# Question from Artin's algebra book

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?

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That looks fine, but you should be explicit about what $\lambda$ is in the last step. – Thomas Andrews Jan 22 '13 at 14:03
Also, in general, $\zeta_k$ is used for a primitive $k$th root of unity. So your choice of notation might be confusing. The answer isn't wrong, really, just a point that might throw somebody off. – Thomas Andrews Jan 22 '13 at 14:08

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 $A^3+4A^2+5A+2I=0$?
roots of $m_A(x)$ are the eigenvalues of $A$ and $A$ has only one eigenvalue and also the roots of $x^r - 1 = 0$ are all unique is it not enough to answer the "hence " part. – jim Jan 22 '13 at 16:22
It is enough, but you never say that the roots of $x^r-1$ are unique. In general, that is all you need - uniqueness of the roots of $x^r-1$. Turns out, you can determine if the roots of a polynomial are all unique without knowing the roots - there is a repeated root to a polynomial $q(x)$ if and only if $q(x)$ and the derivative $q'(x)$ have a common factor, and there are ways to compute the "greatest common divisor" of two polynomials without knowing roots. – Thomas Andrews Jan 22 '13 at 16:27
Well, the definition of separable polynomial is not having repeated roots. The result about the $\gcd$ of $q(x)$ and $q'(x)$ is a theorem - that is, the ability to determine if a polynomial is separable without explicitly knowing its roots is the big trick. – Thomas Andrews Jan 22 '13 at 16:33