TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-28

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The cycle graph $C_{n}$ has adjacency spectrum ${2 \cos (2 π j / n) : 0 \leq j < n}$

Statement

For every integer $n \geq 3$ , the cycle graph $C_{n}$ has adjacency spectrum

${2 \cos (2 π j / n) : 0 \leq j < n} .$

Facts & Assumptions

Given: An integer $n \geq 3$ and the cycle graph $C_{n}$ .

[F1]

The graph $C_{n}$ has vertices $0, \dots, n - 1$ and edges between consecutive residues modulo $n$ (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

[F2]

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

Proof

technique · direct

1.1F1algebra

Let $ω = e^{2 π i / n}$ . For each $0 \leq j < n$ , define the vector $x^{(j)} = (1, ω^{j}, ω^{2 j}, \dots, ω^{(n - 1) j})^{T}$ . If $A$ is the adjacency matrix of $C_{n}$ , then [F1] gives $(A x^{(j)})_{r} = x_{r - 1}^{(j)} + x_{r + 1}^{(j)} = ω^{j r} (ω^{- j} + ω^{j}) = 2 \cos (2 π j / n) x_{r}^{(j)}$ , with indices modulo $n$ . So $x^{(j)}$ is an eigenvector with eigenvalue $2 \cos (2 π j / n)$ .

2.1step 1.1F2algebra∎

The vectors $x^{(0)}, \dots, x^{(n - 1)}$ are linearly independent: they are the columns of a Vandermonde matrix built from the distinct numbers $1, ω, \dots, ω^{n - 1}$ . Therefore step 1.1 already lists $n$ eigenvectors of the $n \times n$ adjacency matrix, so it lists all eigenvalues with multiplicity. By [F2], this is the spectrum of $C_{n}$ .

Depends on

Used by

Two cospectral graphs need not be isomorphic Counterexample
The cycle C₄ has adjacency spectrum {2,0,0,-2} Example

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Sources

Steve Butler, Spectral Graph Theory course notes, lecture 3 (standard reference, not scraped)

The cycle graph CnCn​ has adjacency spectrum {2cos⁡(2πj/n):0≤j<n}{2cos(2πj/n):0≤j<n}