Find the Eigenvalues and its Eigen vectors. Consider the  $ \ n \times n \ $ matrix $$ \begin{pmatrix} a & -1 &  &  & & \\ -1 & a & -1 & & & \\ & -1 & a & -1 & & & \\ & & -1 & a & -1 &  \\ & & ..... & ....  & .... \\  & & && a & -1 \\ & & & &-1  & a  \end{pmatrix} $$ 
Find the Eigenvalues and its Eigen vectors. 
Also determine the values of $ \ a \ $ for which the matrix is positive definite.
Answer:
For our convenience consider the $ \ 3 \times 3 \ $ matrix as follows 
$$ A=\begin{pmatrix} a & -1 & \\ -1 & a & -1  \\ & -1 & a  \end{pmatrix} $$ 
The blank positions must be filled with $ \ 0 \ $. Thus, 
$$ A=\begin{pmatrix} a & -1 & 0 \\ -1 & a & -1  \\ 0& -1 & a  \end{pmatrix} $$
Let $ \ \lambda \ $ be the Eigen value of matrix $ \ A \ $ , then 
$ |A-\lambda I |=0 \\ \Rightarrow \begin{vmatrix} a-\lambda & -1 &0 \\ -1 & a-\lambda & 0 \\ 0 & -1 & a-\lambda \end{vmatrix} =0 \\ \Rightarrow (a-\lambda)^3-(a-\lambda)=0 \\ \Rightarrow (a-\lambda) [(a-\lambda)^2-1)=0 \\ \Rightarrow \lambda=a, \ a-1, \ a+1 $ 
For $ \ 2 \times 2 \ $ such matrix we have 
$$ A'=\begin{pmatrix} a & -1  \\ -1 & a   \end{pmatrix} $$
The Eigen values of $ \ A' \ $  are $ \ a-1 , \ a+1 \ $ 
For $ \ 4 \times 4 \ $ such that
the eigen values are $ \ a-1, a-1 , a+1, a+1 \ $
Thus in general the eigen vlaues are  ($ \ if \ n=even \ $ )
$ a-1, a-1, ......  \frac{n}{2} \ times \  \\ a+1 , a+1, ........ \frac{n}{2} \ times  \ $
If $ n=odd \ $ , then the eigen values of the $ \ n \times n \ $ matrix are
$ a-1 , a-1, ................. \frac{n-1}{2}  \ times \\ a+1,a+1,..............\frac{n-1}{2}  \ times \\ and \ \ a \  \ $
Am I right ?
But how to find the eigen vectors ?
 A: First question: No, it appears you've made a computational error. I'm getting eigenvalues that follow in the example.
Given a matrix $A,$ then one computes eigenvectors with eigenvalue $\lambda$ by computing a basis for $$\ker(A-\lambda I_n).$$

Example:
  $$A=\begin{pmatrix} a & -1 &0 \\ -1 & a & -1  \\0& -1 & a  \end{pmatrix}$$
  then for the eigenvalue $a$
  $$\ker(A-aI_3)=\text{span}\left\{\begin{pmatrix}-1\\0\\-1\end{pmatrix}\right\}$$
  for the eigenvalue $-\sqrt{2}+a$
  $$\ker(A-(-\sqrt{2}+a)I_3)=\text{span}\left\{\begin{pmatrix}1\\\sqrt{2}\\1
\end{pmatrix}\right\}$$
  for the eigenvalue $\sqrt{2}+a$
  $$\ker(A-(\sqrt{2}+a)I_3)=\text{span}\left\{\begin{pmatrix}1\\-\sqrt{2}\\1
\end{pmatrix}\right\}$$

A: You're not quite right about the eigenvalues for the $3\times 3$ matrix since you're missing a $-1$ in the $(2,3)$ position. 
For the higher dimensional cases let $A_n$ be the $n\times n$ matrix with $a-\lambda$ along the diagonal and $-1$s above and below the diagonal. Then $A_n$ has the form 
\begin{pmatrix}
a-\lambda & -1 & 0 & \cdots\\
-1 & & & \\
0 & & A_{n-1}&\\
\vdots & & & 
\end{pmatrix} 
From this you can see that $\det A_n = (a-\lambda)\det A_{n-1} - \det A_{n-2}$. Considering the case $n=1$ the only eigenvalue is $a$. Then if $n$ is odd we can then see that $a$ is always an eigenvalue.
This isn't a complete answer but I hope it helps!
