# The form of the states on an algebra of $n\times n$ matrices with complex entries

I have a question concerning $n \times n$ matrices.

Denote by $M_n(\mathbb{C})$ the algebra of all $n \times n$ matrices with complex entries. Let $\phi$ be a state on $M_n(\mathbb{C})$ i.e. a linear functional $\phi \colon M_n(\mathbb{C}) \rightarrow \mathbb{C}$ such that $\phi(Id)=1$ and $\phi(A^*A) \geq 0$, for each $A \in M_n(\mathbb{C})$.

Show that $\phi$ must be of the form $\phi(A) = \text{tr}(\rho A)$, where $\rho$ is a positive-definite matrix such that $\mbox{tr}(\rho)=1$.

Thank you for any help.

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This looks like a (homework) question. –  user2468 Mar 5 '12 at 2:50

If you fix matrix units $\{E_{kj}\}$, let $$\rho=\sum_{k,j}\phi(E_{jk})E_{kj}.$$ Then, for any $A=\sum_{kj}a_{kj}E_{kj}$, $$\text{tr}(\rho A)=\sum_{h}(\rho A)_{hh}=\sum_{h,k}\rho_{hk}a_{kh}=\sum_{h,k}\phi(E_{kh})\alpha_{kh}=\phi(A).$$ The positivity of $\phi$ implies that $\rho$ is selfadjoint (i.e. hermitian). We also have $\text{tr}(\rho)=\phi(\text{Id})=1$.
It only remains to see that $\rho$ has non-negative eigenvalues. Since $\rho$ is selfadjoint, we have $$\rho=\sum_{k}\lambda_kP_k,$$ where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $\rho$, and $P_k$ are rank-one pairwise orthogonal projections with sum $\text{Id}$. Since $\text{tr}(P_k)=1$ for all $k$, we have $$\lambda_k=\text{tr}(\rho P_k)=\phi(P_k)=\phi(P_k^*P_k)\geq0$$ by the positivity of $\phi$.