 Solution

The linear least-squares problem is to minimize $\Vert Ax-b{\Vert }_2^2$ over $x\in {\mathbb{R}}^n$, allowing an overdetermined system to have a nonzero residual. For a full QR decomposition, $Q$ is $m\times m$ orthogonal and $R$ is $m\times n$ with an upper-triangular top block. Orthogonal transformations preserve the Euclidean norm, soIf $A$ has full column rank, the top triangular block is nonsingular; setting its residual to zero gives the unique minimizer, while the remaining residual is independent of $x$. Without full column rank, minimizers still exist but need not be unique.

For the specified matrix, one Householder reflection already produces an upper-triangular matrix. Take $v=(-1,1,1,1{)}^T$, the difference between the first column and $2e_1$, and setThe formula gives $H^T=H$, $H^2=I$. Direct multiplication yields the Householder QR decompositionAll subdiagonal entries in the remaining columns are already zero, so no further reflection is needed. For the supplied right-hand side, $Q^{T}b=(2,0,2,-2{)}^T$. Back substitution gives $x_3=1$, $x_2=-1$, $x_1=2$, henceThe residual $A{x}^{\ast }-b=(1,-1,-1,1{)}^T$ is orthogonal to every column of $A$, independently confirming the normal equations $A^T(A{x}^{\ast }-b)=0$.