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

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.

• math.stackexchange.com/questions/311580/… Commented Apr 12, 2018 at 20:18
• The proof in Albert and Muckenhoupt's 1957, paper, On matrices of trace zeros, e.g., made use of rational canonical form. It is not "coordinate-free", but the basis they chose was somewhat intrinsic and wasn't entirely arbitrary. So, that at least qualifies as a "basis-independent" proof. Commented Apr 12, 2018 at 21:30
• @user1551 Thank you that's a good reference, I was aware of the ratoinal canonical form proof and I see what you mean if the basis is natural then perhaps you can argue it's coordinate free. But I really want a proof that does not appeal to a basis in any way. Perhaps no such proof is possible but it seems any statement about transformations on a vector space that has nothing to do with a basis, should be provable without appealing to the concept of a basis. But perhaps that's not the case. Commented Apr 12, 2018 at 22:02
• If we knew that the lie algebra $\mathfrak{gl}(V)=: L$ does not have any proper ideals greater than $L^{(1)}$, we would be finished: Regarding our field $k$ as an abelian lie algebra, the trace is a lie algebra homomorphism ($tr([AB]) = 0 = [tr(A) tr(B)]$). Its kernel must thus be an ideal containing $L^{(1)}$, but cannot be the whole space, since the trace is nontrivial. But I have no idea how to prove the premise in a “coordinate-free” manner. Commented Jan 16, 2019 at 0:20
• Implication is not true for $\mathfrak{sl}(2, \mathbb{F}_2),$ traceless $2\times 2$ matrices over field $\mathbb{F}_2=\{0, 1\}.$ Which part of Lie algebra theory (directly related to question) can be developed "coordinate-free" is disputable matter. For field $k$ of characteristic 0, the basic fact for semisimple $\mathfrak{sl}(n, k)$ finite dimensional Lie algebra is that $[\mathfrak{sl}(n, k), \mathfrak{sl}(n, k)] = \mathfrak{sl}(n, k)$ relying on Cartan's criterion, Casimir element and enveloping algebra. Shortcut proof would be appreciated.
– dsh
Commented Jan 10 at 14:55

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.
• (Fixed) You do not need scalar product $(A,B)\mapsto\operatorname{trace}(A^\top B),$ non-degenerate bilinear form $(A,B)\mapsto\operatorname{trace}(AB)$ is sufficient.