Eigenvalues and eigenvectors of non-symmetrical tridiagonal matrix

Question

The question is the following: given a matrix $$A=\begin{pmatrix} 1& 2 & & & & \\ 1& 0& 1 & & & \\ & 1& 0& 1 & &\\ & & \ddots & \ddots & \ddots & \\ & & & 1& 0 & 1\\ & & & & 1 &0 \end{pmatrix}.$$ Is it possible to give analytic expressions for the eigenvalues and eigenvectors of $A$?

Wang et. al [1] show that if the elements on the main diagonal are all 0, the eigenvalues and eigenvectors of $A$ can be expressed in trigonometric functions.

Thanks for your answer.

References

[1] W. Wang, C. M. Wang and S. L. Guo, On the walk matrix of the Dynkin graph $D_n$, Linear Algebra Appl. 653 (2022) 193-206.

Answer 1

Defining the $2 \times 2$ transfer matrix \begin{align}\tag{1} Q = \begin{pmatrix} -\lambda & 1 \\ -1 & 0 \end{pmatrix}, \end{align} the characteristic polynomial (CP) of the $M \times M$ matrix $A_M$ is given by \begin{align} P_M(\lambda) &= \det(A_M -\lambda \, I)\tag{2a}\\ &=\langle 1{-}\lambda, 2| \, Q^{M-1} \,|1,0\rangle \tag{2b}\\ &= 2 T_M\left(-\frac \lambda 2 \right) + U_{M-1}\left(-\frac \lambda 2 \right) \tag{2c}\\ &= 2\cos(M \varphi) + \frac{\sin(M \varphi)}{\sin(\varphi)}\tag{2d}, \end{align} with Chebyshev polynomials $T_M,U_M$, and with $\lambda=-2\cos\varphi$.

The unnormalized right eigenvectors $A_M x_\mu=\lambda_\mu x_\mu$ have the elements \begin{align} x_{\mu,m} = \langle 1, 0| \, (-Q)^m \, |1,0\rangle,\quad m=0,\ldots,M-1.\tag{3} \end{align} The eigenvector normalization can be related to the derivative $P_M'(\lambda)$, see, e.g., https://arxiv.org/abs/2103.10776 for details.

Regarding to your question, I don't think that a closed form expression exist for $\lambda_\mu$ if $M>5$, as due to the left boundary term, the CP does not factorize in this case (up to one trivial eigenvalue $\lambda=\pm1$ if $M=3n\pm1$). However, it is often not necessary to explicitly calculate the eigenvalues, see the cited paper, as the CP together with the eigenvectors contains enough, or even more, information.

Note added (22.11.22,11:22):

If $(A_M)_{11}=a_0$, then \begin{align} P_M(\lambda) &= 2 T_M\left(-\frac \lambda 2 \right) + a_0 \, U_{M-1}\left(-\frac \lambda 2 \right) \tag{4a} \\ &=2\cos(M \varphi) +a_0 \frac{\sin(M \varphi)}{\sin(\varphi)}\tag{4b}, \end{align} such that for $a_0=0$ the eigenvalues are the well known zeroes of the Chebyshev polynomial of the first kind $T_M$. This is the case in the paper [1] cited by the OP.

Answer 2

I don't know about a general form, but at least for each of the cases of dimension $6n-1$ or $6n+1$, $n\in \mathbb{N} $, there is an eigenvalue $-1$ or $1$, respectively. Denoting $v_3 \equiv 1,0,-1$,

for dimension $6n-1$, $$ (-1, v_3, v_3, v_3, \ldots ,v_3, 1) $$ has eigenvalue $-1$, and for dimension $6n+1$, $$ (v_3, -v_3, v_3, -v_3, \ldots ,v_3, -v_3, 1) $$ has eigenvalue $1$.