Prove that $A^5 \neq I$ $A \in M_{5}(\mathbb{C})$, $\operatorname{trace}(A) = 0$, $I-A$ is invertible. Prove that $A^5 \neq I$
I think this problem has something to do with eigenvalue and the fact that trace of a matrix = sum of the eigenvalues. But, I have no idea how to proceed!
Thanks! 
 A: Assume, by the way of contradiction, that $A^5 = I$ holds. In this case, as $A^5 - I = 0$, we can factorize this expression in $(A - I)(A^4 + A^3 + A^2 + A + I) =0 $. 
As $A - I $ is invertible, the first factor is nonzero, hence the last one is and the minimal polynomial of $A$ divides it, forcing  $A$ to have all eigenvalues being fifth roots of 1, but distinct from 1 because  $A-I $ is invertible. Now, since  $tr (A) = 0$, the sum of all eigenvalues must be zero. Can you see now that, no matter what is the multiplicity of each eigenvalue, if no eigenvalue is 1 it is impossible to sum them all and have 0? It leads to a contradiction, then.
A: Suppose you had $A^5 = I$. Then the eigenvalues of $A$ are all fifth roots of unity. The fact that $I - A$ is invertible means that $1$ is not an eigenvalue. So you're left with the other $4$ fifth roots of unity. 
You're given that the trace of $A$ is zero, so the sum of all eigenvalues of zero. So if you can verify that there's no way of adding the remaining fifth roots of unity to get zero then you're done. But any such way would result in an equation
$$c_3e^{8\pi i \over 5} + c_2e^{6\pi i \over 5} +c_1e^{4\pi i \over 5} + c_0e^{2\pi i \over 5} = 0$$
Here the $c_i$ are integers. So $z = e^{2\pi i \over 5}$ satisfies 
$$c_3z^3 + c_2z^2 + c_1z + c_0 = 0 \tag{*}$$
But $x^4 + x^3 + x^2 + x + 1$ is the minimal polynomial of $z$ ( since cyclotomic polynomials are irreducible.) Hence $(*)$ cannot happen and thus the assumption that $A^5 = I$ leads to a contradiction.
