How to find the eigenstates of this Toeplitz matrix?

Question

Now I have a $N\times N$ matrix $H$ in hand, which takes this form: $$ { \begin{pmatrix} 0 & a & a & a &... & a \\ b & 0 & a & a &... & a \\ b & b & 0 & a &... & a \\ \vdots & & & & \ddots & \vdots\\ b & b & b & b &... & 0 \\ \end{pmatrix}}_{N\times N} $$ It's upper triangle only contains element $a$ and lower triangular contains $b$, where both $a$ and $b$ are real and $a\ne b$. I'm trying to solve this eigen function $$ H\boldsymbol{X}(\lambda)=\lambda\boldsymbol{X}(\lambda), $$ where $\boldsymbol{X}(\lambda)=\left(x_1(\lambda),x_2(\lambda),...,x_N(\lambda)\right)^T$. Now I've already known that all the eigenstates $\boldsymbol{X}(\lambda)$ decay exponentially, so I can give a guess about the expression of $\boldsymbol{X}(\lambda)$, i.e., $$ \boldsymbol{X}(\lambda) = \left(\beta(\lambda),\beta(\lambda)^2,...,\beta(\lambda)^N\right)^T. $$ Now inserting this guess solution into eigen function gives: $$ \begin{pmatrix} -\lambda & a & a & a &... & a \\ b & -\lambda & a & a &... & a \\ b & b & -\lambda & a &... & a \\ \vdots & & & & \ddots & \vdots\\ b & b & b & b &... & -\lambda \\ \end{pmatrix} \begin{pmatrix} \beta(\lambda) \\ \beta(\lambda)^2 \\ \beta(\lambda)^3 \\ \vdots \\ \beta(\lambda)^N \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}. $$ My question is: Is it possible to give the explicit expresstion of $\beta$ written with $\beta(a,b,\lambda,N)$?

Update 1: Given $(a,b)$ and order $N$, I can numerically verify that the decay length satisfies the relation$$\left|\beta(\lambda_1)\right|=\left|\beta(\lambda_2)\right|=...=\left|\beta(\lambda_N)\right|$$and$$\left|\beta(\lambda)\right|\propto\frac{1}{N}$$