Is there any elegant way to calculate the eigenvalues of the Cartan matrix of type $A_n$? I was solving a physics problem which involved decoupling oscillators. This required me to calculate the eigenvalues of the the Cartan matrix of $A_n$ (the ultimate goal is to diagonalise $A_n$). I am not sure if this is a standard result in Lie algebra, as I know only about the standard procedure for finding eigenvalues (equating $det|A-\lambda{I}|$ to $0$, and solving for $\lambda$). Even pointing me in the right direction would be highly appreciated.
 A: The characteristic polynomial, and hence the eigenvalues, of Cartan matrices have been determined, e.g., in the article On the characteristic polynomial of Cartan matrices and Chebyshev polynomials by P.A. Damianu.
A: The  idea is to use the infinite Cartan matrix, find some eigenvectors, and then get by truncation eigenvectors for the finite Cartan matrix. 
Using the simple identity;
$$\sin(k-1) \theta +\sin(k+1) \theta = 2 \cos \theta\cdot \sin k \theta $$
we see that the vector $(\sin k \theta)_k$ is an eigenvector for the infinite Cartan matrix and eigenvalue $2-2\cos\theta$. Now, if we have moreover $\sin (n+1)\theta=0$ then $(\sin k \theta)_{k=1}^n$ is an eigenvector for the Cartan matrix $A_n$, eigenvalue $2 - 2 \cos \theta$. So we get the eigenvalues of $A_n$ to be $2- 2\cos (\frac{l\pi}{n+1})$, $l=1, \ldots, n$, and the corresponding eigenvector $v_{l}=(\sin (\frac{k l \pi}{n+1}))$. One checks easily that the norm of $v_l$ is $\sqrt{\frac{n+1}{2}}$ for all $1\le l \le n$.
The matrix formed by $v_l$ is up to a constant the matrix of a discrete sine transform (see DST-I )
If we consider numeric functions on a lattice like $\mathbb{Z}$ the eigenvalues of the laplacian are like above $2 \cos \theta$, for eigenfunctions of form 
$$\cos ( k \theta + \phi)$$
 One can consider a finite problem for the domain $\{1,2, \ldots, n\}$ where we introduce linear conditions like $\alpha f(0) + \beta f(1)=0$, $\gamma f(n+1) + \delta f(n)=0$ (discrete boundary value problems). So with the same method one can say diagonalize the matrices $A_n'$, $A_n''$, where one or two extreme diagonal $2$ is replaced by $1$. Again, the matrices of eigenvectors are interesting, providing a discrete transform. 
