Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 4

The prescribed form is a sum of products of real polynomial evaluations. It is a symmetric bilinear form, since each evaluation is a linear map. AlsoEquality forces $f(-1)=f(0)=f(1)=0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, by the root bound for a polynomial. Thus equality holds only for the zero polynomial, establishing positive definiteness. Therefore the form is a real inner product, and $V$ is an inner product space. This is a discrete sampling inner product on polynomial spaces.

The relevant inner products areHence $1$ and $x$ are orthogonal vectors. Dividing by their norms gives the orthonormal basisof their linear span $U$.

For any orthonormal basis $f,g$ of $U$, bilinearity givesTogether with $⟨f,g⟩=0$, these identities prove pairwise orthogonality. The construction subtracts the orthogonal projection onto a finite-dimensional subspace $U$ from $x^2$, leaving the orthogonal projection residual.

Use $f,g$ from i. Their inner products with $x^2$ areThus $h=x^2-2/3$. Its values at $-1,0,1$ are $1/3,-2/3,1/3$, soBy ii the three polynomials are orthogonal vectors, and normalizing $h$ givesThis orthonormal set is a basis of the three-dimensional vector space $V$. It is the result of the Gram-Schmidt process applied to $1,x,x^2$.

If $T$ is an orthogonal transformation and $e_1,e_2,e_3$ is an orthonormal basis, thenThe images form an orthonormal set, hence are linearly independent and form an orthonormal basis in dimension three.

Conversely, suppose the images $T{e}_1,T{e}_2,T{e}_3$ of one chosen orthonormal basis form an orthonormal basis. Write $u={\sum }_i{a}_i{e}_i$ and $v={\sum }_j{b}_j{e}_j$. By linearity of $T$ and bilinearity,Thus mapping one orthonormal basis to an orthonormal basis is equivalent to preserving every inner product. In the converse, a single chosen orthonormal basis suffices.

Put $p=(3x^2-2)/\sqrt{2}$ and use the orthonormal basis $\mathcal{B}=(1,\sqrt{3/2}x,p)$ from iii. The columns of the given matrix giveConsequentlyThe images of $\mathcal{B}$ are therefore $(p,\sqrt{3/2}x,-1)$, again an orthonormal basis. By i, $T$ is an orthogonal transformation. Its matrix in this orthonormal basis is

$A$ is not an orthogonal matrix. Its first column has squared Euclidean normso $A^{T}A\ne I$. The reason this is compatible with ii is that $1,x,x^2$ is not an orthonormal basis for the given inner product. Its Gram matrix isThe correct condition on this matrix representation of a linear map iswhich follows either by direct multiplication or by the inner product preservation just proved. This is the rule for an orthogonal transformation in a nonorthonormal basis.