From my limited understanding, strictly positive elements in a $C^*$-algebra $\mathcal{A}$ can be defined in one of two (equivalent) ways:
(1) $x \in \mathcal{A}_+$ is strictly positive provided $\overline{x \mathcal{A}x} = \mathcal{A}$
(2) $x^\ast = x$ is strictly positive provided $\phi (x) > 0$ for every state $\phi$ on $\mathcal{A}$.
In what follows, let's assume that $\mathcal{A}$ is unital and has tracial states. I am wondering if there is any connection between tracial states, strictly positive elements, and (sums) of commutators. E.g., perhaps something like the following holds:
Let $x^\ast = x$. If $\tau (x) > 0$ for every tracial state $\tau$ on $\mathcal{A}$, then $x = p + c$ for some strictly positive element $p$ and some $c$ which is a (sum) of commutator(s).
Of course, I am not committed to the above being true, although it would be nice if it were; I am just wondering if something like it is true. `
EDIT
Perhaps this MO post is relevant: https://mathoverflow.net/questions/66343/commutators-in-the-reduced-c-algebra-of-the-free-group The answers there suggest that a sum of commutators should be replaced a limit of sums of commutators (no?).
Also, I should add that I am also interested in case of tracial von Neumann Algebras (vNAs); e.g., $II_1$ factors. I know all vNAs are $C^\ast$-algebras, but I've been told that, in general, it isn't particularly helpful to think of vNAs as $C^\ast$-algebras. So perhaps the case of vNAs makes the question more manageable or easier to interpret (I know my question is somewhat vague, and I apologize for that)
EDIT:
A day or two ago someone posted a comment on this question that went roughly like the following:
The answer is, unfortunately, unexciting: let $p$ be any strictly positive element. Then $x = \frac{\tau (x)}{\tau (p)} p + (x - \frac{\tau(x)}{\tau (p)}p)$
My question is, why was this comment deleted? I can't see anything wrong with it. The above decomposition shows that $x$ can be written as (any) strictly positive element plus a norm limit of sums of commutators.