Uniqueness of f(AB) = f(BA)
Introduction It is easily verified that the trace of a matrix has the property $\operatorname{tr}(AB) = \operatorname{tr}(BA)$. Now I will attempt to prove that trace is the unique linear functional upto scaling which satisfy this property. Proof Assume $f$ is a linear functional on $M_n(\mathbb{C})$ such that $f(AB) = f(BA)$. Consider $E_{ij} \in M_n(\mathbb{C})$, the matrix with $e_{ij} =1$ and all the rest of elements $0$. It can be shown that $E_{ii} E_{ij} = E_{ij}$ and $E_{ij}E_{ii} = 0$, the latter when $i \neq j$. Hence $f(E_{ij}) = f(E_{ii}E_{ij}) = f(E_{ij}E_{ii}) = f(0) = 0$ whenever $i \neq j$. ...