[proofplan] We compute the $(j,k)$-entry of the matrix of $\alpha^*$ with respect to the dual bases by evaluating $\alpha^*(\eta_k)$ at basis vectors, and show it equals $A_{kj}$, the $(k,j)$-entry of $A$. This gives the matrix of $\alpha^*$ as $A^\top$. [/proofplan] [step:Compute the matrix entries of $\alpha^*$ in dual bases] Let $\alpha(e_j) = \sum_{k=1}^m A_{kj}\,f_k$, so $A = (A_{kj})$ is the matrix of $\alpha$ with respect to $(e_j)$ and $(f_k)$. The matrix $B$ of $\alpha^*$ with respect to $(\eta_k)$ and $(\varepsilon_j)$ is defined by $\alpha^*(\eta_k) = \sum_{j=1}^n B_{jk}\,\varepsilon_j$. The coefficient $B_{jk}$ equals $\alpha^*(\eta_k)(e_j)$ (since the dual basis extracts coordinates). Compute: \begin{align*} B_{jk} = \alpha^*(\eta_k)(e_j) = \eta_k(\alpha(e_j)) = \eta_k\Bigl(\sum_{\ell=1}^m A_{\ell j}\,f_\ell\Bigr) = \sum_{\ell=1}^m A_{\ell j}\,\delta_{k\ell} = A_{kj}. \end{align*} Therefore $B_{jk} = A_{kj}$, which means $B = A^\top$. [/step]