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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 ?

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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 $\sqrt{2}+a$ $$\ker(A-(\sqrt{2}+a)I_3)=\text{span}\left\{\begin{pmatrix}1\\-\sqrt{2}\\1 \end{pmatrix}\right\}$$

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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!

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