Probably it's very simple to show, but I'd like to have a proof of the fact that a $n$ x $n$ circulant matrix with the following structure:
$$A= \begin{bmatrix} (n-1)/n & -1/n & \cdots & -1/n\\ -1/n & (n-1)/n & \cdots & -1/n\\ \vdots & \vdots &\ddots&\vdots \\ -1/n & -1/n &\cdots&(n-1)/n \end{bmatrix}$$
has all eigenvalues but one equal to $1$ . (The remaining one is clearly zero since the matrix is not invertible).
On
If you closely look at $A$, then you see that it can be written as $\mathbf{A}= \mathbf{I}_n - \frac{1}{n}\mathbf{ee}^T$ where $\mathbf{e}$ is the all one vector.
Part-A: Eigenvalues of $\mathbf{I}_n$ are all $1$ and eigenvectors span the whole $\mathbb{R}^n$, so any set of an orthonormal basis of $\mathbb{R}^n$ can be considered as its eigenvectors.
Part-B: It is a symmetric matrix of rank one, so the eigenvectors of this matrix are also orthogonal. $\big(1,\mathbf{e}\big)$ is an eigenpair of this matrix.
$\mathbf{e}$ is an eigenvector of $A$ and the corresponding eigenvalue is $0$. For any other vectors $ \mathbf{v} \in \mathcal{V}$ where $\mathcal{V}=\left\{ \mathbf{v}|\mathbf{v}^T \mathbf{e}=0, \mathbf{v} \in \mathbb{R}^n\right\}$, $\mathbf{Av=v}.$ These proves that other eigen values are $1$.
This is $I-\frac1nE$, where $I$ is the identity and $E$ is the matrix with all entries $1$. All vectors orthogonal to the vector with all entries $1$ are eigenvalues of $E$ with value $0$, so the $(n-1)$-dimensional orthogonal complement of the $1$-dimensional space spanned by that vector is an eigenspace of $E$ with value $0$ and thus of $I-\frac1nE$ with eigenvalue $1$.