TITLE: Question from Artin's algebra book
QUESTION [6 upvotes]: Let $A$ be an $n\times n$ matrix such that $A^r =I$ and $A$ has exactly one eigenvalue ,then $A= \lambda I$.
My answe: 
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?

REPLY [3 votes]: It would be better to be explicit about what the $\lambda$ is in the last step.
You say "hence, ..." but you don't say the reason this step is true. What property do you need for the $\zeta_i$ to be able to make this step?
Indeed, what is the property of $x^r-1$ that you are using here, really? If, instead, the question said, $A^r-A-I=0$, would it still be true? If $A^3+4A^2+5A+2I=0$?