Sum of Eigenvectors Entries of an Adjacency Matrix

Question

I have a question regarding the sums $\sum_{i=1}^{n}v_{j}\left(i\right)$ where $v_j$ are eigenvectors of adjacency matrix $A$ which have been normalized to unit length.

Ordering the eigenvectors by their eigenvalues and computing these sums on both random and real-world graphs I noticed that $\sum_{i=1}^{n}v_{1}\left(i\right)>\sum_{i=1}^{n}v_{j}\left(i\right)$ for $j\geq2$.

It should be easy to prove that this is always the case but I am currently at a dead end. Are people aware of any results which may be useful?

I am also interested in putting bounds on this difference for random matrices.

Thanks in advance!

P.S. We should assume that the graph is non-bipartite.

Answer 1

This is not true in general. Consider first a graph with two components, one a complete graph on $4$ vertices and the other a cycle on $n-4$ vertices.

Then $v_1$ is the eigenvector for the largest eigenvalue of the complete graph, and sums to $2$, while $v_2$ is the eigenvector for the largest eigenvalue of the cycle, and adds to $\sqrt{n-4}$.

This graph is disconnected, but we can perturb it to a connected graph by adding a single edge without changing things too much. When I tried this with $n=24$, the largest eigenvalue was $3.10$ and its eigenvector was still mostly supported on the complete graph, with the components adding to $2.26$. Meanwhile the second largest eigenvalue $\lambda_2=1.98$ had an eigenvector adding to $4.07$.