Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 2 / 18D (Numerical Analysis)

Let $P=u{u}^T/(u^{T}u)$, well defined because $u\ne 0$. It is symmetric and satisfies $P^2=P$, so for the Householder reflection $H=I-2P$,Thus $H$ is orthogonal; it reverses the component parallel to $u$ and fixes the perpendicular hyperplane.

An orthogonal matrix preserves the Euclidean norm: $\Vert Qz{\Vert }^2=z^T{Q}^{T}Qz=z^{T}z$. For every candidate $x$,The two objective functions agree pointwise, so their minimizers coincide, whether or not the minimizer is unique.

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.