$T^3=\frac{1}{2}(T+T^*) \rightarrow$ 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!

$T^3=\frac{1}{2}(T+T^*)\Rightarrow T^4=\frac{1}{2}(T^2+TT^*)=\frac{1}{2}(T^2+T^*T)\text{ implies }TT^*=TT^*.$
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 $T=T^*.$
