Spectrum of the n-cycle graph $C_n$, $n\ge3$. I am looking for the spectrum of a cycle graph i.e, eigenvalues of adjacency matrix of $C_n$ and their multiplicities.
I know that the adjacency matrix of $C_n$ is always a circulant matrix. Hence, using the fact that the eigenvalues of any circulant matrix with its first row as $(c_o,c_{n-1},c_{n-2},....,c_1)$ is given by $\lambda_j = c_o+c_{n-1}w_j+c_{n-2}w_j^2+....+c_1w_j^{n-1}$,    $j=0,1,....,n-1$, where $w_j=exp\big(i\frac{2\pi j}{n}\big)$, we can compute all the eigenvalues.
But I am looking for a way where we could find the spectrum of $C_n$, independent of the concept of circulant matrices.
Is there any other way to find the spectrum of $C_n$?
I am looking for another way other than the one I mentioned above, please.
This is't a duplicate question.
It is easy to say that 2 is always an eigenvalue of the adjacency matrix of any $C_n, n\ge3$, with multiplicity 1, but what about other eigenvalues? 
 A: Think of a cycle permutation on $n$ vertices, or a cyclic directed graph with edges pointing only one way. For example, such a cycle matrix $M_4$ on four vertices would look like
$$ M_4 = \begin{pmatrix} & & & 1 \\ 1 & & & \\ & 1 & & \\ & & 1 & \end{pmatrix} $$
You can see that for this matrix $M_n$ that we have $M_n^n = I$, and furthermore $M_n^k \neq I$ for any $1 \leq k < n$. So the characteristic polynomial of this matrix is the same as the minimal polynomial, and is $(x^n - 1)$. Therefore the eigenvalues of $M_n$ are the $n$th roots of unity, $1, \omega, \omega^2, \ldots, \omega^{n-1}$ where $\omega = \exp(2 \pi i / n)$.
Now you need to somehow see that the eigenvectors of this matrix $M_n$ are of the form $(1, 1, \ldots, 1)$, $(1, \omega, \omega^2, \ldots, \omega^{n-1})$, $(1, \omega^2, \omega^4, \ldots, \omega^{2n - 2})$ with eigenvalues $1, \omega, \omega^2$ and so on. (This is not hard to do if you explicitly write out a vector, declare it to be an eigenvector with a certain eigenvalue and just apply the matrix to it and check the conditions).
Now comes the last part. The matrix $M_n^{-1}$ has the same eigenvectors, with inverse eigenvalues. So the vector $(1, \omega^2, \omega^4, \ldots, \omega^{2n - 2})$ is an $M_n$-eigenvector with eigenvalue $\omega^2$, and an $M_n^{-1}$-eigenvector with eigenvalue $\omega^{-2}$. Therefore their sum $M_n + M_n^{-1}$ has eigenvalues $1 + 1 = 2$, $\omega + \omega^{-1} = 2 \cos (2 \pi / n)$, $\omega^2 + \omega^{-2} = 2 \cos(4 \pi / n)$, and so on. Furthermore, $M_n + M_n^{-1}$ is the adjacency matrix of the cycle graph.
