Givens rotation

In numerical linear algebra, a Givens rotation is a rotation in the plane spanned by two coordinates axes. Givens rotations are named after Wallace Givens, who introduced them to numerical analysts in the 1950s while he was working at Argonne National Laboratory.

Matrix representation
A Givens rotation is represented by a matrix of the form


 * $$G(i, j, \theta) =

\begin{bmatrix}  1   & \cdots &    0   & \cdots &    0   & \cdots &    0   \\ \vdots & \ddots & \vdots &       & \vdots &        & \vdots \\ 0  & \cdots &    c   & \cdots &   -s   & \cdots &    0   \\ \vdots &       & \vdots & \ddots & \vdots &        & \vdots \\ 0  & \cdots &    s   & \cdots &    c   & \cdots &    0   \\ \vdots &       & \vdots &        & \vdots & \ddots & \vdots \\ 0  & \cdots &    0   & \cdots &    0   & \cdots &    1 \end{bmatrix},$$

where $c = cos θ$ and $s = sin θ$ appear at the intersections $i$th and $j$th rows and columns. That is, for fixed $i$ $>$ $j$, the non-zero elements of Givens matrix are given by:
 * $$\begin{align}

g_{kk} &{}= 1 \qquad \text{for} \ k \ne i,\,j\\ g_{kk} &{}= c \qquad \text{for} \ k = i,\,j\\ g_{ji} &{} = -g_{ij}= -s\\ \end{align}$$

The product $G(i, j, θ)x$ represents a counterclockwise rotation of the vector $x$ in the $(i, j)$ plane of $θ$ radians, hence the name Givens rotation.

The main use of Givens rotations in numerical linear algebra is to transform vectors or matrices into a special form with zeros in certain coefficients. This effect can, for example, be employed for computing the QR decomposition of a matrix. One advantage over Householder transformations is that they can easily be parallelised, and another is that often for very sparse matrices they have a lower operation count.

Stable calculation
When a Givens rotation matrix, $G(i, j, θ)$, multiplies another matrix, $A$, from the left, $G A$, only rows $i$ and $j$ of $A$ are affected. Thus we restrict attention to the following counterclockwise problem. Given $a$ and $b$, find $c = cos θ$ and $s = sin θ$ such that
 * $$ \begin{bmatrix} c & -s \\ s & c \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} r \\ 0 \end{bmatrix}, $$

where $$ r = \sqrt{a^2 + b^2} $$ is the length of the vector $$(a,b)$$. Explicit calculation of $θ$ is rarely necessary or desirable. Instead we directly seek $c$ and $s$. An obvious solution would be
 * $$\begin{align}

c &{}\larr a / r \\ s &{}\larr -b / r. \end{align}$$

However, the computation for $r$ may overflow or underflow. An alternative formulation avoiding this problem is implemented as the hypot function in many programming languages.

The following Fortran code is a minimalistic implementation of Givens rotation for real numbers. If the input values 'a' or 'b' are frequently zero, the code may be optimized to handle these cases as presented here.

Furthermore, as Edward Anderson discovered in improving LAPACK, a previously overlooked numerical consideration is continuity. To achieve this, we require $r$ to be positive. The following MATLAB/GNU Octave code illustrates the algorithm.

The IEEE 754  function, provides a safe and cheap way to copy the sign of   to. If that is not available, $|x|⋅sgn(y)$, using the abs and sgn functions, is an alternative as done above.

Triangularization
Given the following 3 ×  3 Matrix:


 * $$ A_1 =

\begin{bmatrix}  6    &    5    &    0   \\ 5   &    1    &    4     \\                         0    &    4    &    3     \\       \end{bmatrix},$$

two iterations of the Givens rotation (note that the Givens rotation algorithm used here differs slightly from above) yield an upper triangular matrix in order to compute the QR decomposition.

In order to form the desired matrix, zeroing elements (2, 1) and (3,  2) is required; element (2,  1) is zeroed first, using a rotation matrix of:
 * $$G_{1} =

\begin{bmatrix}  c    &    -s    &    0   \\ s  &    c    &    0     \\ 0   &    0    &    1     \\       \end{bmatrix}.$$

The following matrix multiplication results:

\begin{align} G_1 A_1 &{}= A_2 \\ &{} = \begin{bmatrix}  c    &    -s    &    0   \\ s  &    c    &    0     \\ 0   &    0    &    1     \\       \end{bmatrix} \begin{bmatrix}  6    &    5    &    0   \\ 5   &    1    &    4     \\                         0    &    4    &    3     \\       \end{bmatrix}, \end{align} $$

where
 * $$\begin{align}

r &{}= \sqrt{6^2 + 5^2} \approx 7.8102 \\ c &{}= 6 / r \approx 0.7682 \\ s &{}= -5 / r \approx -0.6402. \end{align} $$

Using these values for $c$ and $s$ and performing the matrix multiplication above yields $A_{2}$:
 * $$A_2 \approx \begin{bmatrix}  7.8102    &     4.4813    &    2.5607   \\

0   &    -2.4327    &    3.0729   \\                                 0    &          4    &         3   \\       \end{bmatrix}$$

Zeroing element (3, 2) finishes off the process. Using the same idea as before, the rotation matrix is:
 * $$G_{2} =

\begin{bmatrix}  1    &    0    &    0   \\ 0  &    c    &    -s     \\ 0   &    s   &    c     \\ \end{bmatrix}$$

Afterwards, the following matrix multiplication is:

\begin{align} G_2 A_2 &{}= A_3 \\ &{}\approx \begin{bmatrix}  1    &    0    &    0   \\ 0  &    c    &    -s     \\ 0   &    s   &    c     \\ \end{bmatrix} \begin{bmatrix}  7.8102    &    4.4813    &    2.5607   \\ 0  &    -2.4327    &    3.0729     \\                         0    &    4    &    3     \\       \end{bmatrix}, \end{align} $$

where
 * $$\begin{align}

r &{}\approx \sqrt{(-2.4327)^2 + 4^2} \approx 4.6817 \\ c &{}\approx -2.4327 / r \approx -0.5196 \\ s &{}\approx -4 / r \approx -0.8544. \end{align} $$ Using these values for $c$ and $s$ and performing the multiplications results in $A_{3}$:
 * $$A_3 \approx

\begin{bmatrix}  7.8102    &    4.4813    &    2.5607   \\ 0  &    4.6817    &    0.9665     \\                         0    &    0    &    -4.1843     \\       \end{bmatrix}. $$

This new matrix $A_{3}$ is the upper triangular matrix needed to perform an iteration of the QR decomposition. $Q$ is now formed using the transpose of the rotation matrices in the following manner:


 * $$Q = G_{1}^T\, G_{2}^T.

$$

Performing this matrix multiplication yields:
 * $$Q \approx

\begin{bmatrix}  0.7682    &   0.3327    &    0.5470   \\ 0.6402  &     -0.3992    &    -0.6564     \\                         0    &    0.8544    &     -0.5196     \\       \end{bmatrix}.$$

This completes two iterations of the Givens Rotation and calculating the QR decomposition can now be done.

Complex matrices
Another method can extend Givens rotations to complex matrices. A diagonal matrix whose diagonal elements have unit magnitudes but arbitrary phases is unitary. Let A be a matrix for which it is desired to make the ji element be zero using the rows and columns i and j>i. Let D be a diagonal matrix whose diagonal elements are one except the ii and jj diagonal elements which also have unit magnitude but have phases which are to be determined. The phases of the ii and jj elements of D can be chosen so as to make the ii and ji elements of the product matrix D A be real. Then a Givens rotation G can be chosen using the i and j>i rows and columns so as to make the ji element of the product matrix G D A be zero. Since a product of unitary matrices is unitary, the product matrix G D is unitary and so is any product of such matrix pair products.

In Clifford algebra
In Clifford algebra and its child structures such as geometric algebra, rotations are represented by bivectors. Givens rotations are represented by the exterior product of the basis vectors. Given any pair of basis vectors $$\mathbf e_i, \mathbf e_j$$ Givens rotations bivectors are:


 * $$B_{ij} = \mathbf e_i \wedge \mathbf e_j.$$

Their action on any vector is written:


 * $$v=e^{-(\theta/2)(\mathbf e_i \wedge \mathbf e_j)}u e^{(\theta/2)(\mathbf e_i \wedge \mathbf e_j)},$$

where


 * $$e^{(\theta/2)(\mathbf e_i \wedge \mathbf e_j)}= \cos(\theta/2)+ \sin(\theta/2) \mathbf e_i \wedge \mathbf e_j.$$

Dimension 3
There are three Givens rotations in dimension 3:



R_X(\theta) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{bmatrix}. $$


 * $$\begin{align} \\

R_Y(\theta) = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix} \end{align} $$


 * $$\begin{align} \\

R_Z(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{align} $$

Given that they are endomorphisms they can be composed with each other as many times as desired, keeping in mind that $g ∘ f ≠ f ∘ g$.

These three Givens rotations composed can generate any rotation matrix according to Davenport's chained rotation theorem. This means that they can transform the standard basis of the space to any other frame in the space.

When rotations are performed in the right order, the values of the rotation angles of the final frame will be equal to the Euler angles of the final frame in the corresponding convention. For example, an operator $$R = R_Y(\theta_3)\cdot R_X(\theta_2)\cdot R_Z(\theta_1)$$ transforms the basis of the space into a frame with angles roll, pitch and yaw $$YPR = (\theta_3,\theta_2,\theta_1)$$ in the Tait–Bryan convention z-x-y (convention in which the line of nodes is perpendicular to z and Y axes, also named Y-X′-Z″).

For the same reason, any rotation matrix in 3D can be decomposed in a product of three of these rotation operators.

The meaning of the composition of two Givens rotations $g ∘ f$ is an operator that transforms vectors first by $f$ and then by $g$, being $f$ and $g$ rotations about one axis of basis of the space. This is similar to the extrinsic rotation equivalence for Euler angles.

Table of composed rotations
The following table shows the three Givens rotations equivalent to the different Euler angles conventions using extrinsic composition (composition of rotations about the basis axes) of active rotations and the right-handed rule for the positive sign of the angles.

The notation has been simplified in such a way that $c_{1}$ means $cos θ_{1}$ and $s_{2}$ means $sin θ_{2})$. The subindexes of the angles are the order in which they are applied using extrinsic composition (1 for intrinsic rotation, 2 for nutation, 3 for precession)

As rotations are applied just in the opposite order of the Euler angles table of rotations, this table is the same but swapping indexes 1 and 3 in the angles associated with the corresponding entry. An entry like zxy means to apply first the y rotation, then x, and finally z, in the basis axes.

All the compositions assume the right hand convention for the matrices that are multiplied, yielding the following results.


 * {| class="wikitable" style="background-color:white; font-weight:500"

!xzx c_2 & - c_1 s_2 & s_1 s_2 \\ c_3 s_2  & c_3 c_2 c_1 - s_3 s_1  & - c_2 c_3 s_1 - c_1 s_3 \\ s_2 s_3  & c_3 s_1 + c_1 c_2 s_3  & c_3 c_1 - c_2 s_3 s_1 \end{bmatrix}$$ !xzy c_2 c_3    & - c_3 s_2 c_1 + s_3 s_1   &  c_3 s_2 s_1 + s_3 c_1 \\ s_2        & c_1 c_2                  & - c_2 s_1 \\ - s_3 c_2  & s_3 s_2 c_1+c_3 s_1     & -s_3 s_2 s_1 + c_3 c_1 \end{bmatrix}$$ !xyx c_2      & s_1 s_2    & c_1 s_2 \\ s_2 s_3  & c_3 c_1 - c_2 s_3 s_1 & - c_3 s_1 - c_1 c_2 s_3 \\ -c_3 s_2 & c_3 c_2 s_1 + c_1 s_3 & c_3 c_2 c_1 - s_3 s_1 \end{bmatrix}$$ !xyz c_3 c_2 &	-s_3 c_1 + c_3 s_2 s_1 &	s_3 s_1 + c_3 s_2 c_1 \\ s_3 c_2 &	c_3 c_1 + s_3 s_2 s_1 &	-c_3 s_1 + s_3 s_2 c_1 \\ -s_2 &	c_2 s_1 &	c_2 c_1 \end{bmatrix}$$ !yxy c_3 c_1 - c_2 s_3 s_1 & s_2 s_3 & c_3 s_1 + s_3 c_2 c_1 \\ s_1 s_2 & c_2 & - c_1 s_2 \\ -c_2 c_3 s_1 - c_1 s_3 & c_3 s_2 & c_3 c_2 c_1 - s_3 s_1 \end{bmatrix}$$ !yxz c_3 c_1-s_3 s_2 s_1 &	-s_3 c_2 &	c_3 s_1+s_3 s_2 c_1 \\ s_3 c_1+c_3 s_2 s_1 &	c_3 c_2 &	s_3 s_1-c_3 s_2 c_1 \\ -c_2 s_1 &	s_2 &	c_2 c_1 \end{bmatrix}$$ !yzy c_3 c_2 c_1 - s_3 s_1 & - c_3 s_2 & c_2 c_3 s_1 + c_1 s_3 \\ c_1 s_2 & c_2 & s_1 s_2 \\ -c_3 s_1 - c_1 c_2 s_3 & s_2 s_3 & c_3 c_1 - c_2 s_3 s_1 \end{bmatrix}$$ !yzx c_2 c_1 &	-s_2 &	c_2 s_1 \\ c_3 s_2 c_1+s_3 s_1 &	c_3 c_2 &	c_3 s_2 s_1-s_3 c_1 \\ s_3 s_2 c_1-c_3 s_1 &	s_3 c_2 &	s_3 s_2 s_1+c_3 c_1 \end{bmatrix}$$ !zyz c_3 c_2 c_1 - s_3 s_1 & - c_2 s_1 c_3 - c_1 s_3 & c_3 s_2 \\ c_3 s_1 + c_1 c_2 s_3 &  c_3 c_1 - c_2 s_3 s_1 & s_2 s_3 \\ -c_1 s_2 & s_1 s_2 & c_2 \end{bmatrix}$$ !zyx c_2 c_1 	    &           -c_2 s_1    &   s_2 \\ s_3 s_2 c_1+c_3 s_1 &	-s_3 s_2 s_1+c_3 c_1 &	-s_3 c_2 \\ -c_3 s_2 c_1+s_3 s_1 &	 c_3 s_2 s_1+s_3 c_1 &	 c_3 c_2 \end{bmatrix}$$ !zxz c_3 c_1 - c_2 s_1 s_3 & - c_3 s_1 - c_1 c_2 s_3 & s_2 s_3 \\ c_2 c_3 s_1 + c_1 s_3 & c_3 c_2 c_1 - s_3 s_1 & - c_3 s_2 \\ s_1 s_2 & c_1 s_2 & c_2 \end{bmatrix}$$ !zxy c_3 c_1+s_3 s_2 s_1 &	-c_3 s_1+s_3 s_2 c_1 &	s_3 c_2 \\ c_2 s_1         &	c_2 c_1 	   &      -s_2 \\ -s_3 c_1+c_3 s_2 s_1 &	s_3 s_1+c_3 s_2 c_1 &  	c_3 c_2 \end{bmatrix}$$
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