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

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

I know that the Gram matrix $$$$G=\begin{pmatrix} \langle v_1,v_1\rangle&\cdots&\langle v_1,v_n\rangle\\ \vdots&&\vdots\\ \langle v_n,v_1\rangle&\cdots&\langle v_n,v_n\rangle \end{pmatrix}=\begin{pmatrix} g_{11}&\cdots&g_{1n}\\ \vdots&&\vdots\\ g_{n1}&\cdots&g_{nn} \end{pmatrix}$$$$ 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).

• math.stackexchange.com/questions/3085301/… Nov 15, 2020 at 10:19
• Construct the dual basis $\{u^1,...,u^n\}$. Then $$g^{i,j}=\langle u^i,u^j\rangle$$ Nov 15, 2020 at 12:35
• @K.defaoite Thank you, I suspected that :) Nov 15, 2020 at 13:33
• @Physor Thank you, you are right that this is essentially the same question (it did not answer my question, but it was interesting to learn about the Gram Matrix, so thank you for sharing the link). Nov 15, 2020 at 13:40
• @Filippo You're welcome! Nov 15, 2020 at 13:40

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 $$$$\langle\omega,\eta\rangle=\left\langle\phi\,^{-1}(\omega),\phi\,^{-1}(\eta)\right\rangle$$$$ for $$\omega,\eta\in V^{*}$$. With this definition, you can easily prove that $$$$\begin{pmatrix} g^{11}&\cdots&g^{1n}\\ \vdots&&\vdots\\ g^{n1}&\cdots&g^{nn} \end{pmatrix}:= \begin{pmatrix} \langle v^1,v^1\rangle&\cdots&\langle v^1,v^n\rangle\\ \vdots&&\vdots\\ \langle v^n,v^1\rangle&\cdots&\langle v^n,v^n\rangle \end{pmatrix}$$$$ is the inverse of $$G$$.
If $$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}$$.