TITLE: Eigenvectors and eigenvalues of a tridiagonal Toeplitz matrix
QUESTION [13 upvotes]: Is it possible to analytically evaluate the eigenvectors and eigenvalues of the following $n \times n$ tridiagonal matrix
$$
 \mathcal{T}^{a}_n(p,q) = \begin{pmatrix}
  0 & q & 0 & 0 &\cdots & 0 & 0 & 0 \\\
  p & 0 & q & 0 &\cdots & 0 & 0 & 0 \\\
  0 & p & 0 & q &\cdots & 0 & 0 & 0 \\\
  \vdots  & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots& \vdots \\\
  0 & 0 & 0 & 0 &\cdots & p & 0 & q \\\
  0 & 0 & 0 & 0 & \cdots & 0 & p & 0
 \end{pmatrix} 
$$
where $p>0$ and $q > 0$?
Furthermore, is it possible to do the same for the following tridiagonal circulant matrix?
$$
 \mathcal{T}^{b}_n(p,q) = \begin{pmatrix}
  0 & q & 0 & 0 &\cdots & 0 & 0 & p \\\
  p & 0 & q & 0 &\cdots & 0 & 0 & 0 \\\
  0 & p & 0 & q &\cdots & 0 & 0 & 0 \\\
  \vdots  & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots& \vdots \\\
  0 & 0 & 0 & 0 &\cdots & p & 0 & q \\\
  q & 0 & 0 & 0 & \cdots & 0 & p & 0
 \end{pmatrix} 
$$

REPLY [5 votes]: @ Emilio: Could you please show how you get $\det(\mathcal{T}_n(p,q)-\lambda)=2(pq)^n T_n(-\lambda/2)$?
Sorry I wanted to comment another answer. I do not know how to delete this post.
Edit
I think the first problem can be reduced to Chebyshev polynomials of the second kind  ${U_n}(x)$ because  
$$\det \left( {{T_n}(p,q) - \lambda } \right) =  - \lambda \det \left( {{T_{n - 1}}(p,q) - \lambda } \right) - pq\det \left( {{T_{n - 2}}(p,q) - \lambda } \right).$$
implies
$$\det \left( {{T_n}(p,q) - \lambda } \right) ={(\sqrt {pq} )^n}{U_n}\left( {-\frac{\lambda }{{2\sqrt {pq} }}} \right). $$
From the fact that the zeros of ${U_n}(x)$ are  $\cos \frac{{k\pi }}{{n + 1}}$ the eigenvalues are
$2\sqrt {pq} \cos \frac{{k\pi }}{{n + 1}}.$