Let \(V\) be a finite-dimensional vector space with a non-degenerate symmetric bilinear form \(f\). Let \[ \mathfrak B=\{\alpha_1,\ldots,\alpha_n\} \] be a basis of \(V\), and let \[ \mathfrak B'=\{\alpha'_1,\ldots,\alpha'_n\} \] be the \(f\)-dual basis from Exercise 13, so that \[ f(\alpha_i,\alpha'_j)=\delta_{ij}. \] Let \(T\) be a linear operator on \(V\), and let \(T'\) be its adjoint relative to \(f\) as in Exercise 11.
Show that \[ [T']_{\mathfrak B'}=[T]_{\mathfrak B}^{\,t}. \]
Show that \[ \operatorname{tr}T=\operatorname{tr}T' =\sum_{i=1}^n f(T\alpha_i,\alpha'_i). \]
Solution
Write \[ [T]_{\mathfrak B}=(a_{ij}), \qquad [T']_{\mathfrak B'}=(c_{ij}). \] Thus \[ T\alpha_j=\sum_i a_{ij}\alpha_i, \qquad T'\alpha'_k=\sum_i c_{ik}\alpha'_i. \]
1. Compare the matrix coefficients of \(T\) and \(T'\).
Using the defining identity for the relative adjoint and the \(f\)-dual-basis relations, \[ \begin{aligned} c_{jk} &=f(\alpha_j,T'\alpha'_k)\\ &=f(T\alpha_j,\alpha'_k)\\ &=f\left(\sum_i a_{ij}\alpha_i,\alpha'_k\right)\\ &=\sum_i a_{ij}\delta_{ik}\\ &=a_{kj}. \end{aligned} \] Therefore \[ \boxed{[T']_{\mathfrak B'}=[T]_{\mathfrak B}^{\,t}.} \]
2. Deduce the trace identities.
Taking traces in step 1 gives \[ \operatorname{tr}T' =\operatorname{tr}([T']_{\mathfrak B'}) =\operatorname{tr}([T]_{\mathfrak B}^t) =\operatorname{tr}T. \]
Also, for each \(i\), \[ f(T\alpha_i,\alpha'_i) =f\left(\sum_j a_{ji}\alpha_j,\alpha'_i\right) =a_{ii}. \] Hence \[ \sum_i f(T\alpha_i,\alpha'_i) =\sum_i a_{ii} =\operatorname{tr}T. \] Combining the two equalities gives \[ \boxed{ \operatorname{tr}T=\operatorname{tr}T' =\sum_{i=1}^n f(T\alpha_i,\alpha'_i).} \]