Given that the eigen values of matrix $A_{5\times 5}$ are $1, \alpha, \alpha^2, \alpha^3, \alpha^4$ such that $\alpha=e^{i{2\pi}/3}$
then the trace of matrix $I+A+A^2$ is
a) $\ 0\quad$ b) $\ 3\quad$ c) $\ 5\quad$ d) $\ 15\quad$
My try:
Given that
$\alpha=e^{i2\pi/3}\implies \alpha^2=e^{-i2\pi/3}, \ \ \alpha^3=1$ &
$1+\alpha+\alpha^2=0$
Eigen value of $I+A+A^2$ corresponding to $1$ :
$=1+1+1^2$
$=3$
Eigen value of $I+A+A^2$ corresponding to $\alpha$ :
$=1+\alpha+\alpha^2$
$=0$
Eigen value of $I+A+A^2$ corresponding to $\alpha^2$ :
$=1+\alpha^2+\alpha^4$
$=1+\alpha^2+\alpha$
$=0$
Eigen value of $I+A+A^2$ corresponding to $\alpha^3$ :
$=1+\alpha^3+\alpha^6$
$=1+1+1$
$=3$
Eigen value of $I+A+A^2$ corresponding to $\alpha^4$ :
$=1+\alpha^4+\alpha^8$
$=1+\alpha+\alpha^2$
$=0$
Now,the trace of $I+A+A^2$ is
$\text{sum of eigen values}$
$=3+0+0+3+0$
$=6$
But, my answer does not match any option. My teacher says that option (c) $5$ is correct, but i don't know how. Please suggest me if i am wrong or help me solve this problem.
thanks