showing that map $f$ is positive 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?
 A: 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.
