CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Cayley's formula follows again from the matrix-tree theorem

Statement

For every integer $n \geq 2$ , the complete graph $K_{n}$ has exactly $n^{n - 2}$ spanning trees.

Facts & Assumptions

Given: An integer $n \geq 2$ .

[L1]

The complete graph $K_{n}$ has adjacency eigenvalues $n - 1$ and $- 1$ with multiplicity $n - 1$ (The complete graph $K_{n}$ has adjacency spectrum ${n - 1, (- 1)^{n - 1}}$ ).

[L2]

For a regular graph, the matrix-tree theorem gives the product formula $τ (G) = \frac{1}{∣ V (G) ∣} \prod_{j = 2}^{∣ V (G) ∣} (d - λ_{j})$ (The matrix-tree theorem becomes an eigenvalue product formula).

[L3]

Cayley's formula already states that $K_{n}$ has $n^{n - 2}$ spanning trees (Cayley's formula: $τ (K_{n}) = n^{n - 2}$ for $n \geq 2$ , with $τ (K_{1}) = 1$ and $τ (K_{0}) = 0$ ).

Proof

technique · direct

1.1

The graph $K_{n}$ is $(n - 1)$ -regular, so [L2] applies with $d = n - 1$ . By [L1], its nontrivial adjacency eigenvalues are all $- 1$ , hence $τ (K_{n}) = \frac{1}{n} \prod_{j = 2}^{n} ((n - 1) - (- 1)) = \frac{1}{n} \prod_{j = 2}^{n} n = n^{n - 2}$ .

L1L2algebra

2.1

This matches the earlier Prüfer-code count in [L3], so the matrix-tree theorem gives a second proof of Cayley's formula.

step 1.1L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Richard P. Stanley, MIT 18.314 handout, Example 1.11 (standard reference, not scraped)