Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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The matrix-tree theorem becomes an eigenvalue product formula

Statement

Let G be a finite simple graph on n2 vertices, and let

0=μ1μ2μn

be the eigenvalues of its Laplacian matrix. Then

τ(G)=1nj=2nμj.

If G is d-regular with adjacency eigenvalues d=λ1λ2λn, then

τ(G)=1nj=2n(dλj).

Facts & Assumptions

Given: A finite simple graph G on n2 vertices with Laplacian L and spanning-tree count τ(G).

[L1]

Every principal cofactor of L equals τ(G) (Kirchhoff's matrix-tree theorem).

[F1]

The adjacency spectrum records the eigenvalues of the adjacency matrix (Adjacency spectrum, spectral radius, and cospectral graphs).

Proof

technique · direct
1.1

Expanding the characteristic polynomial det(tIL) at t=0 shows that the coefficient of t is (1)n1i=1ndetL(i). By [L1], this is (1)n1nτ(G). On the other hand, [L2] gives det(tIL)=j=1n(tμj)=tj=2n(tμj), so the same coefficient of t is (1)n1j=2nμj. Therefore nτ(G)=j=2nμj.

L1L2algebra
2.1

If G is d-regular, then L=dIA. Applying [L3] to the polynomial p(x)=dx shows that the Laplacian eigenvalues are dλ1,,dλn. Since λ1=d, the zero Laplacian eigenvalue is dλ1, and substituting the others into step 1.1 gives τ(G)=1nj=2n(dλj).

step 1.1L3F1
3.1

Steps 1.1 and 2.1 are exactly the two displayed formulas.

step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources