On the third largest eigenvalue of graphs

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• 作者：Mohammad Reza Oboudi^a ; ^b ; ^{mr_oboudi@yahoo.com" class="auth_mail" title="E-mail the corresponding author} ; ^{mr_oboudi@shirazu.ac.ir" class="auth_mail" title="E-mail the corresponding author}
• 关键词：05C31 ; 05C50
• 刊名：Linear Algebra and its Applications
• 出版年：2016
• 出版时间：15 August 2016
• 年：2016
• 卷：503
• 期：Complete
• 页码：164-179
• 全文大小：416 K

文摘

Let G   be a graph with eigenvalues λ₁(G)≥⋯≥λ_n(G)${\lambda }_1(G)\ge \cdots \ge {\lambda }_n(G)$. In this paper we investigate the value of λ₃(G)${\lambda }_3(G)$. We show that if the multiplicity of −1 as an eigenvalue of G   is at most n−13$n-13$, then λ₃(G)≥0${\lambda }_3(G)\ge 0$. We prove that ${\lambda }_3(G)\in \{-\sqrt{2},-1,\frac{1-\sqrt{5}}{2}\}$ or −0.59<λ₃(G)<−0.5$-0.59<{\lambda }_3(G)<-0.5$ or λ₃(G)>−0.496${\lambda }_3(G)>-0.496$. We find that ${\lambda }_3(G)=-\sqrt{2}$ if and only if G≅P₃$G\cong P_3$ and ${\lambda }_3(G)=\frac{1-\sqrt{5}}{2}$ if and only if G≅P₄$G\cong P_4$, where P_n$P_n$ is the path on n vertices. In addition we characterize the graphs whose third largest eigenvalue equals −1. We find all graphs G   with −0.59<λ₃(G)<−0.5$-0.59<{\lambda }_3(G)<-0.5$. Finally we investigate the limit points of the set $\{{\lambda }_3(G):\text{\hspace{0.17em}G\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}a\hspace{0.17em}\hspace{0.17em}graph\hspace{0.17em}\hspace{0.17em}such\hspace{0.17em}\hspace{0.17em}that}{\lambda }_3(G)<0\}$ and show that 0 and −0.5 are two limit points of this set.