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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 $x\in M_n(\mathbb C)$.

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Is $x*x$ just multiplication of matrices? Is $(\alpha_i,\alpha_i)$ inner product on ${\bf C}^n$? – Gerry Myerson May 20 '11 at 12:46
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I believe $x*x$ wants to be $x^*x$ and $x(\alpha_i,\alpha_i)$ wants to be $(x\alpha_i,\alpha_i)$. The function $f$ is probably supposed to be linear. Any linear $f$ is of the form $f(X)=Tr(XA)$ for some matrix $A$ and the inequality forces $A$ to be positive semidefinite, hence the statement is true. The condition $f(1)=1$ is not used. – user8268 May 20 '11 at 13:09
    
yes,absolutely right... – TIMP May 20 '11 at 13:36
    
x star is the complex conjugate of x and t that is just inner product. – TIMP May 20 '11 at 13:39
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@user10805: No problem :) no need to feel bad. I edited accordingly. Is it okay now? By the way: write @username if you want to notify someone. – t.b. May 20 '11 at 13:43

Slightly expanding the comment-answer by user8268:

  1. 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.)
  2. 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.)
  3. 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.
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