eigenvector of a symmetric matrix has $i$ and $j$ such that $x_i\ge 0$ and $x_j\le 0$

Question

Consider a symmetric matrix $B$ with the orthonormal eigenvectors $v_1,\ldots,v_n$ of the corresponding eigenvalues $0 = \lambda_n \le \ldots \le \lambda_1 = 1$ and $v_1= \frac{1}{\sqrt n} (1,\ldots ,1)$

Let's denote $x:=v_2$.

I need to prove that there're $i$ and $j$ such that $x_i\ge 0$ and $x_j \le 0$.

I already proved that there's an $i'$ such that $|x_{i'}| \ge \frac{1}{\sqrt n}$.

This is part of a bigger proof regarding normalized adjacency matrix $A$ of a $d$-regular graph $G$, and $B = \frac{1}{2}I + \frac{1}{2}A$, so I'm not sure if more details should be added to the question.

Answer 1

From $x\perp v_1$, we get $\sum_{k=1}^n x_k=\sqrt{n}\langle x,v_1\rangle=0$. So, $x$ can neither be a positive vector nor a negative vector, i.e. it must have some non-positive entries and some non-negative entries.