Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Notation: $\lambda_{\max}(G)$ is the largest eigenvalue of the Laplacian matrix of the graph $G$ (aka the Laplacian index of $G$).

Now suppose $G$ is a Hamiltonian graph with Hamiltonian cycle $C$.

Is there a non-trivial lower bound on $\lambda_{\max}(G)-\lambda_{\max}(G-C)$?

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.