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.



Sum of Eigenvectors Entries of an Adjacency Matrix
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 $4.07$. 


