Linear functional on matrix space, nonnegative on positive semidefinite matrices 
Let $f:M_n(\mathbb C) \to \mathbb C$ be a linear function such that $f(x^* x)\ge0$ for all  $x$ and $f(1)=1$. Show that there exist $\alpha_1,...,\alpha_k\in \mathbb C^n$ such that $f(x)=\sum_{i=1}^{k}\langle x\alpha_i,\alpha_i \rangle$ for all $x\in M_n(\mathbb C)$.

 A: Slightly expanding the comment-answer by user8268:


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*Every linear function $f:M_n(\mathbb C)\to \mathbb C$ is of the form $f(X)=\operatorname{tr }(XA)$ for some matrix $A\in M_n(\mathbb C)$. (Observe that $\operatorname{tr }(XA) = \sum_{ij}X_{ij}A_{ji}$ and recall the general form of linear functional on a finite dimensional vector space.) 

*Take $\alpha\in \mathbb C^n$. The matrix $\alpha\otimes \alpha^*$ (outer product) is positive definite, hence $f$ is nonnegative on it. Since $\operatorname{tr }(\alpha\otimes \alpha^* A) = \alpha^*A\alpha$, the nonnegativity of this expression for all $\alpha\in \mathbb C^n$ implies that $A$ is positive semidefinite. (Here it is important that we work over complex numbers; in the real case  $A$ could be non-symmetric.)

*A positive definite matrix can be diagonalized and therefore written in the form $A=\sum_{j=1}^n \alpha_j\otimes \alpha_j^*$. Hence $f(X)=\sum_{j=1}^n\operatorname{tr }(X\alpha_j\otimes \alpha_j^*) = \sum_{j=1}^n \alpha_j^* X\alpha_j$ as required.

