Sources