For a polynomail $p(x) = a_k x^k + \cdots + a_1 x +a_0 \in \Bbb C[x] $ let $\bar {p}(x) = \bar {a_k} x^k + \cdots +\bar{ a_1} x +\bar{a_0}$ Suppose $ T\in \mathcal L(V) $
Prove that for every polynomail $p(x) \in \Bbb C [x] $ that $p(x) \in \operatorname{Ann} (T) \iff \bar{p}(x) \in \operatorname{Ann} (T^{*}) $ we have $$ p(T)=a_k T^k+ \cdots a_1 T + a_0 I =0 \iff \bar{p}(T)=\overline {a_k T^k+ \cdots a_1 T + a_0 I} =0 $$ so $\bar{p}(T)=\bar {a_k} \bar T^k + \cdots +\bar{ a_1} \bar T +\bar{a_0} =0 $ now $\bar T= (T^{*})^{\intercal} $ so we have something like $\bar{p}(T)=\bar {a_k} ((T^{*})^{\intercal})^k + \cdots +\bar{ a_1} (T^{*})^{\intercal} +\bar{a_0} =0 $
i kinda want to take the transpose of both side but im not sure that makes anysense? also i think i need to write T as w.r.t an orthonormal basis but that shouldnt be a big deal.
Hint: $p(T)^* = \bar p(T^*)$ follows immediately from:
$(T^k)^* = (T^*)^k$ because $(AB)^* = B^* A^*$
$(aT^k)^* = \bar a(T^*)^k$ because $(aA)^* = \bar a A^*$
$(A+B)^* = A^* + B^*$