Type of the Coxeter matrix for $\tilde{A_n}$

Question

Consider the $(n+1) \times (n+1)$ Coxetermatrix $$ A = \begin{pmatrix} 1 & -\frac{1}{2} & 0 & \cdots & -\frac{1}{2}\\ -\frac{1}{2} & 1 & -\frac{1}{2} & \ddots & \vdots \\ 0 & -\frac{1}{2} & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & 1 & -\frac{1}{2} \\ -\frac{1}{2} & \cdots & 0 & -\frac{1}{2} & 1 \\ \end{pmatrix}$$ of the Coxeter group $\tilde{A}_n, n \geq 2$. (Note that only the three main diagonals as well as the entries $A_{1,n+1},A_{n+1,1}$ are nonzero.)

I am trying to determine the eigenvalues of this matrix and prove that $A$ is positive semidefinite (i.e. has nonnegative eigenvalues and zero is one of them). Is there a way to do this by induction or recursively? (Clearly for 3x3 or 4x4 one can simply calculate the characteristic polynomial and find its zeroes, but the calculation gets messy and long as $n$ grows)

Thank you for any input.:)

Answer 1

Because your matrix is a circulant matrix, it has the complex vector $(1, \zeta, \zeta^2, \ldots, \zeta^{n - 1})$ as an eigenvector for each root of unity $\zeta$. It's not hard to see that these are linearly independent, and it's easy to compute explicitly the eigenvalue associated to each eigenvector, and to see that it is a nonnegative real number. (Let me know if you'd like more details.)

Answer 2

I am not sure what the eigenvalues of $A$ are, but it isn't hard to prove that $A$ is positive semidefinite.

Let $C$ be the permutation matrix for a cyclic shift. Then $A=\frac12\left[(I-C)+(I-C)^T\right]$. To prove that $A$ is PSD, we may show that $u^TAu\ge0$ for every unit vector $u$. This is equivalent to $u^T(I-C)u\ge0$ or $u^TCu\le1$. Yet the latter is a direct consequence of Cauchy-Schwarz: $u^TCu=\langle u,Cu\rangle\le\|u\|_2\|Cu\|_2=1$.

Remark. Since the vector of ones lies inside the null space of $A$, the matrix $A$ is singular.