Discussion
Let \(\alpha = e^{2\pi i/5}\) and the matrix
\[ M = \begin{bmatrix} 1 & \alpha & \alpha^{2} & \alpha^{3} & \alpha^{4} \\ 0 & \alpha & \alpha^{2} & \alpha^{3} & \alpha^{4} \\ 0 & 0 & \alpha^{2} & \alpha^{3} & \alpha^{4} \\ 0 & 0 & 0 & \alpha^{3} & \alpha^{4} \\ 0 & 0 & 0 & 0 & \alpha^{4} \end{bmatrix}. \]
Then the trace of the matrix \(I + M + M^{2}\) is
Answer: D
The trace is 5.
\(M\) is upper triangular, so its diagonal entries \(1, \alpha, \alpha^{2}, \alpha^{3}, \alpha^{4}\) are its eigenvalues, and \(M^{2}\) is upper triangular with diagonal \(1, \alpha^{2}, \alpha^{4}, \alpha^{6}, \alpha^{8}\).
\[ \begin{aligned} \operatorname{tr} I &= 5 \\ \operatorname{tr} M &= 1 + \alpha + \alpha^{2} + \alpha^{3} + \alpha^{4} = 0 \\ \operatorname{tr} M^{2} &= 1 + \alpha^{2} + \alpha^{4} + \alpha^{6} + \alpha^{8} = 0 \end{aligned} \]
Both sums vanish because \(\alpha\) is a primitive fifth root of unity: the five fifth roots of unity sum to zero, and since 2 is prime to 5, the powers \(\alpha^{2k}\) run through the same five roots. So \(\operatorname{tr}(I + M + M^{2}) = 5 + 0 + 0 = 5\). The values 0 and 3 forget the identity or count only part of it, and -5 has the wrong sign.
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