Let $R$ be a commutative ring and $M_n(R)$ denote the ring of $n\times n$ matrices with coefficients in $R$. For $X,Y\in M_n(R)$, their commutator $[X,Y]$ is defined by

$$[X,Y] := XY – YX.$$ The trace of any matrix is defined as the sum of its diagonal entries.

If $X$ and $Y$ are any matrices, what is the trace of $[X,Y]$? It's zero! That's because the trace of $XY$ is the same as the trace of $YX$. Therefore:

Any commutator has trace zero.

What about the converse? Is any trace zero matrix also a commutator? In other words, given a trace zero matrix $Z\in M_n(R)$, can we find matrices $X$ and $Y$ such that $Z = [X,Y]$? Albert and Muckenhoupt proved that you can, assuming that $R$ is a field.

What happens if you also want $X$ and $Y$ to have trace zero?

Good question. In general, this is not possible. For example, let's consider the simplest field of all, the field with two elements denoted by $\F_2$. Okay, it's actually debatable whether $\F_2$ really is the simplest field, because so many problems happen in characteristic two. For example, this problem we've been considering: in $\F_2$, if $X$ and $Y$ are $2\times 2$ matrices of trace zero, then $[X,Y]$ will have zero off-diagonal entries. So for example, the matrix

$$\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}\in M_2(\F_2)$$ cannot be written as a commutator of two matrices with trace zero.

It seems that characteristic two is the only obstruction, in the case of $2\times 2$ matrices. In fact, Alexander Stasinski proved in his paper [1] the following:

**Theorem.** Let $R$ be a principal ideal domain. If $n\geq 3$ then any matrix in $M_n(R)$ of trace zero can be written as the commutator of two matrices in $M_n(R)$, each having trace zero. The same holds for $n=2$ if two is invertible in $R$.

Notice how the characteristic two problem only happens in the $2\times 2$ case.

[1] Stasinski, A. Isr. J. Math. (2018). https://doi.org/10.1007/s11856-018-1762-5