Is there any elegant way to calculate the eigenvalues of the Cartan matrix of type $A_n$?

Question

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.

Answer 1

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.

Answer 2

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.