Given $T^2=\frac{T+T^{\ast \:}}{2}$. Find all eigenvalues of $T$ and show that $T^2 = T$ I need help to show that $T^2 = T$
$T:V\to V\:$ is linear transformation over $F=\mathbb{R}\:or\:F=\mathbb{C}$
$V$ is a finite dimensional vector space.
First I think that it will help to show that $T$ is normal:
$T^2=\frac{T+T^{\ast \:\:}}{2}\:\Rightarrow \:T^{\ast }=2T^2-T\Rightarrow \:T^{\ast }T=2T^3-T^2=T\left(2T^2-T\right)=\:TT^{\ast }$
Now I manages to show that $\left(T^{\ast \:}\right)^2=T^2\:\:and\:\left(T^{\ast \:\:}\right)^3=T^3$
this helped me to observe that 
$T^3 - T^4 = 0 \implies T^3(T-I)=0$
Now, $if\:q\left(t\right)=t^3\left(t-1\right)\:then\:q\left(T\right)=0\:$ Since it is known that the minimal polynomial of T divides $q(t)$
therefore the possible eigenvalues of $T$ exists and equals to zero or one.
How I managed to show what are the eigenvalues of $T$:
let $v\in V$ be an eigenvector of $T$, assuming $\lambda $ is the corresponding 
eigenvalue.
Therefore 
$T^2v=\lambda \:^2v\:=\:\frac{1}{2}\left(T+T^{\ast }\right)v=\:\frac{1}{2}\left(\lambda \:v\:+\:\bar{\lambda \:}v\right)$
the last equality is because if $T$ is normal then $\lambda \:eigenvalue\:of\:T\:\Rightarrow \:\bar{\lambda }\:eigenvalue\:of\:T^{\ast }$
Now $\lambda \:\in \:\left\{0,1\right\}\:or\:\left\{1\right\}\:or\:\left\{0\right\}\:\Rightarrow \:\lambda \:\:\in \:\mathbb{R}\:\Rightarrow \lambda =\bar{\lambda }$
So we get that:
$\lambda ^2v=\lambda v\:\Rightarrow \lambda \left(\lambda -1\right)=0\:\Rightarrow \lambda \in \left\{0,1\right\}$
Now I struggle to show that $T^2 = T$ and need some help with that.

$$
         T^2 = \frac{T+T^*}{2}.
$$
Therefore $T^2=(T^*)^2$ follows from taking adjoints of both sides. Using this,
$$
       2T^2-T=T^* \\
     4T^4-4T^3+T^2=T^2 \\
      T^4-T^3=0 \\
      T^3(T-1)=0.
$$
$T$ is normal because $T$ commutes with $T^*=2T^2-T$. Therefore,
$$
       \mathcal{N}(T)=\mathcal{N}(T^2).
$$
Therefore, $T^3(T-I)=0$ implies $T(T-I)=0$, which is the desired result that $T^2=T$.
