For $0<\lambda<1$, consider the operator $$A=(S_L+\lambda)(S_R+\lambda),$$ where $S_L,S_R$ are the left-shift and right-shift operators on $\mathbb{C}^n$ with nontrivial kernel (open boundary conditions). I'm looking for an eigenvector of $A$ with eigenvalue exponentially decaying to $0$ as $n\to \infty$.
How do I solve for this special eigenvector?
Let $M$ be the matrix given by $M = (1+\lambda^2) I + \lambda(S + S^T)$, with $S$ as in my comment above. We note that $M$ is tridiagonal and Toeplitz, which means its eigenvalues are given nicely as $$ \mu_k = (1+\lambda^2) + 2\lambda\cos(k\pi/(n+1)), \quad k =1 , \dots ,n $$ Let $e$ denote the column vector $e = (0,\dots,0,1)$, so that $A = M - ee^T$.
By the matrix determinant lemma, we may state that for any $\nu$: $$ \det(A - \nu I) = \det((M - \nu I) - ee^T) = (1 - e^T(M - \nu I)^{-1}e)\det(M - \nu I) $$ In other words: suppose that $\nu$ is not an eigenvalue of $M$. Then $\nu$ will be an eigenvalue of $A$ if and only if the $n,n$ entry if $(M - \nu I)^{-1}$ is equal to $1$. Alternatively, the adjugate version yields $$ \det((M - \nu I) - ee^T) = \det(M - \nu I) - e^TAdj(M - \nu I)e $$ which yields an interesting condition: $\nu$ is an eigenvalue of $A$ if and only if $$ \det(M - \nu I) = \det([M - \nu I][1,\dots,n-1;1,\dots,n-1]) $$ where $[M - \nu I][1,\dots,n-1;1,\dots,n-1]$ denotes the submatrix in which the last row and column of $M$ are deleted.
This will eventually lead to the characteristic polynomial of $A$, but that might not be so helpful.