Eigenvalues of an $n$-path

Question

Show that the eigenvalues of the Laplacian matrix of an $n$-path graph are $$ 2 - 2 \cos \left( \frac{i\pi}{n} \right), \qquad i=0,1,\dots,n-1$$

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I know that the characteristic polynomial of the Laplacian of the $n-$path satisfies

$$\chi_n(x)=x\chi_{n-1}(x)-\chi_{n-2}(x)$$

and thus $\chi_n(x) = U_n(x/2)$ is a Chebyshev polynomial of second kind because Chebyshev's polynomials of second kind satisfy the recurrence:

$$U_n(x) = 2xU_{n-1}(x)-U_{n-2}(x)$$

Wiki says $U_n(x)$ has roots $\cos\left (\frac{i\pi}{n+1}\right ), i=1,..,n$. Thus the roots of $\chi_n(x) = U_n(x/2)$ are $2\cos\left (\frac{i\pi}{n+1}\right )$ and so am wondering where the additional $2$ is coming from? Thanks.

Answer 1

I am not familiar with what $n$-path is, but the Laplacian with the eigenvalues $2-2\cos(i\pi/n), i=0,\ldots,n-1$ is the matrix of order $n$ of the form,

\begin{bmatrix} 1&-1\\ -1&2&-1\\ &\ddots&\ddots&\ddots\\ &&-1&2&-1\\ &&&-1&1 \end{bmatrix}