Specify the inverse of the matrix $g_{ij}=langle v_i,v_jrangle$, 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_1rangle&cdots&langle v_1,v_nrangle\
vdots&&vdots\
langle v_n,v_1rangle&cdots&langle v_n,v_nrangle
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).

Filippo · Accepted Answer

First of all, we know that
begin{align*}
    phicolon V&to V^*\
    v&mapstolangle v|,cdot,rangle
end{align*}
is an isomorphism. We can construct an inner product on $V^*$ by defining
begin{equation}
langleomega,etarangle=leftlanglephi,^{-1}(omega),phi,^{-1}(eta)rightrangle
end{equation}
for $omega,etain V^{*}$.
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^1rangle&cdots&langle v^1,v^nrangle\
vdots&&vdots\
langle v^n,v^1rangle&cdots&langle v^n,v^nrangle
end{pmatrix}
end{equation}
is the inverse of $G$.

Yiorgos S. Smyrlis · Answer

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

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

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