Linear operator $T$ whose square is the average of $T$ and $T^*$ Having trouble with a question and I will be thankful for a hint.
Let $V$ be a an inner-product space with finite dimension over $\Bbb C$, $\space$ $T : V \to V$ linear transformation such that $T^2 =\frac12(T+T^*)$
A) Prove that $T$ is normal
B) Prove that $T^2-T=0$
So A was ok, since $T^*=2T^2-T$, and $T$ commutes with any polynomial in $T$, so $T$ is normal.
Having trouble with solving B, I've tried to use the fact that $T$ is diagonalizable and tried also some algebra techniques, but I feel like I'm missing something.
Sorry for my horrible English, I would appreciate any help.
Thanks!

Hint: applying the given equation to any eigenvector of eigenvalue $\lambda\in\Bbb C$, gives that the eigenvalue satisfies $\lambda^2=\frac12(\lambda+\overline\lambda)=\Re\lambda$. It is easy to see that the only solutions to that equation are real, and then $\lambda^2-\lambda=0$ follows.
And a diagonalisable operator, all of whose eigenvalues are roots of a given polynomial (here $X^2-X$), is itself annihilated by that polynomial.
