$T^3=\frac{1}{2}(T+T^*) \rightarrow$ 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!
It is not true that if $T$ is normal and $T^3$ is selfadjoint, then $T$ is selfadjoint. Example:
$$
T=\begin{bmatrix}e^{i\pi/3}&amp;0\\0&amp;e^{i\pi/3}\end{bmatrix}.
$$
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 $r^2&gt;0$, 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.
We have $2A^3 =A+A^*$ hence $A+A^* =(2A^3)^*=2(AAA)^* =2(A^* )^3$ thus the operator $A^3 $ is self-adjont. And therefore the operator $A=f(A^3 ) $ where $f(x) =\sqrt[3]{x} $ is self - adjont.