Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 2 / 18D / c / Solution

The first column already has zeros below the top entry. In the last three rows of the second column, reflect $w=(4,2,4{)}^T$ to $(-6,0,0{)}^T$ using $u=w+6e_1=(10,2,4{)}^T$. Since $u^{T}u=120$, the Householder reflection isWith $Q=\mathrm{diag}(1,H)$, direct multiplication givesThe transformed least-squares objective is $(x_1+2x_2-1{)}^2+(-6x_2+2/3{)}^2+50/9$. Its first two squares can both vanish, so the unique answer isUniqueness follows because the triangular leading block is invertible; this calculation implements Householder QR decomposition without squaring the condition number through normal equations.