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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 ${\mathcal{U}}_n$ byThen $(u_i,{\hat{u}}_k)={\sum }_l{G}_{il}{b}_{lk}={\delta }_{ik}$. Its own Gram entries areHence the inverse Gram matrix is the Gram matrix of the dual vectors: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.