= 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 put
$$
\widehat u_k=\sum_{\ell=1}^n b_{\ell k}u_\ell.
$$
The proposed vectors satisfy
$$
(u_i,\widehat u_k)=\sum_\ell G_{i\ell}b_{\ell k}=(GB)_{ik}=\delta_{ik}.
$$
They therefore give the unique dual family within $\mathcal U_n$. Their own <Gram matrix> is
$$
\bigl((\widehat u_j,\widehat u_k)\bigr)_{j,k}=B^*GB.
$$
The inverse of a Hermitian positive-definite <matrix> is Hermitian, so $B^*=B$. Using $GB=I$ gives $B^*GB=B$. Consequently
$$
\boxed{(\widehat u_j,\widehat u_k)=b_{jk}.}
$$
For a real space this is the familiar identity $B^TGB=B$. 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.
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