Let \(n\) be a positive integer, and let \(V\) be the space of all \(n \times n\) matrices over the field of complex numbers. Show that the equation \[ f(A, B) = n\operatorname{tr}(AB) - \operatorname{tr}(A)\operatorname{tr}(B) \] defines a bilinear form \(f\) on \(V\). Is it true that \(f(A, B) = f(B, A)\) for all \(A, B\)?
Solution
Goal. Show \(f(A,B) = n\operatorname{tr}(AB) - \operatorname{tr}(A)\operatorname{tr}(B)\) is bilinear, and check symmetry.
1. \(f\) is bilinear.
1.1. \(\operatorname{tr}(AB)\) is bilinear in \((A, B)\).
\(\operatorname{tr}((A_1 + A_2)B) = \operatorname{tr}(A_1 B) + \operatorname{tr}(A_2 B)\) and \(\operatorname{tr}(A(B_1 + B_2)) = \operatorname{tr}(AB_1) + \operatorname{tr}(AB_2)\), and \(\operatorname{tr}(cAB) = c\operatorname{tr}(AB)\).
1.2. \(\operatorname{tr}(A)\operatorname{tr}(B)\) is bilinear.
\(\operatorname{tr}\) is linear, and the product of two linear functionals is bilinear.
1.3. Hence \(f = n\operatorname{tr}(AB) - \operatorname{tr}(A)\operatorname{tr}(B)\) is bilinear.
a linear combination of bilinear forms is bilinear.
2. \(f\) is symmetric.
2.1. \(\operatorname{tr}(AB) = \operatorname{tr}(BA)\).
the trace is cyclic: \(\operatorname{tr}(AB) = \sum_i (AB)_{ii} = \sum_{i,j} A_{ij} B_{ji} = \sum_{j,i} B_{ji} A_{ij} = \operatorname{tr}(BA)\).
2.2. \(\operatorname{tr}(A)\operatorname{tr}(B) = \operatorname{tr}(B)\operatorname{tr}(A)\).
multiplication of scalars is commutative.
2.3. Hence \(f(A, B) = n\operatorname{tr}(AB) - \operatorname{tr}(A)\operatorname{tr}(B) = n\operatorname{tr}(BA) - \operatorname{tr}(B)\operatorname{tr}(A) = f(B, A)\).
2.1 and 2.2.
3. Q.E.D.
1 shows bilinearity; 2 shows \(f(A,B) = f(B,A)\) (yes, it is symmetric).