Show that the trace functional on \(n \times n\) matrices is unique in the following sense. If \(W\) is the space of \(n \times n\) matrices over the field \(F\) and if \(f\) is a linear functional on \(W\) such that \(f(AB) = f(BA)\) for each \(A\) and \(B\) in \(W\), then \(f\) is a scalar multiple of the trace function. If, in addition, \(f(I) = n\), then \(f\) is the trace function.
Solution
Let \(E_{ij}\) denote the standard matrix unit. Put \[ c=f(E_{11}). \] We prove that \(f(A)=c\operatorname{tr}(A)\) for every \(A\in W\).
1. The functional vanishes on every off-diagonal matrix unit.
Fix \(i\ne j\). Since \[ E_{ii}E_{ij}=E_{ij}, \qquad E_{ij}E_{ii}=0, \] the hypothesis \(f(AB)=f(BA)\) gives \[ f(E_{ij})=f(0)=0. \] Thus \(f(E_{ij})=0\) whenever \(i\ne j\).
2. The functional has the same value on every diagonal matrix unit.
For any \(i,j\), \[ E_{ij}E_{ji}=E_{ii}, \qquad E_{ji}E_{ij}=E_{jj}. \] Applying the defining identity to \(A=E_{ij}\) and \(B=E_{ji}\) yields \[ f(E_{ii})=f(E_{jj}). \] Hence every diagonal matrix unit has value \(c\).
3. Therefore \(f\) is a scalar multiple of trace.
Every matrix \(A=(a_{ij})\) has the expansion \[ A=\sum_{i,j}a_{ij}E_{ij}. \] By linearity and steps <1>1–<1>2, \[ f(A)=\sum_{i,j}a_{ij}f(E_{ij}) =c\sum_i a_{ii} =c\operatorname{tr}(A). \] This proves the asserted uniqueness up to scalar multiple [@HK71].
4. The normalization \(f(I)=n\) forces \(c=1\).
Since \(I=\sum_{i=1}^nE_{ii}\), \[ f(I)=nc. \] The hypothesis says \(nc=n\). Hoffman–Kunze works over a subfield of \(\mathbb C\) unless otherwise specified, so the characteristic is zero and \(n\ne0\) in \(F\) [@HK71]. Hence \(c=1\), and step 3 gives \[ f=\operatorname{tr}. \]