Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 1 / 6F / Solution

The first matrix column is already the unit vector $e_1$, so the first QR elimination needs no reflection. In the remaining three coordinates, the second column is $v=(2,2,1{)}^T$, with norm three. To map it to $(-3,0,0{)}^T$, use the Householder transformationIt is symmetric and orthogonal and satisfies $Hv=(-3,0,0{)}^T$. Extend it to $Q^T=\mathrm{diag}(1,H)$. The QR decomposition and transformed right side areFor the lower right-side block, $w^T(1,4,-1{)}^T=12$, so its transformed value is $(1,4,-1{)}^T-(12/15)w=(-3,12/5,-9/5{)}^T$.

Orthogonality preserves the least-squares residual norm. Thus the first two squared residual terms can and must be set to zero, while the last two are independent of the unknowns:The minimum squared residual is $(12/5{)}^2+(-9/5{)}^2=9$. The solution is unique since the triangular factor has nonzero diagonal entries. No normal equations are used.