Suppose $V$ is an inner product space and $T$ $\in \mathcal L(V)$. Suppose $$a_0+a_1z+a_2z^2+\cdots+a_{m-1}z^{m-1}+z^m$$ is the minimal polynomial of $T$. Prove that$$\overline{a_0}+\overline{a_1}z+\overline{a_2}z^2+\cdots+\overline{a_{m-1}}z^{m-1}+z^m$$is the minimal polynomial of $T^*$.
Since the first polynomial is a minimal polynomial, we have$$a_0+a_1T+a_2T^2+\cdots+a_{m-1}T^{m-1}+T^m=0$$ I try to use the fact that $M(T)=(\overline{M(T^*)})^t$.
Can anyone give me a hint or tell me what should I consider next? I am stuck here..
Recall that the minimal polynomial of $X\in\mathcal L(V)$ is the monic polynomial $\mu_X(t)$ of least degree that annihilates $X$.
Also recall that the computation $$ \overline{p}(X) = \overline{a_0}I+\overline{a_1} X+\dotsb+\overline{a_n}X^n = (a_0I+a_1X^*+\dotsb+a_n(X^*)^n)^* = p(X^*)^* $$ proves that $\overline{p}(X)=p(X^*)^*$ for any $X\in\mathcal L(V)$ and any polynomial $p(t)$.
Now, you wish to show that $\mu_{T^*}(t)=\overline{\mu_T}(t)$. To do so, suppose $p(t)$ annihilates $T^*$. Then $$ \overline{p}(T)=p(T^*)^*=0^*=0 $$ so $\mu_T(t)$ divides $\overline{p}(t)$. It follows that $\overline{\mu_T}(t)$ divides $p(t)$.
That is, $\overline{\mu_T}(t)$ is a monic polynomial that annihilates $T^*$ and divides every other polynomial that annihilates $T^*$. Hence $\mu_{T^*}(t)=\overline{\mu_T}(t)$ as desired.