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

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