Is this a state on $M_n(\mathbb{C})?$

Question

Let $T \in M_n(\mathbb{C})$ be a positive matrix (i.e. $T=T^*$ and $\langle \xi, T\xi \rangle\geq 0$ for all $\xi \in \mathbb{C}^n$) with $Tr(T)=1$. Is the linear functional $$\omega_T: M_n(\mathbb{C}) \to \mathbb{C}: X \mapsto Tr(TX)$$

a state on $M_n(\mathbb{C})$?

I.e., if $X$ is a positive matrix, do we have $\omega_T(X) \geq 0$?

I tried calculating

$$\omega_T(X) = Tr(TX) = \sum_i \sum_j T_{ij}X_{ji}$$

and I want to somehow conclude that this last double sum is a positive real number, but I can't see how exactly this follows.

Answer 1

The answer is yes. As a hint as to showing that $\omega_T$ is indeed a state, there are two approaches. One uses the spectral decomposition of $T$, and another uses the fact that $T$ can be decomposed in to a product $T = M^*M$ for some matrix $M$.

It can also be shown, by the way, that every state over $M_n(\Bbb C)$ can be written in this form.