If $A \in M_2(\mathbb R)$ non identical with $A^3=I $ then $\text{tr}(A)=-1$ 
Let $A \in M_2(\mathbb R)$ a $2\times 2$ matrix with real coefficient, such that $A \ne I$ and
  $$
A^3=I
$$
  Then $\text{tr}(A)=-1$. What if we consider $M_n(\mathbb R)$? Is the statement still true?

I didn't manage to solve it, but I have a question: can we say $A$ is non singular? Indeed, 
$$
A^3=I \Rightarrow AA^2=A^2A=I\Leftrightarrow A^{-1} = A^2.
$$
Is it right? How can we prove the statement? 
 A: The minimal polynomial of $A$ must divide $X^3-1=(X-1)(X^2+X+1)$; since $A\neq I$ we have that the minimal polynomial of $A$ is $X^2+X+1$ (the degree of the minimal polynomial is smaller or equal to $2$).Because it has degree $2$, it must be equal to its characteristic polynomial, therefore $\operatorname{tr}(A)=-\text{ coefficient of }X=-1$ 
I don't think that it's true in general.For instance if $n=3$, the minimal polynomial of $A$ must be $X^3-1$ (why?), so $\operatorname{tr}(A)=0$.
A: You're right in that it is nonsingular. Otherwise, it would have a zero eigenvalue, so it would not be annihilated by a polynomial with a constant term.
$A$'s minimal polynomial divides $x^3-1$, so its eigenvalues are all third roots of unity.
If they were both $1$, $A$ would be identity (it can't have nontrivial Jordan blocks, because $x^3-1$ is squarefree), so at least one of them is nonreal.
$A$'s entries are real, so nonreal complex roots occur in conjugate pairs, and the sum of the two conjugate nonreal third roots of unity is $-1$.
In general, for an $n\times n$ matrix, the trace of such a matrix will be an integer of the form $n-3k$ with $0< k\leq \lfloor n/2\rfloor$ (and any such integer can be found as a trace of such a matrix, consider block matrices with diagonal blocks equal to $A$ or a $1\times 1$ matrix $(1)$ and $0$ elsewhere).
