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).
Best Answer
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 $G$.