Probability eigenvectors of discrete random matrix are orthogonal to discrete random vector

Question

Let $W=(w_{ij})_{1 \leq i, j \leq N}$ and $\textbf{v}=(v_j)_{1 \leq j \leq N}$ be a random $N\times N$ matrix and N-vector, respectively, where all $w_{ij}$ are jointly independent and have discrete distributions with non-zero variance. Likewise all $v_j$ are jointly independent (and independent of $w_{ij}$) and have discrete distributions with non-zero variance. Let $\textbf{q}_1, \ldots, \textbf{q}_N$ be the eigenvectors (or generalized eigenvectors) of $W$.

Can we show that as $N \rightarrow \infty$ the probability that $P\left(\bigwedge_i \,\textbf{v} \cdot \textbf{q}_i \neq 0 \right) \rightarrow 1$, or else show that this is not the case?

Answer 1

If there is no connection between the distributions for different $N$, it's not true. Suppose $w_{ij}$ and $v_i$ all have Bernoulli distributions with parameter $1 - 2^{-N}$. Then with high probability, all entries of $\bf W$ and $\bf v$ are $1$, and $\bf v$ is orthogonal to all but one of the eigenvectors of $\bf W$.