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If so, could you provide examples and specify the conditions under which this occurs? Thank you in advance

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  • $\begingroup$ Hi Dante, this website is for research questions in mathematics, see MSE for general questions in mathematics. The adjacency matrix is (most of the time) a symmetric matrix, so its eigenvalues are all real. Note that the trace of a matrix is the sum of its eigenvalues. Since (unless you take loops into account) the trace is zero, and positive semi-definite means the eigenvalues are positive, this will [almost] never happen. I would advise you make the setup of your question clearer. $\endgroup$
    – ARG
    Commented May 21, 2020 at 20:49

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If you don't allow self-loops in the graph, then the trace is $0$. If the adjacency matrix is PSD then $0$ is the only eigenvalue. Since your adjacency matrix is symmetric, it must equal to $0$. So you essentially have the empty graph.

If you allow self-loops then you can also get a PSD adjacency matrix by adding some diagonal matrix. This corresponds to adding some self-loops in your original graph.

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    $\begingroup$ No problem. I suspect that MO might not be the right place for this question. Next time you can try math stack exchange first. $\endgroup$
    – 谁家的鸡
    Commented May 20, 2020 at 9:07

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