I have managed to show that $\chi_{T^*}=\overline{\chi_{T}}$ and now I want to show the same for the minimal polynomial i.e : $m_{T^*}=\overline{m_{T}}$ but Im having trouble doing so, any help will be appreciated!
Edit: Where $T^*$ is the adjoint of $T$
On
Suppose $f(X)=a_0+a_1X+\dots+a_nX^n$ is a polynomial and that $f(T)=0$; then, by applying $(-)^*$ we get $$ \overline{a_0}I+\overline{a_1}T^*+\dots+\overline{a_n}(T^*)^n=0 $$ If we denote $$ \bar{f}(X)=\overline{a_0}+\overline{a_1}X+\dots+\overline{a_n}X^n $$ we have proved that if $f(X)$ is satisfied by $T$ (that is, $f(T)=0$), then $\bar{f}$ is satisfied by $T^*$. Similarly, if $g(X)$ is satisfied by $T^*$, then $\bar{g}(X)$ is satisfied by $T$.
The minimal polynomial $m_T(X)$ for $T$ is the monic polynomial of minimal degree $f(X)$ satisfied by $T$. The above proves that $\overline{m_T}$ is a monic polynomial satisfied by $T^*$ and that no lower degree monic polynomial can be satisfied by $T^*$.
If $m_T=a_0+a_1x+\cdots+a_nx^n$, then$$a_0+a_1T+\cdots+a_nT^n=0$$and therefore$$(a_0+a_1T+\cdots+a_nT^n)^t=0.$$This means that$$a_0+a_1T^t+\cdots+a_n(T^t)^n=0.$$Applying connjugation, one gets that$$\overline{a_0}+\overline{a_1}T^*+\cdots+\overline{a_n}(T^*)^n=0.$$So, $T^*$ is a root of $\overline{m_T}$. So, the minimal polynomial of $T^*$ divides $\overline{m_T}$. But now you can apply the same argument starting from $m_{T^*}$. So, $m_{T^*}=\overline{m_T}$.