trace of the matrix $I + M + M^2$ is 
Let $ \alpha = e^{\frac{2\pi \iota}{5}}$  and the matrix 
  $$ M= \begin{pmatrix}1 & \alpha & \alpha^2 & \alpha^3 & \alpha^4\\ 
0 & \alpha & \alpha^2 & \alpha^3 & \alpha^4\\
0 & 0 & \alpha^2 & \alpha^3 & \alpha^4 \\
0 & 0 & 0 & \alpha^3 & \alpha^4\\
 0 & 0 & 0 & 0 & \alpha^4 \end{pmatrix}$$
Then the trace of the  matrix $I + M + M^2$ is
  
  
*
  
*$-5$;
  
*$0$;
  
*$3$;
  
*$5$.
  

I am stuck on this problem. Can anyone help me please?
I got trace of the matrix $$\operatorname{tr}(I+M+M^2) = 7 + \alpha + 2  \alpha^2   + \alpha^3 + 2 \alpha^4 + \alpha^6 +\alpha^8.$$ Now what to do?

Note that the trace of $M$ is $0$, since $1+\alpha+\alpha^2+\alpha^3+\alpha^4= 0$.
Also $M$ is upper triangular so that $M^2$ has diagonal elements which are just the square of the diagonal elements of $M$, i.e. $1,\alpha^2, \alpha^4, \alpha^6, \alpha^8$.
Using the fact that $\alpha^5 = 1$ we see that the trace of $M^2$ is again $0$.
Thus tr$(I+M+M^2)$ = tr$(I)$ = 5.
