Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 4 / 1F / Solution

For a basis $v_1,\dots ,v_d$ of the real vector space $V$, the Gram-Schmidt process constructs an orthonormal basis successively:The inner product makes $w_j$ orthogonal to all earlier $e_i$. Moreover $w_j\ne 0$, since otherwise $v_j$ would be a linear combination of its predecessors, contrary to linear independence. Induction gives $\mathrm{span}(e_1,\dots ,e_j)=\mathrm{span}(v_1,\dots ,v_j)$. Starting with a basis of a linear subspace $U$ and extending it to a basis of $V$ gives the same construction with the first $r=\dim U$ vectors spanning $U$.

The orthogonal complement is $U^{\perp }=\{v\in V:⟨v,u⟩=0\text{ for every }u\in U\}$. For every $v\in V$, set $u={\sum }_{i=1}^r⟨v,e_i⟩e_i$. Then $u\in U$ and $v-u\in U^{\perp }$. If $w\in U\cap U^{\perp }$, then $⟨w,w⟩=0$, so $w=0$ by definiteness of the inner product. Thus the direct sum isThis also covers $U=0$ and $U=V$.

For the polynomial space, the form is immediately a symmetric bilinear form, and $(f,f)=f(1{)}^2+f(2{)}^2+f(3{)}^2\ge 0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, so this form is an inner product for $n=1,2$. For $n\ge 3$, the nonzero polynomial $(x-1)(x-2)(x-3)$ has $(f,f)=0$. Therefore, among the stipulated positive integers,