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

Mathematics Asked by Filippo on December 1, 2020

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).

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 $$G$$.

Correct answer by Filippo on December 1, 2020

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

Answered by Yiorgos S. Smyrlis on December 1, 2020