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Rows two and three of $C$ both have zero in column one. Any rotation of these rows therefore preserves those zeros, so $\boxed{d_{21}=d_{31}=0}$. In column two, $d_{32}=c_{22}\sin {\theta }_3+c_{32}\cos {\theta }_3$. For $r_3=\sqrt{c_{22}^2+c_{32}^2}>0$, chooseThis gives $d_{32}=0$. Again choose ${\theta }_3=0$ if both relevant entries are zero. Thus the successive Givens rotations have eliminated all entries below the diagonal without recreating earlier ones.

The second Givens rotation leaves row three unchanged, so $\boxed{c_{31}=b_{31}=0}$. Its other relevant entry is $c_{21}=b_{11}\sin {\theta }_2+b_{21}\cos {\theta }_2$. For $r_2=\sqrt{b_{11}^2+b_{21}^2}>0$, takeThen $c_{21}=0$ and $c_{11}=r_2$. If both entries vanish, choose ${\theta }_2=0$. The first column now has zeros below its diagonal.

Left multiplication by the given Givens rotation mixes rows two and three, so $b_{31}=a_{21}\sin {\theta }_1+a_{31}\cos {\theta }_1$. If $r_1=\sqrt{a_{21}^2+a_{31}^2}>0$, chooseEquivalently ${\theta }_1=atan2(-a_{31},a_{21})$ modulo $2\pi$. Substitution gives $b_{31}=0$ and $b_{21}=r_1$. If $a_{21}=a_{31}=0$, take ${\theta }_1=0$; the desired zero is already present. This also covers singular matrices without division by zero.

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.

Use the genuine planar Givens rotationIn the second component the cosine multiplies $v_j$: the repeated $v_i$ in the printed formula is an error. With that printed expression, at $\theta =0$ the pair becomes $(v_i,v_i)$, which does not preserve length or measure. The rotation-based claims require the corrected expression. Also take $N\ge 2$, since the normalization by $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ is undefined for $N=1$.

Let $C_N=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ and $U_{ij,\theta }F=F\circ R_{ij,\theta }$. The change of variables formula and determinant one give $\Vert U_{ij,\theta }F{\Vert }_2=\Vert F{\Vert }_2$. Thus each is a unitary operator, with adjoint $U_{ij,-\theta }$. The Kac collision operator is the averageThe Minkowski integral inequality gives $\Vert QF{\Vert }_2\le \Vert F{\Vert }_2$, so $Q$ is bounded. For the Hilbert space inner product, integration and the angular change $\theta \mapsto -\theta$ giveHence $Q=Q^{\ast }$ and $\Vert Q\Vert \le 1$. In fact a nonzero radial Gaussian function is fixed by every rotation, showing $\Vert Q\Vert =1$. Angular averages can be understood as strong Bochner integrals; continuity of rotations in $L^2$ follows first for smooth compactly supported functions, then by density.

For each rotation, unitarity givesAveraging and using self-adjointness of $Q$ yields the Dirichlet form of the Kac collision operatorThe velocity integral is necessary: its omission from the printed right-hand side would leave a function of $\mathbf{v}$ rather than a scalar. This identity applies to every $L^2$ function, with complex modulus when necessary, and is nonnegative.

If $(I-Q)F=0$, every nonnegative angular integral is zero, so $\Vert U_{ij,\theta }F-F{\Vert }_2=0$ for almost every angle. Strong continuity in angle extends equality to every angle. The coordinate-plane Givens rotations generate the special orthogonal group $SO(N)$, hence $F$ is invariant in $L^2$ under every element of this group. To identify its shape rigorously despite almost-everywhere representatives, average over the normalized Haar measure of $SO(N)$. This averaging leaves $F$ unchanged, while transitivity of the rotation group on each sphere makes the average a radial function. Thus $F(\mathbf{v})=\varphi (|\mathbf{v}|)$ almost everywhere.

Conversely, every radial function is fixed by every coordinate-plane rotation, and so by $Q$. ThereforeThis is the radial kernel of the Kac collision operator. The rotation correction is essential to this conclusion: with the literal printed map, even $F(\mathbf{v})=e^{-|\mathbf{v}{|}^2}$ in dimension two is not fixed. At $\mathbf{v}=(0,1)$ its printed-map angular average is the average of $e^{-{\sin }^2\theta }$, strictly greater than its value $e^{-1}$.

A Givens rotation is the identity outside rows and columns $p,q$, with the $2\times 2$ blockThe block's transpose times itself is the identity, so the whole matrix satisfies $\boxed{{\Omega }^T\Omega =I}$ and is an orthogonal matrix. In particular it preserves the Euclidean norm. To annihilate the lower entry of a column pair $(a,b{)}^T$, take $c=a/r$, $s=b/r$, $r=\sqrt{a^2+b^2}$; the rotated pair is $(r,0{)}^T$. If both entries are zero, take the identity rotation.

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.

A Givens rotation ${\Omega }^{[p,q]}$ is the identity except in rows and columns $p,q$, where it has the blockFor $a=A_{pj}$ and $b=A_{qj}$, choosewhen $(a,b)\ne (0,0)$. ThenIf both entries vanish, any angle works.

For the given matrix, first useIt givesThen useThus $R={\Omega }^{[2,3]}{\Omega }^{[1,3]}A$ isAll leading row entries are positive. The resulting QR decomposition by Givens rotations iswhereand $R$ is the matrix above. Since it is a product of transposed rotations, $Q$ is orthogonal.

A Givens rotation is identity except for a $2\times 2$ block $\left(\begin{array}{cc}c& s\\ -s& c\end{array}\right)$ in rows $p,q$. Choose $p=j,q=i$ and $c=a_{jj}/r$, $s=a_{ij}/r$, $r=(a_{jj}^2+a_{ij}^2{)}^{1/2}$, with sign adjusted to the block convention; this mixes only those rows and zeros $a_{ij}$. For $3\times 3$, successively apply rotations $(1,2)$, $(1,3)$, $(2,3)$, producing patterns $\ast \ast \ast ;0\ast \ast ;\ast \ast \ast$, then $\ast \ast \ast ;0\ast \ast ;0\ast \ast$, then an upper triangular $R$. The product of transposed rotations is $Q$.