Coordinate free proof that $\operatorname{trace}(A)=0\:\Longrightarrow\:A=BC-CB$

Question

As you probably know, the trace function on square matrices has the property that $$\operatorname{trace}(AB-BA)=0\,.$$ You might also know that the converse is true:
$$\operatorname{trace}(A)=0\;\text{ implies } A=BC-CB\:\text{ for some matrices } B\text{ and }C.$$

This is in fact true of linear operators on a vector space, it's a coordinate free fact.

BUT all proofs I'm aware of fix a basis and give a proof using coordinates.

So the question is, does anybody know a basis-free proof that does not use coordinates?

Thanks for any information - even if the information is that everything I've said here is wrong and I'm a complete idiot.

Answer 1

Using that $(A,B)\mapsto\operatorname{trace}(A^\top B)$ is a scalar product (or as suggested by @dsh that $(A,B)\mapsto\operatorname{trace}(A B)$ is a non-degenerate bilinear form) we can prove coordinate free that $$ \mathbf{span} \left\{ AB-BA : A,B\in\mathbb{R}^{d\times d} \right\} = \ker \operatorname{trace}.$$

Let $X\in\mathbb{R}^{d\times d}$ be orthogonal to all matrix commutators $AB-BA$ and $M\in\mathbb{R}^{d\times d}$. Then, for all $N\in\mathbb{R}^{d\times d}$, \begin{align*} \langle XM-MX, N\rangle &= \langle XM, N\rangle - \langle MX, N\rangle \\ &= \operatorname{trace}(N^\top XM) - \operatorname{trace}(N^\top MX) \\ &= \operatorname{trace}(M N^\top X) - \operatorname{trace}(N^\top MX) \\ &= \langle NM^\top - M^\top N,X\rangle \\ &= 0. \end{align*} This proves $XM=MX$. Since $X$ commutes with all matrices it must be a scalar multiple of the identity (see here for a coordinate free proof). In other words, \begin{align} \left\{ AB-BA : A,B\in\mathbb{R}^{d\times d} \right\}^\perp = \mathbb{R} I. \end{align} This extends to the generated subspace, which shows that it is a hyperplan. Being contained in the kernel of the trace it must be equal to it.

Disclaimer: this answer is incomplete since it remains to prove that the set of matrix commutators is a linear subspace, and in particular closed under addition. As a remark, this cannot be proved directly with some algebraic manipulations that only use the definition of the commutator: indeed, this does not necessarily hold for commutator rings (combine for instance Theorem 15 and Proposition 19 from Commutator rings, Mesyan, 2008).