Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 61 / 7 / 1

Write $r=x-u^{\ast }$. If $u^{\ast }$ is a best approximation in the finite-dimensional linear subspace, then for every $v$ in it and every real $t$,A nonzero real linear coefficient would make this negative for sufficiently small $t$ of the appropriate sign. Hence $\mathrm{Re}(r,v)=0$. In a complex inner product space, apply the argument to $iv$ too; the imaginary part also vanishes. Thus $(x-u^{\ast },v)=0$ for every $v$ in the subspace.

Conversely, if that orthogonality holds, every other candidate has the form $u^{\ast }+v$, and the Pythagorean theorem in the inner product norm givesEquality requires $v=0$. ThereforeThe next part constructs this orthogonal projection explicitly. No completeness of the whole inner product space is required for projection onto a finite-dimensional subspace.

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.