Prove that $f: x \to Tx$ is positive on $\mathbb{C}^n$ iff $T$ has ony non-negative eigen values, for a complex $n\times n$ Hermitian matrix $T$.
To prove that $f$ is positive I need to show that $\langle fx, x\rangle \geq 0, \forall x\in \mathbb{C^n}$. It is equivalent to show that $ \langle Tx, x\rangle \geq 0$, so I think that this means that $ x' Tx \geq 0$ ($x'$ is the transpose of $x$), but I am not sure how to argue it. Then it will be easy because I just need to show that $T$ is positive semi-definite. Can someone help me to get this conclusion?
Let's do both directions separate.
Let us first note that every hermitian matrix has only real eigenvalues. If you don't know this fact, I will add a prove here.
If $f:x\mapsto Tx$ is positive, consider an eigenvalue $\lambda$ for an eigenvector $v$. We have $$0\leq\langle fv,v\rangle=\langle Tv,v\rangle=\langle \lambda v,v\rangle=\lambda \langle v,v\rangle.$$ Since $v\neq 0$, this shows that $\lambda \geq 0$.
Now let us assume that all eigenvalues are greater than or equal to zero. We can use the fact that every hermitian matrix is diagonalisable by a hermitian matrix. Thus take $A$ such that $A^{-1} T A=\overline{A}^T T A=D$, where $D$ is the diagonal matrix with entries the eigenvalues of $T$. $$\langle Dv,v\rangle=\langle \overline{A}^T T Av,v\rangle=\langle TAv,Av\rangle=\langle Tw,w\rangle $$ for $w=Av$. Since $A$ is invertible, it is bijective and we can write every vector $w$ as $Av$. So this shows that we can calculate the scalar product by using $D$ instead of $T$. But since $D$ has only real and positive eigenvalues, $\langle Dv,v\rangle \geq 0$.
If something remained unclear, feel free to comment.