TITLE: A be a $3\times3$ real valued matrix such that $A^{3}=I$ but $A \neq I$ .Then trace(A)=?
QUESTION [6 upvotes]: I was thinking about the following problem:
Let A be a $3\times3$ real valued matrix such that $A^{3}=I$ but $A \neq I$ . Then trace of A must be 

(a)0,
  (b)1,
  (c)-1,
  (d)3.

My attempts: I take A to be $$\begin{pmatrix}
x &0  &0 \\ 
 0& x &0 \\ 
 0&0  & x
\end{pmatrix}.$$ 
Now we see $A^{3}=I$ gives $x^{3}=1$ which gives $x=1,w,w^{2}$ where $w$ being the cube root of unity. Thus we see that trace of A is $1+w+w^{2}=0$. So ,i think that (a) is the  right choice.Am i going in the right direction? Please give your valuable opinion. Thanks in advance for your time.

REPLY [6 votes]: Since $A^3=I \Longrightarrow$  the minimal polynomial of $A$ divides $x^3-1$. Thus the eigenvalues of $A$ are in $\{1,\omega,\omega^2\}$ ($\omega$ is a primitive third root of unity). Since $A$ is real if $\omega$ is an eigenvalue of $A$ then $\bar{\omega}=\omega^2$ is an eigenvalue of $A$. The same if $\omega^2$ is an eigenvalue. Therefore we have only two possibilities for the eigenvalues of $A$. Either $1,1,1$ or $1,\omega,\omega^2$. If the eigenvalues are $1,1,1$ then the minimal polynomial of $A$ is $x-1$, thus $A=I$ (why?). 
Therefore if $A\neq I$ the eigenvalues are $1,\omega,\omega^2$ and the trace is $0$.