# Weighted eigenvector-entry sums of bipartite graphs

Let $\lambda_m$ denote the $m$-th eigenvalue of the adjacency matrix $A$ of a bipartite graph $G$ of order $N$ and ${\bf v}_m$ the normalized eigenvector corresponding to $\lambda_m$. I am looking for possible ways to evaluate or bound $\sum_{m=1}^N {\bf v}_m(i)/\lambda_m$ for a node $i$ without a priori knowledge of the spectrum and eigenvectors. One may ignore $m$ for which $\lambda_m=0$ in the sum. I am hoping that I can exploit the symmetry of the eigenvalue spectrum and the properties of eigenvectors of bipartite graphs (like the ones mentioned in the answer to this question) to achieve this. Any ideas on how to proceed?

• What is the corresponding eigenvector? Feb 25, 2015 at 17:59
• ${\bf v}_m$ is the eigenvector corresponding to $\lambda_m$. Feb 25, 2015 at 18:07
• You missed the the in my question. First, you have a whole eigenspace. Second, even if each eigenvalue is simple (which is often not the case), how do you normalize $v$? The bound that you are seeking obviously depends on this normalization. (And, if $\lambda$ is multiple, on the choice of a basis.) Feb 25, 2015 at 19:13
• As for normalization, assume that ${\bf v}_m \cdot {\bf v}_m = 1$. Feb 25, 2015 at 20:16
• And if $\lambda_m=0$? Feb 25, 2015 at 21:43