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LetThenThus ${\overline{Q}}_k{\overline{R}}_k$ is a QR decomposition of the matrix power $A^{k+1}$. In particular, the first $r$ columns of ${\overline{Q}}_k$ span the same subspace as the first $r$ columns of $A^{k+1}$ whenever those columns are independent.

For a full-column-rank $n\times k$ matrix $V$, extend the thin QR decomposition $V=Q_1{R}_1$ to an orthogonal matrix $Q=(Q_1{Q}_2)$. Thenso $Q^T$ maps the column space of $V$ onto the first $k$ coordinate directions. Successive Householder transformations construct the same reduction without first forming a full orthogonal basis.

Each plane rotation satisfies $R_j^T{R}_j=I$. Hence $U=R_3{R}_2{R}_1$ is an orthogonal matrix and $D=UA$ is an upper triangular matrix. Multiplication by $U^T$ gives the QR decompositionThe zero-pair cases already handled make this construction valid for every real $3\times 3$ matrix, including rank-deficient matrices. Orthogonality follows from the product identity, not from any invertibility assumption on $A$.

For a full-column-rank matrix, a thin QR decomposition has $A=Q_{1}R$, where $Q_1^T{Q}_1=I_n$ and $R$ is nonsingular upper triangular. Extend $Q_1$ to an orthogonal matrix $Q=(Q_1,Q_2)$. Orthogonality givesThus the unique least-squares solution satisfies $R{x}^{\ast }=Q_1^{T}b$, with residual norm $\Vert Q_2^{T}b\Vert$. In a rank-deficient problem $R$ is singular and uniqueness needs separate treatment.

A Givens rotation is the identity except for the rows/columns $p,q$, whose block is $\left(\begin{array}{cc}c& s\\ -s& c\end{array}\right)$ with $c^2+s^2=1$. Its transpose times itself has that block equal to the identity, so it is orthogonal. Choose $c=4/5,s=3/5$ in rows two and three. ThenBack substitution in the first three rows givesThe fourth row of the rotated residual is the unavoidable value $-2$, while the other three rows vanish. Thus this rotation also directly exhibits the least-squares optimality.

A QR decomposition of an $m\times n$ matrix, $m\ge n$, is $A=QR$, where $Q$ is an $m\times m$ orthogonal matrix and $R$ is $m\times n$ and upper trapezoidal: $R_{ij}=0$ for $i>j$. In particular its bottom $m-n$ rows vanish.

Successive left Givens rotations annihilate the entries below the diagonal, column by column. When working on column $j$, rotate rows $j$ and $i>j$; their earlier columns are already zero, so the earlier eliminations persist. If $G_k\cdots G_{1}A=R$, thenA product of orthogonal matrices is orthogonal. The construction remains valid for a rank-deficient matrix, although uniqueness is not asserted. This is QR decomposition by Givens rotations.

First rotate rows $2,3$ with $c=4/5$, $s=3/5$. The second column pair $(4,3)$ becomes $(5,0)$ and the corresponding third column pair $(1,2)$ becomes $(2,1)$. Next rotate rows $3,4$ with $c=4/5$, $s=3/5$, sending the third column pair $(1,3/4)$ to $(5/4,0)$. Thus $R=G_{34}{G}_{23}A$ and $Q=G_{23}^T{G}_{34}^T$. One exact QR decomposition isThe construction proves $Q^{T}Q=I$, and direct multiplication checks $QR=A$.

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

Let $W=\mathrm{span}\{\mathbf{v},\mathbf{w}\}=\mathrm{col}V$. Since the two columns are linearly independent, the leading $2\times 2$ block $R_1$ inis invertible. Hence $S(W)=\mathrm{span}\{{\mathbf{e}}^{(1)},{\mathbf{e}}^{(2)}\}$. The hypothesis that $W$ is an invariant subspace of $A$ implies that this coordinate plane is invariant under $\hat{A}=SA{S}^{-1}$. Consequently the first two columns of $\hat{A}$ have no entries below row two, andwhere $B$ is the top-left $2\times 2$ block and $D$ is the bottom-right $(n-2)\times (n-2)$ block. Eigenvalue deflation by an invariant subspace now givessoagain counting algebraic multiplicities.

For the required orthogonal reduction, take a thin QR decomposition $V=Q_1{R}_1$, extend the two orthonormal columns of $Q_1$ to an orthogonal matrix $Q$, and set $S=Q^T$. ThenEquivalently, apply one Householder transformation to annihilate entries $2,\dots ,n$ of $\mathbf{v}$, followed by a second Householder transformation acting only on coordinates $2,\dots ,n$ to annihilate entries $3,\dots ,n$ of the transformed $\mathbf{w}$.

The linear least-squares problem is to choose $x\in {\mathbb{R}}^n$ minimizing the Euclidean norm of the residual:For a QR decomposition $A=QR$, orthogonality of $Q$ gives $\Vert Ax-b{\Vert }_2=\Vert Rx-Q^{T}b{\Vert }_2$. If $A$ has full column rank and $R=\left(\genfrac{}{}{0ex}{}{R_1}{0}\right)$ is in standard form with invertible upper-triangular $R_1\in {\mathbb{R}}^{n\times n}$, the minimizer solves $R_{1}x=(Q^{T}b{)}_{1:n}$.

Applying the Gram-Schmidt process to the columns of the given matrix yieldsIndeed, $Q$ is an orthogonal matrix and $A=QR$. Moreover,Back substitution in the leading triangular system givesso the unique least-squares solution isThe unused final transformed residual component is $1$, so the minimum residual norm is $1$.

Each QR decomposition $A_k=Q_k{R}_k$ givesInduction therefore yields the first Accumulated QR factorization identity:

We next prove the corresponding factorization of a matrix power. The case $k=0$ is just $A=Q_0{R}_0$. Ifthen the similarity formula above impliesConsequently,The product ${\overline{Q}}_k$ is orthogonal, and the product ${\overline{R}}_k$ is upper triangular. Henceis a QR decomposition.

Let the columns of $A$ be $a_1,a_2,a_3$. The Gram-Schmidt process givesNext,soFor the final column,andThus $q_3=\frac{1}{2}(-1,-1,1,-1{)}^T$ and $r_{33}=2$. The reduced QR decomposition isThe columns of $Q$ are orthonormal, and direct multiplication gives $QR=A$.