Let $A$ be an $n\times n$ matrix such that $A^r =I$ and $A$ has exactly one eigenvalue ,then $A= \lambda I$.
As $A$ is a $n\times n$ matrix then characteristic polynomial has degree n and also exactly one root so $p(x) = (x-a)^n$ ($p(x)$ is the char. polynomial.)
Now the minimal polynomial $m_A(x)|p(x) $ also $m_A(x) | (x^r-1) =(x-\zeta_1)...(x-\zeta_r)$ (where $\zeta_i $ are the rth roots of unity) hence $m_A(x)$ is $(x-\zeta_i)$ for some $i$. As a result $A$ is diagonalizable and $A$ is of the form $A = \lambda I$
Is this correct?