Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2011 / 2

The Gram-Schmidt process subtracts projections onto directions already constructed and then normalizes. Define recursivelywhere the empty sum for $k=1$ is zero. Assuming the preceding $e_j$ are orthonormal, taking the inner product of $w_k$ with $e_i$, $i<k$, gives zero. Moreover $w_k\ne 0$: otherwise $v_k$ would belong to the span of $v_1,\dots ,v_{k-1}$, contrary to linear independence. Normalization is therefore legitimate and produces an orthonormal basis. At each stage the first $k$ constructed vectors have the same span as the first $k$ original vectors.

A real inner product must be symmetric. Here ${\beta }_a(e_1,e_2)=-2$ and ${\beta }_a(e_2,e_1)=-a$, forcing $a=2$. For this value,This is strictly positive for every nonzero vector. Bilinearity is already given, so symmetry and positive definiteness prove

Use the inner product with $a=2$. The required first vector $v_1=(1/2,0{)}^T$ already has squared norm $4(1/2{)}^2=1$. Apply the Gram-Schmidt process to $v_1$ followed by $e_2=(0,1{)}^T$. Since $\beta (e_2,v_1)=-1$, its orthogonal remainder is $w=e_2+v_1=(1/2,1{)}^T$. Its squared norm is $\beta (w,w)=1-2+3=2$. Therefore an orthonormal basis containing the specified vector isThe second vector may also be replaced by its negative.