 Solution

The cyclic matrix acts by $(v_1,v_2,v_3,v_4)\mapsto (v_2,v_3,v_4,v_1)$. For each fourth root of unity $\lambda$, the vector $v_{\lambda }=(1,\lambda ,{\lambda }^2,{\lambda }^3{)}^T$ satisfies $A{v}_{\lambda }=\lambda v_{\lambda }$, because ${\lambda }^4=1$. This supplies the four distinct eigenvalues and nonzero eigenvectors:Eigenvectors for distinct eigenvalues are linearly independent. Thus, with these columns in the indicated order,They form a basis over $\mathbb{C}$, so there are no other eigenvalues. Equivalently the characteristic polynomial is ${\lambda }^4-1$.