$T^2 = T$ and $T$ is normal implies $T$ is hermitian 
Let $T:V\to V$ ($V$ is finite dimensional), a normal linear-operator such that $T^2=T$. Show that $T$ is hermitian.

So I know that if $T$ is normal then $T$ is hermitian iff the roots of $f_T(x)$ are real.
I also figured out that $T^2 = T$ implies $T(c)\in\mathbb{R}.\forall c\in\mathbb{C}$.
How can I deduce that indeed the roots are real?

If $T^2=T$, then the only eigenvalues of $T$ are $0$ and/or $1$. If $T$ is normal, then it is unitarily diagonalizable. If $T$ is unitarily diagonalizable with real eigenvalues, then it is Hermitian.
This might be an overkill proof, but it gets the job done. Note that this requires $V$ to be finite dimensional.
