$T^3=\frac{1}{2}(T+T^*) \rightarrow$ T is self adjoint 
Let $T$ be a normal transformation on a finite-dimensional Hilbert space; that is, $TT^*=T^*T$, where $T^*$ is the adjoint of $T$.

Prove that if $T^3=\frac{1}{2}(T+T^*)$, then $T$ is self adjoint.
I have tried to do some math on $(Tv,u)$ but I was not successful in proving the following: $(Tu,v)=(u,Tv)$ which is what I need for self-adjoint transformation.
Edit:
$(T^3,v)=\frac{1}{2}\left(\left(T+T^*\right),u \right)= \left( Tu,v\right)+  \left( T^*u,v\right)=\left( u,T^*v\right)+\left( u,Tv\right)=\left( u,T^3v\right)$
Therefore, $T^3$ is self adjoint. Does it mean that $T$ is self adjoint?
Thanks!
 A: It is not true that if $T$ is normal and $T^3$ is selfadjoint, then $T$ is selfadjoint. Example:
$$
T=\begin{bmatrix}e^{i\pi/3}&0\\0&e^{i\pi/3}\end{bmatrix}.
$$
Then $T^3=I$ (or, even when $n=1$ and $T=e^{i\pi/3}$).  
But here we know more. Since $T$ is normal, it is unitarily diagonalizable. So we may assume that $T$ is diagonal. We can then see that the equality $T^3=\text{Re}\,T$ will be satisfied by the eigenvalues of $T$. If 
$$\tag1
\lambda^3=\text{Re}\,\lambda
$$
and we write $\lambda=re^{it}$, we immediately get 
$$\tag2
r^3e^{3it}=r\cos t.
$$
There is the obvious solution $r=0$, i.e. $\lambda=0$. For $r\ne0$, the equality $(2)$ becomes
$$\tag3
r^2\cos 3t=\cos t$$ and $$\tag4 r^2\sin 3t=0.
$$
The equality $(4)$ gives us $t=k\pi/3$. And now we have 
$$\tag5
r^2=\frac{\cos k\pi/3}{\cos k\pi}=(-1)^k\cos k\pi/3.
$$For $k=0,1,2,3,4,5$, the right-hand-side in $(5)$ is 
$$
0,-\frac12,-\frac12,1,-\frac12,-\frac12.
$$
As $r^2>0$, the only acceptable values of $k$ are $0$ and $3$, which give $\lambda=1$ and $\lambda=-1$ respectively. 
In summary, the only possible eigenvalues of $T$ are $0,1,-1$, and thus $T$ (being normal) is selfadjoint.
A: $T^3=\frac{1}{2}(T+T^*)\Rightarrow T^4=\frac{1}{2}(T^2+TT^*)=\frac{1}{2}(T^2+T^*T)\text{ implies }TT^*=TT^*.$
Hence T is normal. Since $T$ is normal, there exists an orthonormal basis consists of its eigenvectors such that $T=UDU^*$, where $D$ is diagonal and $U$'s column vectors are $T$'s eigenvectors. $T$ is normal also implies $U^*(T^3)U=\frac{1}{2}(U^*TU+U^*T^*U)$ is a real diagonal matrix.
Let $\lambda=a+bi$ be one of the $T$'s eigenvalue, then $\lambda^3=\frac{1}{2}(\lambda+\bar{\lambda})\Rightarrow a^3-3ab^2+(3a^2b-b^3)i=a.$ Suppose $b\neq 0,$ then $b^2=3a^2\neq 0\Rightarrow 1=a^2-9a^2=-8a^2\Rightarrow a=\pm\frac{1}{2\sqrt{2}}i,$ leads to a contradiction. Hence $b=0$, i.e. $\lambda\in\mathbb{R}$, which means $T=T^*.$
