Let $A\in M_{n\times n}(\Bbb R)$ s.t. $A^3 = I_n$. Show that $\operatorname{tr}(A) \in \Bbb Z$.
I know that $P(A) = 0$, where $ P(x) = x^3 - 1 = (x-1)(x^2+x+1)$, that is, $1$ is an eigenvalue of $A$. Also, trace is the sums of the eigevalues and in $\Bbb C$ have $1, \omega, \omega^2$ are the eigenvalues of $A$. However I'm stuck in $\Bbb R$. I guess, it's not true that the trace in $\Bbb R$ is equal to the trace in $\Bbb C$. Can you help me?
Hint: use the fact that all irreducible factors of the characteristic polynomial of $A$ divide its minimal polynomial. (See Minimal and characteristic polynomial have same set of irreducible factors)
The minimal polynomial of $A$ divides $x^3-1$, so the irreducible factors of the minimal polynomial have all integer coefficients. (The only possible factors are $x-1$ and $x^2+x+1$.) Hence the characteristic polynomial of $A$ is a product of polynomials with integer coefficients. In particular, $\operatorname{Tr}(A) \in \mathbb Z$.
Hint If $\lambda$ is an eigenvalue of $A$ then $\lambda \in \{1, \omega, \overline{\omega} \}$ where $\omega^2+\omega+1=0$.
Let $m,n,k$ be the multiplicities of these eigenvalues (which could be 0).
Then $$\mbox{tr}(A)=m+n \omega+k \bar{\omega}$$
Use the fact that $A$ has real entries (or that $\mbox{tr}(A)$ is real) to show that $n=k$. Then use the fact that $\omega+\bar{\omega}=-1$.
