The sum of the two largest eigenvalues (Solved)

Importance: Medium ✭✭
Author(s): Gernert, Dieter
Keywords: eigenvalues
spectrum
Recomm. for undergrads: no
Posted by: mdevos
on: June 6th, 2007
Solved by: Vladimir Nikiforov, Linear Combinations of Graph Eigenvalues, Electronic Journal of Linear Algebra 15 pp 329-336
Problem   Let $ G $ be a graph on $ n $ vertices and let $ \lambda_1 \ge \lambda_2 \ge \ldots \ge \lambda_n $ be the eigenvalues of $ G $. Is $ \lambda_1 + \lambda_2 \le n $?

This property does hold for all regular graphs $ G $. If $ G $ is $ k $-regular, then $ \lambda_1 = k $. Further, if we let $ \bar{G} $ denote the complement of $ G $ and let $ \bar{\lambda_1} \ge \bar{\lambda_2}, \ldots \ge \bar{\lambda_n} $ denote its eigenvalues, then $ \lambda_2 = -1 - \bar{\lambda_n} \le -1 + (n-k-1) $ (the second inequality here follows from the observation that $ \bar{G} $ is $ (n-k-1) $-regular).

Bibliography

Open Problems in Spectral Graph Theory (a list maintained by Dragan Stevanović).


* indicates original appearance(s) of problem.

Solved!

About 5 months ago, I was shown a counterexample to this conjecture by Bojan Mohar (I believe in joint work with two of his grad. students - Javad Ebrahimi and Azhvan Sheikh). However, thanks to Gordon Royle, I have just learned that this problem was resolved earlier by Vladimir Nikiforov. See Linear combinations of graph eigenvalues.

Is this really solved?

I'm confused how the paper's result (which you have posted) solves this question. I didn't read the paper, but the abstract only gave a looser bound that the sum of the first two eigenvalues is <= 2/sqrt(3) * n (and not just n).

a negative solution

The paper shows that the sum of the first two eigenvalues exceeds 1.125n - 25, so exceeds n when n is sufficiently large. Thus Nikiforov has given a negative solution to the problem. (Note: I have not checked the claims of the paper thoroughly.)