Let $\mathcal{A}$, $\mathcal{B}$ be two unital (non-commutative) $C^{*}$-Algebras. Then it is a well-known fact that in general not every positive element $\rho \in \mathcal{A} \otimes \mathcal{B}$ (i.e. there exists $x \in \mathcal{A} \otimes \mathcal{B}$ s.t. $\rho = x^{*} x$) admits a representation $$ \rho = \sum_{i=1}^{m} a_i^{*} a_i \otimes b_i^{*} b_i$$ where $a_i \in \mathcal{A}$, $b_i \in \mathcal{B}$, i.e. a sum of positive elements. Now I have read that if we restrict to the case where at least one of the two spaces is commutative, then, every positive element can be decomposed in the above way. I have already tried the following naive approach:
Assume $\mathcal{A}$ is commutative. For $\rho \geq 0$ there exists an element $x \in \mathcal{A} \otimes \mathcal{B}$ such that $\rho = x^{*} x$ with decomposition $$ x = \sum_{i=1}^{n} a_i \otimes b_i$$ But I don't see whether the mixed of the elementary tensors in the product $x^{*} x$ are positive and neither how I can use the commutativity of $\mathcal{A}$.
EDIT: Assume that $\mathcal{A}, \mathcal{B}$ are finite-dimensional to guarantee that every element in the tensor product space can be written as a finite sum of elementary tensors.
This can't even be true in the commutative setting. If every positive element of $A\otimes B$ was of the form $\sum a^*_ia_i\otimes b^*_ib_i$, then every $x\in A\otimes B$ is of the form $\sum a_i\otimes b_i$, i.e., $A\odot B=A\otimes B$. But this is not true in general, even in the commutative case.