📝 SummarXiv

Abstract

Two graphs are said to be $Q$-cospectral if they share the same signless Laplacian spectrum. A simple graph is said to be determined by its signless Laplacian spectrum (abbreviated as DQS) if there exists no other non-isomorphic simple graph with the same signless Laplacian spectrum. In this paper, we establish the following results: (1) Let$G \cong K_{1} \vee \bigl(C_{k} \cup qK_{2} \cup sK_{1}\bigr),$ with $q,s \geq 1$, $k \geq 4$, and at least $21$ vertices. If $k$ is odd, then $G$ is DQS. Moreover, if $k$ is even and $F$ is $Q$-cospectral with $G$, then $$F \cong G \quad \text{or} \quad F \cong K_{1} \vee \bigl(C_{4} \cup P_{k-3} \cup P_{3} \cup (q-2)K_{2} \cup sK_{1}\bigr).$$ (2) Let $G\cong K_1\vee (C_{k_1}\cup C_{k_2}\cup\cdots \cup C_{k_t}\cup qK_2\cup sK_1)$ with $t\ge 2$, $q,s\ge 1$, $k_i\ge 4$ and at least $33$ vertices. If each $k_i$ is odd, then $G$ is DQS. (3) The graph $K_{1} \vee \bigl(C_{3} \cup C_{k_{1}} \cup C_{k_{2}} \cup \cdots \cup C_{k_{t-1}} \cup qK_{2} \cup sK_{1}\bigr),$ with $t,q,s \geq 1$ and $k_{i} \geq 3$, is not DQS. Moreover, it is $Q$-cospectral with $K_{1} \vee \bigl(K_{1,3} \cup C_{k_{1}} \cup C_{k_{2}} \cup \cdots \cup C_{k_{t-1}} \cup qK_{2} \cup (s-1)K_{1}\bigr).$ Here $P_{n}$, $C_{n}$, $K_{n}$ and $K_{n-r,r}$ denote the path, the cycle, the complete graph and the complete bipartite graph on $n$ vertices, while $\cup$ and $\vee$ represent the disjoint union and the join of two graphs, respectively. Furthermore, the signless Laplacian spectrum of the graphs under consideration is computed explicitly.