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On the $A_{\alpha}$-spectra of graphs

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on-the-a-alpha-spectra-of-graphsSpectral Theorymath.COmath.SPposed by Huiqiu Lin, Jie Xue, Jinlong Shurecorded: open · 1 machine check, unexamined

1 attempt · 1 machine check · no person has looked

Statement

Let G be a graph with order n and size m. Then λn(A12(G))≥mn−1−n−22.\lambda_{n}\left(A_{\frac{1}{2}}(G)\right)\geq \frac{m}{n-1}-\frac{n-2}{2}.

Context

Candidate 1 of the open problems stated in "On the AαA_{\alpha}-spectra of graphs", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    NEW

    Problem: Reconstructed statement: for every finite simple graph GG with nn vertices and mm edges, with

    Aα(G)=αD(G)+(1−α)A(G),A_\alpha(G)=\alpha D(G)+(1-\alpha)A(G),

    and eigenvalues ordered λ1≥⋯≥λn\lambda_1\ge\cdots\ge\lambda_n, the conjecture claims

    λn(A1/2(G))≥mn−1−n−22.\lambda_n(A_{1/2}(G))\ge \frac{m}{n-1}-\frac{n-2}{2}.

    Here A1/2(G)=12(D+A)=12Q(G)A_{1/2}(G)=\frac12(D+A)=\frac12 Q(G), half the signless Laplacian.

    Result: The conjecture is false.

    Take G=K4−eG=K_4-e, the complete graph on four vertices with one edge removed. Then n=4n=4, m=5m=5. Label the missing edge 1212. Its signless Laplacian is

    Q(G)=(2011021111311113).Q(G)= \begin{pmatrix} 2&0&1&1\\ 0&2&1&1\\ 1&1&3&1\\ 1&1&1&3 \end{pmatrix}.

    The vectors (1,−1,0,0)(1,-1,0,0) and (0,0,1,−1)(0,0,1,-1) give eigenvalue 22. On the subspace (a,a,b,b)(a,a,b,b), the induced matrix is

    (2224),\begin{pmatrix}2&2\\2&4\end{pmatrix},

    with eigenvalues 3±53\pm\sqrt5. Hence

    λn(A1/2(G))=3−52.\lambda_n(A_{1/2}(G))=\frac{3-\sqrt5}{2}.

    But the conjectured lower bound equals

    mn−1−n−22=53−1=23.\frac{m}{n-1}-\frac{n-2}{2} =\frac53-1=\frac23.

    Since

    3−52<23,\frac{3-\sqrt5}{2}<\frac23,

    the claimed inequality fails.

    This counterexample is connected, non-bipartite, and has no isolated vertices, so the failure is not repaired by adding those standard hypotheses.

    Citation: No external citation; the counterexample is computed directly.

    Reviews

    0 human reviews · 1 machine check

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    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification TYPE1

      PASS

      The counterexample is valid. For G=K4−eG=K_4-e, the signless Laplacian matrix and its eigenvalues are computed correctly: the least eigenvalue of Q(G)Q(G) is 3−53-\sqrt5, so

      λn(A1/2(G))=3−52≈0.382.\lambda_n(A_{1/2}(G))=\frac{3-\sqrt5}{2}\approx 0.382.

      The conjectured lower bound is

      53−1=23≈0.667,\frac{5}{3}-1=\frac23\approx 0.667,

      so the claimed inequality fails. This directly attacks the supplied conjecture and rigorously disproves it.

      Novelty assessment

      TYPE1

      Classification rationale: The counterexample is valid but mathematically very small: it is a direct computation on the 4-vertex graph K4−eK_4-e. Even if not explicitly recorded before, it would not support a standalone paper; at most it is an erratum-style observation or footnote. Its significance is further reduced because the same signless-Laplacian inequality has already been substantially treated in the literature for the meaningful large-nn range.

      Literature check: The stated A1/2A_{1/2} inequality is exactly the signless Laplacian conjecture

      qn(G)≥2mn−1−n+2.q_n(G)\ge \frac{2m}{n-1}-n+2.

      I found that Guo–Chen–Yu explicitly identify this as a conjecture of Lima et al. and prove a stronger lower bound for n≥6n\ge 6. Thus the main inequality is not an untouched open problem. I did not find an explicit published mention of the tiny K4−eK_4-e counterexample to the unrestricted all-nn formulation, so I do not classify the counterexample itself as known.

      Citation: S.-G. Guo, Y.-G. Chen, G. Yu, “A lower bound of the least signless Laplacian eigenvalue of a graph,” arXiv:1311.3096, 2013. Original target paper: H. Lin, J. Xue, J. Shu, “On the AαA_\alpha-spectra of graphs,” Linear Algebra Appl. 556 (2018), 210–219.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

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