Consider the $N$ by $N$ matrix


$$M\_N= \begin{pmatrix} 1+3\lambda & -1-2\lambda & - \lambda & 0 & 0 &0 &0\\
-1-2\lambda & 2+3\lambda & -1 & -\lambda & 0 & 0 & 0\\
-\lambda & -1 & 2(1+\lambda) &-1 & -\lambda & 0& 0 \\
0 & \ddots & \ddots & \ddots & \ddots& \ddots & 0 \\
0 & 0 & -\lambda & -1 & 2(1+\lambda) & -1 & -\lambda \\
0&0 & 0& -\lambda & -1& 2+3\lambda & -1-2\lambda \\ 
0 & 0 & 0 &0 & -\lambda & -1-2\lambda& 1+3\lambda \\
\end{pmatrix}.
$$


I want to show that there exists $\varepsilon>0$ such that for all $\lambda \in (-\varepsilon,\varepsilon)$ **and independent of $N$**, the matrix $M\_N$ does not have an eigenvector of the form $u=(0,u\_2,...,u\_N)$,i.e. where the first component vanishes.


Numerically it seems to be true up to $\varepsilon =1/4,$ but it is hard to understand this 5 term recurrence for me.



Support of eigenvectors
This looks to be true for $\varepsilon=1/4$.


First of all, the space of symmetric ($u\_i=u\_{N+1-i}$ for all $i=1,\ldots, N$) vectors is invariant for $M\_N$, and so is the space of antisymmetric ($u\_i=-u\_{N+1-i}$ for all $i=1,\ldots, N$) vectors. Since the sum of these two subspaces consists of all vectors, we conclude that for any eigenvalue there exists either a symmetric or antisymmetric eigenvector. At first, we consider the case when this vector has zero first coordinate (thus zero last coordinate aswell). For a vector $u=(u\_1,\ldots,u\_N)$ we attribute a polynomial $p\_u:=\sum u\_i t^{N+1-i}$. Then $M\_Nu$ corresponds to a polynomial $p\_u(—\lambda/t^2—1/t+2+2\lambda-t-\lambda t^2)+R$, where $R$ is a remainder term which depends on $u\_1,u\_2,u\_{N-1},u\_N$. If $u$ is a symmetric or antisymmetric eigenvector with $u\_1=u\_N=0$ and eigenvalue $\mu$, the remainder $R$ equals $-\lambda u\_2(t-1)^2(t^{n-2}\pm 1/t)$, and we see that $\lambda\ne 0$ and all roots of $—\lambda/t^2—1/t+2+2\lambda-t-\lambda t^2-\mu$ must lie on the unit circle. If they are $e^{\pm ia}, e^{\pm ib}$, then $2\cos a, 2\cos b$ must be roots of $-\lambda x^2-x+2+4\lambda-\mu=0$, and by Vieta theorem $|1/\lambda|=2|\cos a+\cos b|\leqslant 4$.


So, it remains to consider the case when the same eigenvalue has both a symmetric and antisymmetric eigenvector. In natural coordinates the corresponding matruces differ only by few elements, that may possibly yield some interlacing type claim.


