Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 69 / 6 / 1

Put $e=x-u^{\ast }$. If $u^{\ast }$ minimizes distance, then for every $v\in {\mathcal{U}}_n$ and every real $t$,Both signs of arbitrarily small $t$ force $\mathrm{Re}(e,v)=0$. In a complex inner product space, apply the same argument to $iv$ to obtain the imaginary part too. Conversely, if $e$ is orthogonal to the subspace, the Pythagorean theorem for inner product spaces givesThus orthogonality characterizes the unique best approximation:Finite dimension also guarantees existence by solving the invertible Gram matrix normal equations; completeness of the ambient inner product space is unnecessary.

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.