Matrix of a Hermitian metric for two different basis

Question

Let $\{e_i\}$ be a basis of a $\mathbb{C}$-linear space $E$ and $h$ a Hermitian metric on $E$. Denote by $H=\{h_{ij}\}$ the matrix of $h$ for $\{e_i\}$, that is $$h_{ij}=h(e_i,e_j).$$ Let now $g\in GL(E)$ be a complex automorphism and consider the new basis $\{e'_i\}=\{ge_i\}$. Denote by $H'=\{h'_{ij}\}$ the corresponding matrix.

What is the relation between $H$ and $H'$?

Answer 1

I'll assume that the Hermitian metric is conjugate-linear with respect to the first argument. First, some notation. Set $\mathcal{E} = (e_1,\dots,e_n)$ and denote by $[v]_{\mathcal{E}}$ the coordinate column vector of $v$ with respect to $\mathcal{E}$. Set also $$ [h]^{\mathcal{E}} = \left( h(e_i,e_j) \right)_{i,j} $$ which is the matrix representing $h$ with respect to the basis $\mathcal{E}$. It is the (unique) matrix satisfying $$ h(v,w) = [v]_{\mathcal{E}}^{*} \cdot [h]^{\mathcal{E}} \cdot [w]_{\mathcal{E}} $$ for all $v,w \in E$. Given a different basis $\mathcal{E}'$, the relation between $[h]^{\mathcal{E}}$ and $[h]^{\mathcal{E}'}$ is given by $$ [h]^{\mathcal{E}'} = \left( P^{\mathcal{E}'}_{\mathcal{E}} \right)^{*} \cdot [h]^{\mathcal{E}} \cdot P^{\mathcal{E}'}_{\mathcal{E}} $$ where $P^{\mathcal{E}'}_{\mathcal{E}} = \left[ \operatorname{id} \right]^{\mathcal{E}'}_{\mathcal{E}}$ is the change of basis matrix which is the unique matrix satisfying $P^{\mathcal{E}'}_{\mathcal{E}} [v]_{\mathcal{E}'} = [v]_{\mathcal{E}}$ for all $v \in E$.

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In your case, the new basis $\mathcal{E}'$ is obtained by applying $g$ to each element of $\mathcal{E}$. We have $$ ge_i = e_i' \iff [g]^{\mathcal{E}}_{\mathcal{E}} = P^{\mathcal{E}'}_{\mathcal{E}}$$ so the final relation is $$ [h]^{\mathcal{E}'} = \left( [g]^{\mathcal{E}}_{\mathcal{E}} \right)^{*} \cdot [h]^{\mathcal{E}} \cdot [g]^{\mathcal{E}}_{\mathcal{E}} $$ where $[g]^{\mathcal{E}}_{\mathcal{E}}$ is the matrix representing $g$ with respect to the basis $\mathcal{E}$ (both in the domain and range).

Of course, you can just write everything in terms of coordinates and calculate to deduce the same result but this notation helps you to remember all the change of coordinates rules.

Answer 2

I will assume that $h$ is an inner product. Put an inner product $(\cdot, \cdot)$ on $E$ such that $\{e_1, \dots, e_n\}$ is an orthonormal basis, i.e., $(e_i, e_j) = \delta_{ij}$. Let $H \colon E \to E$ be the linear transformation represented by the matrix $H$. We have by assumption that $$h(e_i, e_j) = H_{ij} = (H^T)_{ji} = (H^Te_i, e_j).$$ By linearity, this identity extends to $h(u, v) = (H^Tu, v)$ for all $u, v \in E$. Thus $$(H')_{ij} = h(ge_i, ge_j) = (H^Tge_i, ge_j) = (g^*H^Tge_i, e_j) = (g^*H^Tg)_{ji},$$ where $g^*$ denotes the adjoint of $g$. Thus $H' = (g^*H^Tg)^T = g^TH(g^{*})^T = g^TH\overline{g}$, where $\overline{g}$ denotes the complex conjugate of the matrix representation of $g$.