 Solution

The dual vectors are the Riesz dual basis in an inner product space, and lie in ${\mathcal{U}}_n$ itself. Take the convention that the inner product is conjugate-linear in its first argument; in a real space the conjugations can simply be omitted. Since the $u_j$ form a basis, their Gram matrix $G$ is positive definite and invertible. Let $B=G^{-1}$ and putThe proposed vectors satisfyThey therefore give the unique dual family within ${\mathcal{U}}_n$. Their own Gram matrix isThe inverse of a Hermitian positive-definite matrix is Hermitian, so $B^{\ast }=B$. Using $GB=I$ gives $B^{\ast }GB=B$. ConsequentlyFor a real space this is the familiar identity $B^{T}GB=B$. If the opposite complex inner-product convention is adopted, conjugate the coefficients in the dual-vector formula; the stated identity remains the same. Requiring the dual vectors to lie in ${\mathcal{U}}_n$ is essential: arbitrary additions perpendicular to ${\mathcal{U}}_n$ preserve the pairings with $u_i$ but alter the Gram matrix of the dual vectors.