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](https://www.sciencedirect.com/science/article/pii/S0024379522002920?via%3Dihub)] 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](https://www.sciencedirect.com/science/article/pii/S0024379522002920?via%3Dihub)] W. Wang, C. M. Wang and S. L. Guo, [On the walk matrix of the Dynkin graph $D\_n$](https://www.sciencedirect.com/science/article/pii/S0024379522002920?via%3Dihub), Linear Algebra Appl. 653 (2022) 193-206.



Eigenvalues and eigenvectors of non-symmetrical tridiagonal matrix
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.


