Let \(V\) be the vector space of \(n \times n\) matrices over the field \(F\).
If \(B\) is a fixed \(n \times n\) matrix, define a function \(f_B\) on \(V\) by \(f_B(A) = \operatorname{trace}(B^t A)\). Show that \(f_B\) is a linear functional on \(V\).
Show that every linear functional on \(V\) is of the above form, i.e., is \(f_B\) for some \(B\).
Show that \(B \to f_B\) is an isomorphism of \(V\) onto \(V^*\).
Solution
For (a), linearity of trace gives, for \(A,C\in V\) and \(c\in F\), \[ f_B(cA+C)=\operatorname{tr}(B^t(cA+C)) =c f_B(A)+f_B(C). \] Hence \(f_B\) is linear.
For (b), let \(E_{ij}\) denote the matrix units. If \(f\in V^*\), define \(B=(b_{ij})\) by \[ b_{ij}=f(E_{ij}). \] For \(A=(a_{ij})\), \[ \operatorname{tr}(B^tA)=\sum_{i,j}b_{ij}a_{ij} =\sum_{i,j}a_{ij}f(E_{ij}) =f(A). \] Thus \(f=f_B\).
For (c), the map \[ \Phi:V\to V^*,\qquad B\mapsto f_B \] is linear by part (a), and it is surjective by part (b). If \(f_B=0\), then evaluating on \(E_{ij}\) gives \[ 0=f_B(E_{ij})=b_{ij} \] for all \(i,j\), so \(B=0\). Hence \(\Phi\) is injective as well. Therefore \(B\mapsto f_B\) is an isomorphism \(V\cong V^*\).