I'm trying to prove $\{ AB - BA : A, B \in \mathbb{F}^{n \times n} \} = \{ A \in \mathbb{F}^{n \times n} : \operatorname{tr} A = 0 \}$.
Showing $\{ AB - BA : A, B \in \mathbb{F}^{n \times n} \} \subseteq \{ A \in \mathbb{F}^{n \times n} : \operatorname{tr} A = 0 \}$ is straightforward but I'm having trouble proving the reverse implication.
Define $M_{ij} = (m_{kl})_{n \times n}$ where $m_{kl} = \begin{cases} 1, & k = i \:\land\: l = j \\ 0, & \text{otherwise}\end{cases}$
We can show $$M_{ik}M_{kj} - M_{kj}M_{ik} = \begin{cases} M_{ij}, & i \ne j \\ M_{ii} - M_{kk}, & \text{otherwise} \end{cases}$$ All matrices with trace zero can be decomposed into a matrix with all diagonal elements being zero and another matrix equal to $\operatorname{diag}(d_1, d_2, \dots, d_n)$ satisfying $d_1 + d_2 + \dots + d_n = 0$ and hence we can construct a basis set for $\{ A \in \mathbb{F}^{n \times n} : \operatorname{tr} A = 0 \}$ as
$$\begin{align}\mathcal{B} & = \{ M_{ij} : 1 \le i, j \le n, i \ne j \} \cup \{ M_{ii} - M_{nn} : 1 \le i \le n-1 \} \\ & = \{ M_{in}M_{nj} - M_{nj}M_{in} : 1 \le i, j \le n \} \\ & \subseteq \{ AB - BA : A, B \in \mathbb{F}^{n \times n} \}\end{align}$$
If I can manage to show $\mathbb{V} = \{ AB - BA : A, B \in \mathbb{F}^{n \times n} \}$ is a vector space, I can further say $\{ A \in \mathbb{F}^{n \times n} : \operatorname{tr} A = 0 \} = \operatorname{span}(\mathcal{B}) \subseteq \mathbb{V}$ but I can't find a way to show $\mathbb{V}$ is closed under addition.
