Spectrum of cycle graph

Question

I am asked to find the spectrum of the cycle graphs $C_n$.

In a previous exercise, I found the eigenvalues of $A_n$, where

$$(A_n)_{ij}=\begin{cases}1 & \text{if }j=i+1\mod n \\ 0 &\text{otherwise.}\end{cases}$$

Those were $\lambda=\exp \left(\frac{2\pi ik}{n}\right)$ for $k \in \{0,\ldots,n-1\}$.

It seems that the adjacency matrix of $C_n$ is equal to $A_n+A_n^\top$, but I have no idea how to proceed.

I also have trouble finding the spectrum of $K_n$. This seems to be the same trick, by noticing that the adjacency matrix of $K_n$ is $J-I$ where $J$ is the matrix with all ones.

Answer 1

For $K_n$ the computation is simple:

$A+I$ has rank 1, hence $-1$ is an eigenvalue of multiplicity at least $n-1$. To find the last eigenvalue use the fact that $$-1+(-1)+...+(-1)+\lambda_n=tr(A)=0$$

For the cycle graph, let $v=(1, \omega, \omega ^2,.., \omega^{n-1})$, where $\omega$ is a nth root of unity. What is $Av$?

Answer 2

The adjacency matrix $B=A_n+A_n^t$ is a circulant matrix. The eigenvectors of a circulant matrix are always $v_k=(1,\zeta^k, \zeta^{2k},\ldots,\zeta^{(n-1)k})$, for integers $k$, where $\zeta=\exp(2\pi i/n)$. So the eigenvalues of $B$ are $\zeta^k+\zeta^{-k}=2\cos(2\pi k/n)$ for $0\le k<n$.