Credit to Sam Hopkins above and to about ten responses in this thread: Geometric interpretation of trace for priming me to this, but I think what I'm looking for is that a linear operator $T$ has trace zero if and only if it is a commutator $T = AB-BA$. It's easy to show that a commutator has trace zero, but the equivalence can come (pribably circularly) from the fact that $\mathfrak{sl}_n$ is semisimple and nonabelian for $n>1$, so its commutator subgroup is a nonzero subalgebra which is then forced to be all of $\mathfrak{sl}_n$. Thus every trace zero matrix is a commutator. This feels like it's categorical enough since it only references other linear maps, even if $A$ and $B$ are kind of non-canonical and auxilliary.

Credit to Sam Hopkins above and to about ten responses in this thread: Geometric interpretation of trace for priming me to this, but I think what I'm looking for is that a linear operator $T$ has trace zero if and only if it is a commutator $T = AB-BA$. It's easy to show that a commutator has trace zero, but the equivalence can come (probably circularly) from the fact that $\mathfrak{sl}_n$ is semisimple and nonabelian for $n>1$, so its commutator subgroup is a nonzero subalgebra which is then forced to be all of $\mathfrak{sl}_n$. Thus every trace zero matrix is a commutator. This feels like it's categorical enough since it only references other linear maps, even if $A$ and $B$ are kind of non-canonical and auxilliary.