CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Two cospectral graphs need not be isomorphic

Statement refuted

If two finite simple graphs are cospectral, then they are isomorphic.

Facts & Assumptions

Given: The star $K_{1, 4}$ and the disjoint union $C_{4} ⊔ K_{1}$ .

[L1]

The graph $K_{1, 4}$ has spectrum ${2, 0, 0, 0, - 2}$ (The complete bipartite graph $K_{m, n}$ has adjacency spectrum ${\sqrt{m n}, 0^{m + n - 2}, - \sqrt{m n}}$ ).

[L2]

The graph $C_{4}$ has spectrum ${2, 0, 0, - 2}$ (The cycle graph $C_{n}$ has adjacency spectrum ${2 \cos (2 π j / n) : 0 \leq j < n}$ ).

[L3]

Isomorphic graphs have the same spectrum (The adjacency spectrum is an isomorphism invariant).

[F1]

Cospectral graphs are those with the same adjacency spectrum (Adjacency spectrum, spectral radius, and cospectral graphs).

Counterexample

technique · direct

1.1

By [L1], the star $K_{1, 4}$ has spectrum ${2, 0, 0, 0, - 2}$ . By [L2], the cycle $C_{4}$ has spectrum ${2, 0, 0, - 2}$ , so adjoining an isolated vertex contributes one more zero eigenvalue and gives the same spectrum for $C_{4} ⊔ K_{1}$ . Hence the two graphs are cospectral by [F1].

L1L2F1

2.1

The graphs are not isomorphic, because $K_{1, 4}$ is connected while $C_{4} ⊔ K_{1}$ is not. Therefore the converse of [L3] fails.

step 1.1L3

3.1

So cospectral graphs need not be isomorphic.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Steve Butler, guest notes on cospectral graphs (standard reference, not scraped)