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Cut test-vector identities for Laplacian mixing

PDF p. 19: Proof of Lemma 3.4

Statement

Let G=(V,E) be a finite undirected graph with nonempty vertex set, at most one edge between any pair of vertices, and possible self-loops. Write n=∣V∣, let A be its adjacency matrix, D the diagonal matrix of row sums of A, and L=D−A. Let I be the identity matrix, J the all-ones matrix, and d∗ a real reference parameter. Set ΔL=L−d∗I+(d∗/n)J. For subsets X,Y⊆V, put e(X,Y)=∑x∈X∑y∈YAxy. The inner product is ⟨u,v⟩=∑x∈Vu(x)‾v(x).

For X⊆V, write Xc=V∖X and let 1X be the function that is 1 on X and 0 elsewhere. Set f=∣Xc∣1X−∣X∣1Xc. Then ⟨f,f⟩=n∣X∣∣Xc∣, ⟨f,Lf⟩=n2e(X,Xc), ⟨f,Jf⟩=0, and ⟨f,ΔLf⟩=n2(e(X,Xc)−(d∗/n)∣X∣∣Xc∣).

Facts & Assumptions

[F1]

The graph adjacency matrix is real, symmetric and nonnegative, and e(X,Y) is the displayed ordered count. A loop contributes once to a row sum (Oriented edge count between sets).

[F2]

For a nonempty finite vertex set and a real symmetric nonnegative matrix, the cut-calculus proposition gives fTf=n∣X∣∣Xc∣, Jf=0, and fTLf=n2e(X,Xc) for the displayed cut vector, with L formed from the row sums. It includes empty cuts and permits diagonal entries (Ordered edge counts and centered cut vectors).

[F3]

The centered Laplacian is ΔL=L−d∗I+(d∗/n)J (Centered graph matrices and reference degree). The displayed Hermitian inner product is linear in its second argument and agrees with transpose multiplication for real vectors (Loewner order, Hermitian inner product, and operator norm).

[F4]

For any vector g, ⟨g,Jg⟩=∣∑x∈Vg(x)∣2 (The constant projection and centering decomposition).

Proof

Given: The graph, real parameter d∗ and subset X in the statement.

Goal: Prove the four displayed identities.

1.1F1F2F3F4given

By [F1], the graph adjacency matrix satisfies the hypotheses of [F2], with the same row-sum Laplacian and ordered cut count. The vector f is real, so [F2] and [F3] give ⟨f,f⟩=n∣X∣∣Xc∣ and ⟨f,Lf⟩=n2e(X,Xc). Also [F2] gives Jf=0, hence ⟨f,Jf⟩=0. Equivalently, its coordinate sum is zero and [F4] gives the same last identity.

2.1F3step 1.1givenalgebra∎

By [F3] and linearity in the second argument, ⟨f,ΔLf⟩=⟨f,Lf⟩−d∗⟨f,f⟩+(d∗/n)⟨f,Jf⟩. Substituting step 1.1 and using n>0 gives n2e(X,Xc)−d∗n∣X∣∣Xc∣=n2(e(X,Xc)−(d∗/n)∣X∣∣Xc∣), as required.

Source notes

The cut-vector calculation is attributed to the proof of Lemma 3.4 in the cited article. The proof above is a local consequence of the stated suppliers.

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