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! 

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.
