If $f_A(x) \ne m_A(x)$ and $A^3=I$ then $A=I$? Suppose $A \in M_{3\times3}(\mathbb R)$ and $f_A(x) \ne m_A(x)$ where $f_A(x)$ is the characteristic polynomial of $A$ and $m_A(x)$ is the minimal polynomial of $A$.
If we were to assume that $A^3=I$, would we get that $A=I$ ?
Would the result be different in case $A$ was $2\times2$ matirx?
If it possible to find such $A$ which isn't $I$, please explain how to do it.
Otherwise, please explain why it isn't possible.
My current work:
I started with the case where $A$ is $3 \times 3 $ matrix.
From the assumption that $A^3=I$ we know that the polynomial $g(x)=x^3-1$ annihilates $A$.
From there I tried to construct $A \in M_{3\times3}(\mathbb R)$ such that $A^3=I$ while keeping in mind that any eigenvalue $c$ of $A$ must satisfies $g(c)=c^3-1=0$ . The only matrices I were able to look for were the rotating matrices and the projection matrices which didn't satisfy  $A^3=I$.
Is there any other way to be able to construct such matrix?
Where $A$ is a $2 \times 2 $ matrix I had the same problem to construct the matrix I want, but I was able to try different combinations. None of them matched the $A$ I was looking for, therefore I'm almost sure it is impossible to find such a matrix when $A$ is $2 \times 2$ matrix.
Thank you
 A: In the $3\times 3$ case we have $$x^3 -1 = (x-1)(x^2+x+1) $$ and $x^2+x+1$ is irreducible in $\mathbb{R}[x]$. 
Thus the minimal polynomial of $A$ can be $x-1$ or $(x^2+x+1) $ because $f_A(x) \neq m_A(x)$. 
But if $m_A(x) = x^2+x+1$ then $f_A(x) \not\in \mathbb{R}[x]$ because the roots of $x^2+x+1$ are complex coniugate, the minimal and the characteristic polynomial have the same roots and $\deg f_A(x) = 3$. Thus $m_A(x) = x-1$ and $A=I$.
In $2 \times 2$ case if $m_A(x) = x^2+x+1$ then $f_A(x) = m_A(x)$. Thus  $m_A(x) = x-1$ and $A=I$ also in this case.
A: Since $m_A(x)\neq f_A(x)$ then the degree of $m_A(x)$ is $1$ or $2$.
But $m_A(x)$ must devide $x^3-1=(x-1)(x^2+x+1)$.
From here show that $m_A(x)=x-1$ and hence $A=I$.
A: The minimal polynomial $m_A$ always divides the characteristic polynomial $f_A$, which always divides a power of $m_A$ (because $f_A$ is the product of the invariant factors associated with $IX-A$). So if $m_A\neq f_A$, then some irredicible factor of $m_A$ divides $f_A$ twice. Given that $X^3-1$ is an annihilating polynomial, the only possible irredicible factors of $m_A$ are $X-1$ and $X^2+X+1$. But the latter squared cannot divide $f_A$ which is of degree$~3$, so $(X-1)^2$ divides $f_A$. But now there is no room (degree) left for a quadratic factor, $X^2+X+1$, so it doesn't divide $f_A$, nor $m_A$. Then $m_A$ is a positive power of $X-1$ that divides $X^3-1$, in other words $m_A=X-1$, which only happens when $A=I$. Also of course $f_A=(X-1)^3$.
In the $2\times2$ case the same argument works, even easier. In the $4\times 4$ case there are easy counterexamples.
