Determinant of $ n \times n$ matrix and its characteristic polynomial.

Question

Suppose, $M_4, M_5,..M_n$ is as follows then determinant and characteristic polynomial of $M_n$. $M_4=\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right),M_5=\left( \begin{array}{ccccc} 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ \end{array} \right),M_6=\left( \begin{array}{cccccc} 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 \\ \end{array} \right), \quad M_8=\left( \begin{array}{cccccccc} 0 & 0 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ \end{array} \right)$

Answer 1

If $J$ is the $n\times n$ matrix with all entries 1, then $M_n=J-I-C_n$, where $C_n$ is the adjacency matrix of the cycle. The all-ones vector is an eigenvector for $C_n$ with eigenvalue 2. Since $C_n$ is symmetric, if $\lambda\ne2$ is an eigenvalue for $C_n$ with eigenvector $z$, then $z$ must be orthogonal to the all-ones vector. Hence $Jz=0$ and therefore $$ M_nz = (J-I-C_n)z = -(1+\lambda)z. $$ Therefore the eigenvalues of $M_n$ are $n-3$ and the numbers $-1-\lambda$, where $\lambda$ runs over the eigenvalues of $C_n$ not equal to 2, i.e., the numbers $2\cos(2\pi k/n)$ for $k=1,\ldots,n-1$.

Answer 2

The matrix $M_n$ is a circulant matrix where $c_0=c_1=c_{n-1}=0$ and other $c_k=1$. Eigenvalues and eigenvectors of circulant matrices can be readily calculated.