Matrix Revolution (2018)

We can find a nonsingular matrix $S$ such that $A = SDS^{-1}$ , where $D$ is the diagonal matrix $D \; = \; \mathrm{diag}(1,\zeta,\zeta^2,...,\zeta^{2017})$ where $\zeta = e^{\frac{2\pi i}{2018}}$ . Thus $A^k=SD^kS^{-1}$ for all integers $k$ , with $D^k \; = \; \mathrm{diag}(1,\zeta^k,\zeta^{2k},...,\zeta^{2017k})$ and hence $B = SES^{-1}$ , where $E$ is diagonal with $E_{jj} \; = \; \sum_{k=0}^{2017}\zeta^{(j-1)k} \; = \; \left\{ \begin{array}{lll} 2018 & \hspace{1cm} & j = 1 \\ 0 & & 2 \le j \le 2018 \end{array} \right.$ so that $E \; = \; \mathrm{diag}(2018,0,0,...,0)$ Thus $\mathrm{Tr}\,(B) = \mathrm{Tr}\,(E) = 2018$ , while $\mathrm{det}\,B = \mathrm{det}\,E = 0$ , making the answer $\boxed{20180}$ .

The characteristic polynomial of $q(f)$ is $\big(x-q(a_1)\big)^{n_1}\big(x-q(a_2)\big)^{n_2} \cdots \big(x-q(a_r)\big)^{n_r}$ , which is proved by finding the Jordan Canonical Form of the matrix representing $f$ .