Positive elements in $M_{n}(C(X))$

Question

Let $X$ be a compact Hausdorff space and $C(X)$ the $C^{*}$-algebra of continuous functions on $X$. Then $M_{n}(C(X))$ is a $C^{*}$-algebra which can be viewed as $M_{n}(\mathbb{C})$-valued functions on $X$. If $F\in M_{n}(C(X))$ is positive then $F=GG^{*}$ for some $G\in M_{n}(C(X))$ and it follows easily that $F(x)\geq 0$ for each $x\in X$. How do I show that the converse is true, i.e., if $F(x)\geq 0$ for all $x\in X$, then $F\geq 0$?

Answer 1

Your question is answered by the following fact:

Let $A$ be a unital $C^*$-algebra and $X$ a compact Hausdorff space. Assume that $F \in C(X,A)$ satisfies $F(x)\in A^+$ for all $x\in X$, then $F\in C(X,A)^+$.

Proof 1: Clearly $F$ is self-adjoint, so it suffices to show that the spectrum of $F$ is positive. Note that $F-\lambda1$ is invertible in $C(X,A)$ if and only if $F(x)-\lambda 1_A$ is invertible in $A$ for every $x\in X$. In other words, $$\sigma_{C(X,A)}(F) = \bigcup_{x\in X} \sigma_A(F(x)).$$ By assumption $\sigma_A(F(x))\subseteq [0, \infty[$ for all $x\in X$, so that $\sigma_{C(X,A)}(F)\subseteq [0, \infty[$ as well.

Proof 2: Define $G: X \to A: x \mapsto F(x)^{1/2}$. Then $G$ is continuous and $F= G^*G\ge 0.$