Consider the $N \times N$ matrix
$$A_{\alpha}=\begin{pmatrix} \lambda_1 & -1 & -\alpha & 0 & \cdots & 0\\ -1 & \lambda_2 & -1 & -\alpha & \cdots & 0\\ -\alpha & -1 & \ddots & \ddots & \cdots & 0\\ 0 & -\alpha & -1 & \lambda_{N-2} & -1 & -\alpha \\ 0 & \cdots & -\alpha & -1 &\lambda_{N-1} & -1 \\ 0 & \cdots & 0 & -\alpha &-1 &\lambda_N \\ \end{pmatrix}$$
The parameters $\lambda$ are defined such that the row sum or equivalently the column sum is equal to zero. This also means that $\lambda_3=\lambda_4 =...=\lambda_{N-2}=2(1+\alpha).$
If $\alpha \equiv 0$ the matrix $A_0$ is just the discrete Neumann Laplacian and it is well-known that all eigenvectors $v=(v_1,...,v_N) \in \mathbb C^N$ of $A_0$ have the property that the first entry $v_1$ and the last entry $v_N$ do not satisfy a Dirichlet condition, i.e. $v_1,v_N$ are not equal to zero, see e.g. here
I conjecture based on numerical experiments that there exists $\varepsilon>0$, independent of $N$ such that the same is true for $A_{\alpha} \in \mathbb C^{N \times N}$ with $\alpha \in (0,\varepsilon).$
To clarify: I do not claim that $v_1$ and $v_N$ are independent of $N$. I claim that there exists $\varepsilon>0$ independent of $N$ such that for all $\alpha \in (0,\varepsilon)$ both $v_1, v_N$ are non-zero!
