2
$\begingroup$

Let $G$ be any graph with at least one edge and let $e$ be any edge of $G$. Let $G-e$ denote the subgraph of $G$ obtained by deletion of the edge $e$. Assume that $G$ has $n$ vertices. Suppose that $\lambda_1(G)\geq\cdots\geq \lambda_n(G)$ and $\lambda_1(G-e)\geq\cdots\geq \lambda_n(G-e)$ are eigenvalues of $G$ and $G-e$, respectively. It is true that $\lambda_i(G)\cdot\lambda_i(G-e)\geq 0$ for all $i=1,\dots,n$?

$\endgroup$
2
$\begingroup$

I think this is false and there are counterexamples.

Here is a sage session.

sage: ed,e=([(0, 2), (0, 4), (0, 5), (1, 3), (1, 5), (2, 4), (3, 5)], (0, 2))
sage: G=Graph(ed);F=Graph(G);F.delete_edge(e)
sage: eg=G.adjacency_matrix().eigenvalues();eg.sort(reverse=1);eg
[2.414213562373095?, 1.732050807568878?, -0.4142135623730951?, -1, -1, -1.732050807568878?]
sage: ef=F.adjacency_matrix().eigenvalues();ef.sort(reverse=1);ef
[2.228328215946636?, 1.360409337131395?, 0.1858853464264596?, -1, -1, -1.774622899504490?]
| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.