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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.

Binxu Wang 王彬旭
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Gregory Grant
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  • https://math.stackexchange.com/questions/311580/coordinate-free-proof-of-operatornametrab-operatornametrba?rq=1 – Bumblebee Apr 12 '18 at 20:18
  • @Bumblebee Thanks but what I'm asking for here is the converse of that. – Gregory Grant Apr 12 '18 at 20:22
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    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. – user1551 Apr 12 '18 at 21:30
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    @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. – Gregory Grant Apr 12 '18 at 22:02
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    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. – Lukas Juhrich Jan 16 '19 at 00:20
  • If we have a coordinate free way to define the trace $\tau$ as a linear function that satisfies $\tau(XY) = \tau(YX)$ then isn't $\ker \tau = {XY - YX : X,Y \in F^{n \times n}}$ and the condition that $\tau(A)=0$ is equivalent to $A \in \ker \tau$? What am I missing? – CyclotomicField Apr 29 '20 at 16:38
  • @CyclotomicField well if we could prove that $\ker \tau = {XY - YX : X,Y \in F^{n \times n}}$ then the problem would be solved wouldn't it? And note that we can define trace as the sum of eigenvalues and prove that the linearity and $\tau(XY) = \tau(YX)$ in a coordinate free way. It is determining the $ker \ \tau$ that is our main issue. – Matin Yousefi Jun 23 '21 at 04:06
  • How do yu define the trace without coordinates? – Paul Frost Jan 10 '24 at 09:11
  • @PaulFrost, one way is the following: Denote the space of linear maps from $V$ to $V$ by $V\otimes V^$. Its dual is naturally isomorphic to $V^\otimes V$, which is the space of linear maps from $V^$ to $V^$. In particular, the identity map $I: V^* \rightarrow V^$ is in the latter space. For any $A \in V\otimes V^$, define the trace of $A$ to be $\langle A, I\rangle.$ – Deane Jan 10 '24 at 14:47
  • @PaulFrost, another way can be found in the Algebra section of this Princeton oral exam: https://web.math.princeton.edu/generals/alpoge_levent. Any rank $1$ $A:V\rightarrow V$ can be written as $A=v\otimes\ell$, where $\ell\in V^$. Define the trace of $A$ to be $\langle\ell,v\rangle$. This can be extended uniquely to a bilinear map of $V\otimes V^$. – Deane Jan 10 '24 at 14:52
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    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 Jan 10 '24 at 14:55
  • @PaulFrost Over $\mathbb{R}$ you can define it as the sum of eigenvalues with multiplicities, or as the differential of the determinant at $I_d$. – reded Jan 11 '24 at 17:35

1 Answers1

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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).

reded
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  • (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. – dsh Jan 10 '24 at 14:33
  • Thanks for your relevant contribution – reded Jan 11 '24 at 13:50