Specify the inverse of the matrix $g_{ij}=\langle v_i,v_j\rangle$, where $(v_1,...,v_n)$ is a basis

Question

Let $V$ be an inner product space with a basis $(v_1,...,v_n)$.

I know that the Gram matrix \begin{equation} G=\begin{pmatrix} \langle v_1,v_1\rangle&\cdots&\langle v_1,v_n\rangle\\ \vdots&&\vdots\\ \langle v_n,v_1\rangle&\cdots&\langle v_n,v_n\rangle \end{pmatrix}=\begin{pmatrix} g_{11}&\cdots&g_{1n}\\ \vdots&&\vdots\\ g_{n1}&\cdots&g_{nn} \end{pmatrix} \end{equation} is invertible, but is it possible to explicitly specify the inverse (maybe involving the dual basis)?

Motivation: For those who have heard about the metric tensor: I'd like to have a formula for $g^{ij}$ (which boils down to the question above).

Answer 1

First of all, we know that \begin{align*} \phi\colon V&\to V^*\\ v&\mapsto\langle v|\,\cdot\,\rangle \end{align*} is an isomorphism. We can construct an inner product on $V^*$ by defining \begin{equation} \langle\omega,\eta\rangle=\left\langle\phi\,^{-1}(\omega),\phi\,^{-1}(\eta)\right\rangle \end{equation} for $\omega,\eta\in V^{*}$. With this definition, you can easily prove that \begin{equation} \begin{pmatrix} g^{11}&\cdots&g^{1n}\\ \vdots&&\vdots\\ g^{n1}&\cdots&g^{nn} \end{pmatrix}:= \begin{pmatrix} \langle v^1,v^1\rangle&\cdots&\langle v^1,v^n\rangle\\ \vdots&&\vdots\\ \langle v^n,v^1\rangle&\cdots&\langle v^n,v^n\rangle \end{pmatrix} \end{equation} is the inverse of $G$.

Answer 2

If $$ v_i=\sum_{j=1}^n a_{ij}e_j $$ with the $e_j$'s forming an orthonormal basis, $$ \langle v_i,v_j\rangle=\sum_{k=1}^n a_{ik} a_{jk} $$ and hence $$ U= \big(\langle v_i,v_j\rangle\big)_{i,j=1}^n=AA^T, $$ where $A=(a_{ij})_{i,j=1}^n$.

So $U^{-1}=A^{-T}A^{-1}$.