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For a nonzero real vector $v$, the Householder transformation isWriting $P=v{v}^T/(v^{T}v)$ gives $P^T=P$, $P^2=P$, so $H^T=H$ and $H^{T}H=(I-2P{)}^2=I$. Thus $H$ is orthogonal. It reverses the component along $v$ and fixes its orthogonal complement.

For Householder QR decomposition, successively choose a reflector on each trailing row block to send its current column to a multiple of the first coordinate vector. Each step annihilates the subdiagonal entries without disturbing earlier columns. The reflector product gives an upper-trapezoidal $R$ and its transpose gives an orthogonal $Q$ with $A=QR$.

Here the first column has norm four. Taking $v=(-1,1,-1,1{)}^T$ gives $H_1=I-v{v}^T/2$ andThe second column already has no unwanted entries. On the last two rows, reflect using $(-1,2{)}^T$:This maps $(9,12{)}^T$ to $(15,0{)}^T$. Therefore a full QR factorization isBoth displayed factors arise from the explicit reflectors; in particular $Q^{T}Q=I_4$ and $QR=A$.

The first three diagonal entries of $R$ are nonzero, so the column rank is three and a consistent system has a unique solution. Consistency requires the last component of $Q^{T}b$ to vanish. For the given right side this component isHenceFor this value the transformed right side is $(3,2,5,0{)}^T$. Back substitution also gives $x=(-5/12,1/2,1/3{)}^T$. This illustrates the Householder reduction of an overdetermined consistent system.

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 $A^T(A{x}^{\ast }-b)=0$.