Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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The complete graph Kn has adjacency spectrum {n−1,(−1)n−1}

Statement

For every integer n≥1, the complete graph Kn has adjacency spectrum

{n−1,(−1)n−1},

that is, the eigenvalue n−1 once and the eigenvalue −1 with multiplicity n−1.

Facts & Assumptions

Given: An integer n≥1 and the complete graph Kn.

[F2]

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

Proof

technique · direct
1.1F1algebra

In the standard vertex order, the adjacency matrix of Kn is J−I, where J is the all-ones matrix, because [F1] makes every off-diagonal entry equal to 1 and every diagonal entry equal to 0. The all-ones vector 1 satisfies J1=n1, so (J−I)1=(n−1)1.

2.1step 1.1algebra

If x is orthogonal to 1, then the coordinates of x sum to 0, so Jx=0. Hence (J−I)x=−x. The subspace 1⊥ has dimension n−1, so −1 is an eigenvalue with multiplicity at least n−1; together with step 1.1 this accounts for all n dimensions.

3.1step 1.1step 2.1F2∎

Therefore the eigenvalues of the adjacency matrix are exactly n−1 and −1 with the stated multiplicities, which is the spectrum by [F2].

Depends on

Used by

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