Given a matrix in the following form

[math]

H

\begin{bmatrix}

0&1 & \\

1&0& 1 \\

&1 &0 & 1 \\

&&\ddots &0& \ddots\\

&&&1&0 & 1 \\

&&&&1&0 \\

\end{bmatrix}_N

[/math]

,how to prove that the eigen value of the matrix H are ?

[math]

E=2\cos\left( \frac{n\pi}{N+1} \right); \quad n = 1, 2, \dots , N

[/math]

Help:confused::confused:

A good start is to construct the characteristic polynomial. Hint: Use recursion.

A good start is to construct the characteristic polynomial. Hint: Use recursion.

I tried the recursion, but I have no idea to connect it to the cosine .

Try using Euler's formula. In particular,

 

[math]2\cos\theta = e^{i\theta}+e^{-i\theta}[/math]

 

 

Addendum

 

aoshima, what have you obtained for the characteristic polynomial?

 

Also, you know the form of the solution. It was handed to you on a platter. This known solution looks a lot like the N+1 roots of unity -- except n starts at 1 rather than 0. The n+1^th roots of unity are the solutions to [math]z^{n+1}-1=0[/math]. Dividing by [math]z-1[/math] removes the trivial solution:

 

[math]\frac{z^{N+1}-1}{z-1} = 1 + z + \cdots + z^N[/math]

 

Some adroit substitutions of the characteristic polynomial will yield something similar to the above.

Edited January 15, 201015 yr by D H