Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 19D / c

The first column is $a=(2,1,2{)}^T$ with length three. Choose $v=a-3e_1=(-1,1,2{)}^T$, so $v^{T}v=6$. Its Householder reflection isDirect multiplication givesTo eliminate the remaining third-row entry in the second column, reflect the last two coordinates using $v_2=(0,1,-1{)}^T$. ThenandBoth steps are Householder reflections; the second happens to exchange two coordinates. Solve the leading triangular equations $3x_1+3x_2=3$ and $3x_2=-3$ to obtainFor an independent check, $A{x}^{\ast }-b=(-1,-2,2{)}^T$, whose squared norm is nine and which is orthogonal to both columns of $A$. Thus $A^T(A{x}^{\ast }-b)=0$, confirming the least-squares minimizer.