If $W=M_{n\times n}(\mathbb F)$ and $f$ is a linear functional on $W$ such that $f(AB)=f(BA)\;\forall A,B\in W$, and $f(I)=n$, then $f$ is the trace function.
I have tried to generate useful matrices that can be represented in the form $AB-BA$ for some $A$ and $B$. My idea was to generate a basis for $W$ of the form $AB-BA$ plus the identity.
Consider the matrix $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. We want to show that necessarily $f(A)=0$. Let $B=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. Now $BA=A$ and $AB=0$, so by linearity of $f$ we have $0=f(0)=f(AB)=f(BA)=f(A)$. The same works for any matrix with a single $1$ off the diagonal in any dimension. Thus, by linearity, $f(A)$ only depends on the diagonal entries of $A$.
Let us then suppose $A$ is invertible and choose $B=A^{-1}C$ for any square matrix $C$. Then the condition $f(AB)=f(BA)$ becomes $f(C)=f(A^{-1}CA)$. If we choose $A$ to be a permutation matrix and let $C$ be a diagonal matrix, we see that $f$ is invariant under permutations of the diagonal elements. By linearity, $f(\text{diag}(c_1,\dots,c_n))=\lambda(c_1+\dots+c_n)$ for some constant $\lambda$. The normalization then implies that $f$ is the trace.