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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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Moreover, please don't order us to show something. If this happens to be homework, then please add the homework tag and tell us what you tried, where you're stuck or at least what kind of thoughts you had on this problem. If it isn't then please add a bit of motivation and background to make it easier for a potential answerer to tell you something useful. –  t.b. May 20 '11 at 12:41
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
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
I am very sorry that I must put this in a homework,and user 8268 is absolutely right..actually I was unable to latex that notation. –  TIMP May 20 '11 at 13:38
@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

1 Answer 1

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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