Opgave 3b. Suppose \(A:U\to V\) with \(U\) and \(V\) finite dimensional vector space with ordered basis \(E:=(e_1,...,e_n)\) for \(U\) and \(F:=(f_1,...,f_m)\) for \(V\). Prove that matrix of \(A^t\) with respect to the basis \(F^*\) and \(E^*\) is the transpose of the matrix of \(A\) with respect to \(E\) and \(F\).
Bewijs.

Observe that \(([A]_E^F)_{ij}=f_i^*Ae_j\) and \(([A^t]_{F^*}^{E^*})_{ij}=e_i^{**}A^tf_j^*\). And that \(([A]_E^F)^\top_{ij}=f_j^*Ae_i\). We need to show that: \[e_i^{**}A^tf_j^*=f_j^*Ae_i\] By definition \(f_j^*Ae_i=A^t(f_j^*)(e_i)\), so we need to show that: \[e_i^{**}(A^tf_j^*)=A^t(f_j^*)(e_i)\] We can prove this equality if we can show that: \[e_i^{**}(\phi)=\phi(e_i) \quad \forall \phi \in E^*\] We only need to check the equality for the basis vectors \((e_1^*,...,e_n^*)\). This equality follows now as: \[e_i^{**}(e_j^*)=\delta_{ij}=\delta_{ij}=e_j^*(e_i)\]