[proofplan] We compute the $(i,j)$-entry of the new matrix $B$ directly from the definition $B_{ij} = \psi(v_i, w_j)$, expand using the change-of-basis relations and bilinearity, substitute $A_{k\ell} = \psi(e_k, f_\ell)$, and recognise the result as the $(i,j)$-entry of the matrix product $P^\top A Q$. [/proofplan] [step:Expand $B_{ij} = \psi(v_i, w_j)$ using bilinearity and the change-of-basis relations] By definition, $B_{ij} = \psi(v_i, w_j)$. Substituting $v_i = \sum_{k=1}^{n} P_{ki}\, e_k$ and $w_j = \sum_{\ell=1}^{m} Q_{\ell j}\, f_\ell$, and using bilinearity of $\psi$: \begin{align*} B_{ij} = \psi\!\left(\sum_{k=1}^{n} P_{ki}\, e_k,\; \sum_{\ell=1}^{m} Q_{\ell j}\, f_\ell\right) = \sum_{k=1}^{n} \sum_{\ell=1}^{m} P_{ki}\, Q_{\ell j}\, \psi(e_k, f_\ell). \end{align*} [guided] We want to express the matrix of $\psi$ in the new bases $(v_i)$, $(w_j)$ in terms of the matrix in the old bases $(e_k)$, $(f_\ell)$. The $(i,j)$-entry of the new matrix is $B_{ij} = \psi(v_i, w_j)$ by definition. Substituting the change-of-basis expressions: \begin{align*} B_{ij} = \psi\!\left(\sum_{k=1}^{n} P_{ki}\, e_k,\; \sum_{\ell=1}^{m} Q_{\ell j}\, f_\ell\right). \end{align*} Bilinearity of $\psi$ allows us to pull the sums and scalars outside: $\psi$ is linear in each argument separately, so we can distribute over the sum in the first argument and then over the sum in the second argument, extracting the scalar coefficients: \begin{align*} B_{ij} = \sum_{k=1}^{n} \sum_{\ell=1}^{m} P_{ki}\, Q_{\ell j}\, \psi(e_k, f_\ell). \end{align*} Note that the coefficient $P_{ki}$ carries index $i$ from the first argument (the new basis vector $v_i$) and index $k$ from the old basis, while $Q_{\ell j}$ carries $j$ from the second argument and $\ell$ from the old basis. [/guided] [/step] [step:Substitute $A_{k\ell} = \psi(e_k, f_\ell)$ and recognise the matrix product $P^\top A Q$] Since $A_{k\ell} = \psi(e_k, f_\ell)$: \begin{align*} B_{ij} = \sum_{k=1}^{n} \sum_{\ell=1}^{m} P_{ki}\, A_{k\ell}\, Q_{\ell j}. \end{align*} The inner sum over $k$ gives $\sum_{k=1}^{n} P_{ki}\, A_{k\ell} = \sum_{k=1}^{n} (P^\top)_{ik}\, A_{k\ell} = (P^\top A)_{i\ell}$. Then \begin{align*} B_{ij} = \sum_{\ell=1}^{m} (P^\top A)_{i\ell}\, Q_{\ell j} = (P^\top A Q)_{ij}. \end{align*} Since this holds for all $i \in \{1, \dots, n\}$ and $j \in \{1, \dots, m\}$, we conclude $B = P^\top A Q$. [/step]