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With zero initial guess, the generalized minimal residual method minimizes the Euclidean norm of the linear-system residual over each Krylov subspace. The Arnoldi decomposition reduces this to ${\min }_y\Vert {\tilde{H}}_{k}y-\Vert b{\Vert }_2{e}_1{\Vert }_2$, followed by $x_k=Q_{k}y$. Givens rotations solve the small problem in $O(k^2)$ operations once the decomposition is available. For an invertible matrix $A$, the same iterate minimizes the error in the weighted norm induced by $A^{T}A$, because $\Vert Ax-b{\Vert }_2=\Vert x-x_{\ast }{\Vert }_{A^{T}A}$.

On a connected transition chain, independently selected resonant pulses realize embedded two-level rotations with tunable angle and phase. Eliminate target amplitudes from one end of the chain using complex Givens rotations, and invert those rotations to prepare the target from a basis state. The pulse angles come from amplitude ratios and the phases from relative complex arguments. Spectator components are unchanged by each embedded rotation. Work in the interaction picture and compensate the known drift phases at the chosen final time when the target is specified in the laboratory frame.

A Givens rotation eliminates one component of a vector by rotating just two coordinates, preserving its Euclidean norm. For example,The length remains five: elimination moves the component into the retained coordinate rather than discarding it. Applied to a matrix, the same operation mixes two rows; QR decomposition by Givens rotations uses these local operations to build an upper triangular matrix.

For indices $i<j$, the matrix $G_{ij}(c,s)$ is the identity matrix except for its block on coordinates $i,j$:This article uses left matrix multiplication on column vectors. If $c=\cos \theta$ and $s=\sin \theta$, the block is a rotation matrix through angle $-\theta$: the minus sign belongs in the lower left. Its transpose reverses the rotation. Direct multiplication gives $G^{T}G=I$ and determinant one, so it is an orthogonal matrix; consequently $\Vert Gx{\Vert }_2^2=x^T{G}^{T}Gx=\Vert x{\Vert }_2^2$. Only two coordinates change, but this Euclidean norm identity holds for the entire vector.

Left matrix multiplication by a Givens rotation replaces rows $i,j$ of a matrix byBoth right-hand sides use the old entries. An in-place algorithm must save the first result temporarily, update the second entry, and only then overwrite the first. The reference BLAS DROT implementation follows this order with its temporary DTEMP.^[ref]

For elimination in column $k$, construct $(c,s)$ from $(A_{ik},A_{jk})$. This sets the lower component to zero in exact arithmetic. In floating-point arithmetic the computed expression can leave a small remainder; after computing the other updated entries, explicitly store $A_{jk}^{\prime }=0$ and $A_{ik}^{\prime }=r$. Previously eliminated columns stay zero if both participating rows are already zero there. This is why the ordering in QR decomposition by Givens rotations matters.

Use geometric quantum state synthesis by adjacent rotations, rather than three full level population swaps. Each resonant pulse implements a complex Givens rotation on one allowed two-state plane. Start with the target amplitude vector, use such rotations to eliminate entries back to the first basis vector, and reverse that sequence. Its angles are fixed by the ratios of target amplitudes, and optical phases supply their relative complex phases.

Here is an explicit real-coordinate construction in the interaction picture. Define a phase-calibrated rotation byA resonant Quantum Hamiltonian proportional to $-i|j⟩⟨k|+i|k⟩⟨j|$ realizes this rotation; the optical phase chooses that quadrature. Let $\beta =\arccos (1/\sqrt{3})$. ThenThe pulse areas equal these rotation parameters in the direct Hamiltonian-coefficient convention of quantum pulse area. A given sign or phase choice for a physical transition depends on its dipole and basis conventions; calibration realizes the specified rotation without assuming every adjacent transition has the same energy ordering.

If the requested amplitudes refer to the laboratory frame at $t_F$, the target for this construction is $U_0(t_F,0{)}^{†}|\Psi ⟩$, not simply $|\Psi ⟩$. Set the rotation phases accordingly, or add the equivalent accessible diagonal phase corrections. This compensates the known drift phases and gives the stated laboratory superposition, not merely the correct level populations. Strong regularity and connected allowed transitions guarantee the required controllability; bandwidth restrictions set how slowly the selective approximation must be implemented.

Choose the Givens rotation convention in which ${\Omega }^{[p,q]}$ equals the identity outside rows and columns $p,q$, and has blockin those coordinates. It is an orthogonal matrix. For any chosen column $j$, write $u=A_{pj}$, $v=A_{qj}$ and $r=\sqrt{u^2+v^2}$. If $r>0$, take $\cos \theta =u/r$, $\sin \theta =v/r$. Then the new entries are $r$ and $-vu/r+uv/r=0$. If both entries vanish, take $\theta =0$. This proves the requested elimination for each chosen $j$; the angle is allowed to depend on $j$, and one angle need not eliminate an entire arbitrary row.

For the given matrix, use $\cos \theta =\sin \theta =1/\sqrt{2}$ first for ${\Omega }^{[1,2]}$, obtainingUse the same cosine and sine for ${\Omega }^{[1,3]}$. It makesThis is already upper triangular, including the zero $(3,2)$ entry, and every leading nonzero row entry is positive. Thus the requested QR decomposition isThe orthogonal matrix property follows from the two rotations, and direct multiplication verifies $A=QR$.

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 $Q$.

A Givens rotation acts as the identity matrix outside two selected coordinates $i,j$, and has the following block on those coordinates:The convention with the opposite signs of the sines is equivalent after replacing $\theta$ by $-\theta$. A Householder reflection with nonzero normal vector $w\in {\mathbb{R}}^n$ isIt reverses the component parallel to $w$ and fixes the perpendicular hyperplane. Both are orthogonal matrices, so their transposes are their inverses and they preserve the Euclidean norm.

Choose the following consistent sign convention. The Givens rotation $J(i,j,\theta )$ equals the identity matrix except for the block on coordinates $i,j$:Thus it changes only the $i$th and $j$th coordinates of a vector. Write $c=\cos \theta$, $s=\sin \theta$. Its two-coordinate block satisfiesAll remaining coordinates are unchanged, so $J^{T}J=I_n$. Hence every such Givens rotation is an orthogonal matrix. With this convention the two-coordinate rotation matrix rotates through angle $-\theta$.

Put $C=Q_{1}A$ and write its second column as $v$. For $n\ge 3$, takewhere each angle is computed from the current second column, in the right-to-left application order. These Givens rotations eliminate its entries in rows $3,\dots ,n$ by accumulating their Euclidean norm in row two.

Each rotation acts only on rows with indices at least two. The first column of $C$ is $\Vert a{\Vert }_2{e}_1$, whose entries in all those rows are zero. Mixing two zeros leaves them zero, and row one is untouched. Thus the first column is unchanged. The resulting matrix has the formThis verifies both required properties, including the case of zero second-column tail. When $n=2$ there are no rows to eliminate: take $Q_2=I_2$, the empty product. No sign change of $v_2$ is required in that case. Each Givens row-pair update preserves previously eliminated columns, which is the central invariant of QR decomposition by Givens rotations.

Apply the product from right to left, choosing every Givens rotation from the current first column. First $J(1,2)$ replaces $(a_1,a_2)$ by $(\sqrt{a_1^2+a_2^2},0)$. Next $J(1,3)$ combines that retained entry with $a_3$ and sets the third entry to zero. Inductively, after eliminating coordinates $2,\dots ,k$, the first coordinate isand coordinates $2,\dots ,k$ are zero. Later Givens rotations involve only coordinate one and a new coordinate, so those zeros remain zero. Thereforewhere $e_1$ is the first standard basis vector. The zero vector is included by choosing the identity matrix for every zero pair. The nonnegative sign follows from the convention fixed in part (c)(ii). This is the first-column stage of QR decomposition by Givens rotations.