# On the third largest eigenvalue of graphs

On the third largest eigenvalue of graphs Let G be a graph with eigenvalues λ1(G)≥⋯≥λn(G). In this paper we investigate the value of λ3(G). We show that if the multiplicity of −1 as an eigenvalue of G is at most n−13, then λ3(G)≥0. We prove that λ3(G)∈{−2,−1,1−52} or −0.59<λ3(G)<−0.5 or λ3(G)>−0.496. We find that λ3(G)=−2 if and only if G≅P3 and λ3(G)=1−52 if and only if G≅P4, where Pn 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. Finally we investigate the limit points of the set {λ3(G): G  is  a  graph  such  thatλ3(G)<0} and show that 0 and −0.5 are two limit points of this set. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Linear Algebra and its Applications Elsevier

# On the third largest eigenvalue of graphs

, Volume 503 – Aug 15, 2016
16 pages

/lp/elsevier/on-the-third-largest-eigenvalue-of-graphs-97glyqaIRU
Publisher
Elsevier
ISSN
0024-3795
eISSN
1873-1856
D.O.I.
10.1016/j.laa.2016.03.037
Publisher site
See Article on Publisher Site

### Abstract

Let G be a graph with eigenvalues λ1(G)≥⋯≥λn(G). In this paper we investigate the value of λ3(G). We show that if the multiplicity of −1 as an eigenvalue of G is at most n−13, then λ3(G)≥0. We prove that λ3(G)∈{−2,−1,1−52} or −0.59<λ3(G)<−0.5 or λ3(G)>−0.496. We find that λ3(G)=−2 if and only if G≅P3 and λ3(G)=1−52 if and only if G≅P4, where Pn 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. Finally we investigate the limit points of the set {λ3(G): G  is  a  graph  such  thatλ3(G)<0} and show that 0 and −0.5 are two limit points of this set.

### Journal

Linear Algebra and its ApplicationsElsevier

Published: Aug 15, 2016

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