[proofplan] We compute $B_{ij} = \phi(f_i, f_j)$ by expanding $f_i$ and $f_j$ in terms of the old basis and using conjugate-linearity in the first argument. The result is $B = P^\dagger A P$ where $P^\dagger = \overline{P}^\top$. [/proofplan] [step:Expand $B_{ij} = \phi(f_i, f_j)$ using the change-of-basis matrix and sesquilinearity] Let $(e_1, \dots, e_n)$ and $(f_1, \dots, f_n)$ be bases for $V$ with $f_j = \sum_{k=1}^n P_{kj}\,e_k$. Let $A_{ij} = \phi(e_i, e_j)$ and $B_{ij} = \phi(f_i, f_j)$. Computing $B_{ij}$: \begin{align*} B_{ij} &= \phi(f_i, f_j) = \phi\Bigl(\sum_k P_{ki}\,e_k,\; \sum_\ell P_{\ell j}\,e_\ell\Bigr). \end{align*} Since $\phi$ is conjugate-linear in the first argument and linear in the second: \begin{align*} B_{ij} &= \sum_{k, \ell} \overline{P_{ki}}\, P_{\ell j}\, \phi(e_k, e_\ell) = \sum_{k, \ell} \overline{P_{ki}}\, A_{k\ell}\, P_{\ell j}. \end{align*} In matrix form, $(\overline{P}^\top)_{ik} = \overline{P_{ki}}$, so the $(i,j)$-entry of $\overline{P}^\top A P$ is $\sum_{k,\ell} \overline{P_{ki}}\, A_{k\ell}\, P_{\ell j} = B_{ij}$. Therefore $B = \overline{P}^\top A P = P^\dagger A P$. [guided] The Hermitian form $\phi$ is conjugate-linear in the first argument: $\phi(cu, v) = \bar{c}\,\phi(u, v)$. This is the source of the conjugate transpose $P^\dagger = \overline{P}^\top$ rather than the ordinary transpose. Expanding with $f_i = \sum_k P_{ki}\,e_k$ and $f_j = \sum_\ell P_{\ell j}\,e_\ell$: \begin{align*} B_{ij} = \phi\Bigl(\sum_k P_{ki}\,e_k,\; \sum_\ell P_{\ell j}\,e_\ell\Bigr) = \sum_{k, \ell} \overline{P_{ki}}\, P_{\ell j}\, \phi(e_k, e_\ell) = \sum_{k, \ell} \overline{P_{ki}}\, A_{k\ell}\, P_{\ell j}. \end{align*} The conjugate $\overline{P_{ki}}$ comes from pulling the scalar $P_{ki}$ out of the first (conjugate-linear) slot; the scalar $P_{\ell j}$ comes out of the second (linear) slot without conjugation. Recognising the sums as matrix multiplication: the outer sum over $k$ contracts $\overline{P_{ki}}$ with $A_{k\ell}$, giving $(\overline{P}^\top A)_{i\ell}$. The inner sum over $\ell$ then contracts with $P_{\ell j}$, giving $(\overline{P}^\top A P)_{ij}$. So $B = \overline{P}^\top A P = P^\dagger A P$. Compare with the real symmetric case, where the transformation is $B = P^\top A P$ (congruence). The Hermitian analogue replaces $P^\top$ by $P^\dagger$, reflecting the conjugate-linearity of the first argument. [/guided] [/step]