Show that the eigenvalues of the cycle $C_n$ are $λ_j=2 \cos(\frac{2πj}{n})$

Question

Show that the eigenvalues of the adjacency matrix for cycle $C_n$ on $n$ vertices are $λ_j=2 \cos(\frac{2πj}{n})$.

I've seen a couple of references that just assume this to be true, but I have not found any sort of prove for it. Any help will be much appreciated.

Answer 1

Let $B$ be the $n\times n$ matrix with columns $a_j = e_{j+1}$ for $j=1,\ldots,n-1$ and $a_n = e_1$, where $e_k$ is the $k$-th standard basis vector. Then the adjacency matrix of the $n$-cycle is $A = B+B^T$.

Now, $B$ is a unitary matrix and thus normal. So, it commutes with $B^T = B^*$. Hence, the eigenvalues of $A = B+B^*$ are given by $\{2\operatorname{Re}\lambda : \lambda\in\sigma(B)\}$, where $\sigma(B)$ is the set of eigenvalues of $B$.

Since $B$ is easily seen to have the eigenvalues $e^{\frac{2\pi ij}n}$, $j=0,\ldots,n-1$ (roots of unity), it follows that the eigenvalues of $A$ are $$ 2\operatorname{Re}e^{\frac{2\pi ij}n} = 2\cos\left(\frac{2\pi j}n\right). $$