Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 61 / 7 / b / Solution

Take the inner product linear in its first argument; for a real space no conjugation convention is needed. Insert $u^{\ast }={\sum }_{j=1}^n{a}_j{u}_j$ into the orthogonality condition against each basis vector $u_i$. The normal equations areEquivalently $Ha=b$, where $H_{ij}=(u_j,u_i)$ and $b_i=(x,u_i)$. This orientation of the Gram matrix matches the stated inner-product convention. It is Hermitian and positive definite: for a nonzero coefficient vector $a$,by linear independence of the basis. Hence $H$ is invertible and $a=H^{-1}b$ gives existence and uniqueness of $u^{\ast }$. For an orthonormal basis this simplifies to $a_i=(x,u_i)$; for a general basis the Gram inverse correctly accounts for interactions between basis vectors.