Support of eigenvectors

Question

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.

Answer 1

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.

Answer 2

For fixed $N$ the answer is yes, the case $\lambda=0$ should be easy, for $\lambda\neq 0$ one can take the candidate eigenvector $v$ whose first entry is zero $v=(0;x_1;x_2;\ldots;x_N)$ $(x_1=1)$ and put the following system to solve $M_{\lambda}v=kv$ in $(k;x_2;\ldots;x_N)$ with $k$ a certain eigenvalue to the given matrix $M_{\lambda}$. It is not hard to see that this is a system of $N$ equations with $N$ unknowns. You see the final equation after reduction is a polynomial in $\lambda$, thus having finite number of roots. Since we excluded the case $\lambda=0$ you can conclude.

For example i tried the case $N=4$, $M_{\lambda}=\begin{pmatrix}1+3\lambda&-1-2\lambda&-\lambda&0\\-1-2\lambda &2+3\lambda&-1&-\lambda\\-\lambda&-1&2+3\lambda&-1-2\lambda\\0&-\lambda&-1-2\lambda&1+3\lambda\end{pmatrix},$ the roots are $-1;\approx-0.26;\approx -0.61$ of the following polynomial $72x^5+211x^4+242x^3+137x^2+38x+4.$

(If i am not mistaken i changed the $(N-1)\times N$ entry to $-1-2\lambda$ ).