Eigenvalues and eigenvectors of non-symmetrical tridiagonal matrix 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.
 A: 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.
A: 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 $1$.
