= Solution
Use the convention that the <inner product> is conjugate-linear in its first argument; the real case is the same without conjugates. <Linear independence> makes the <Gram matrix> $G$ positive definite, so $B=G^{-1}$ exists and is Hermitian. Define the <Riesz dual basis in an inner product space> within $\mathcal U_n$ by
$$
\widehat u_k=\sum_{l=1}^n b_{lk}u_l.
$$
Then $(u_i,\widehat u_k)=\sum_lG_{il}b_{lk}=\delta_{ik}$. Its own Gram entries are
$$
(\widehat u_j,\widehat u_k)=\sum_{l,m}\overline{b_{lj}}G_{lm}b_{mk}
=(B^*GB)_{jk}=B_{jk}.
$$
Hence \b[the inverse <Gram matrix> is the <Gram matrix> of the dual vectors]:
$$
\boxed{b_{jk}=(\widehat u_j,\widehat u_k).}
$$
The dual vectors are required to lie in $\mathcal U_n$; adding arbitrary vectors orthogonal to that subspace would preserve the displayed biorthogonality but would destroy this conclusion.
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