6.3. Rotation, reflection and scaling#
Three of the most common geometrical linear transformations is rotation of vectors about the origin, reflection of vectors about a line and translation of vectors from one position to another.
6.3.1. Rotation#
(Rotation transformation)
The linear transformation \(Rot(\theta): \mathbb{R}^2 \to \mathbb{R}^2\) applied to the vector \(\vec{u}\) rotates \(\vec{u}\) by an angle \(\theta\) anti-clockwise about the origin.
Consider the diagram in Fig. 6.2. Here we have the vector \(\vec{u}\) which is rotated by angle \(\theta\) anti-clockwise about the origin to give the vector \(\vec{v}\). To determine the linear mapping \(Rot(\theta): \vec{u} \mapsto \vec{v}\) we first consider the rotation from the vector \(\vec{e}_1|\vec{u}|\), which points along the \(x\)-axis and has the same magnitude as \(\vec{u}\), by angle \(\phi\) anti-clockwise about the original to \(\vec{u}=(u_x, u_y)^\mathsf{T}\). Forming a right-angled triangle (the blue triangle) with the angle \(\phi\) and hypotenuse of length \(|\vec{u}|\) then
Doing similar for \(\vec{v} = (v_x, v_y)^\mathsf{T}\) (the red triangle) we have
using the angle sum identities
then
Substituting equation (6.6) into equation (6.7)
Therefore the vector \(\vec{u}\) rotated by angle \(\theta\) anti-clockwise about the original is
\(\mathbb{R}^2\))
(Rotation inThe rotation of a vector \(\vec{u} = (u_x, u_y)^\mathsf{T} \in \mathbb{R}^2\) anti-clockwise by an angle \(\theta\) is the linear transformation \(Rot(\theta): \mathbb{R}^2 \to \mathbb{R}^2\) defined by
The transformation matrix for rotation in \(\mathbb{R}^2\) is
6.3.1.1. Inverse rotation transformation#
The inverse transformation of rotating a vector anti-clockwise by an angle \(\theta\) is rotating clockwise by the same angle. This can be achieved by negating \(\theta\), since \(\cos(-\theta) = \cos(\theta)\) and \(\sin(-\theta) = -\sin(\theta)\) then
We could have also calculated the inverse of the rotation matrix to give the same result
Rotate the vector \(\vec{u} = (2, 1)^\mathsf{T}\) by angle \(\theta = \frac{\pi}{2}\) anti-clockwise about the origin.
Solution
The transformation matrix for this rotation is
Applying the transformation to \(\vec{u} = (2, 1)^\mathsf{T}\)
This rotation is illustrated in the diagram below
6.3.2. Reflection#
(Reflection about a line)
The linear transformation \(Re\!f(\theta): \mathbb{R}^2 \to \mathbb{R^2}\) is the reflection of a vector \(\vec{u} \in \mathbb{R}^2\) about the line which passes through the origin and makes an angle of \(\theta\) with the \(x\)-axis such that the distance of the head of the image vector \(\vec{v}\) is the same as that of \(\vec{u}\).
The simplest reflection we can perform is to reflect a vector \(\vec{u} =(u_x, u_y)^\mathsf{T}\) about the \(x\)-axis or \(y\)-axis (Fig. 6.4).
To reflect about the \(x\)-axis we simply change the sign of the \(y\) co-ordinate
and the transformation matrix is
Note that here the line of reflection makes an angle of \(\theta=0\) with the \(x\)-axis.
The determine the reflection about a line that makes and arbitrary angle \(\theta\) with the \(x\)-axis we first perform a rotation by \(-\theta\) so that the line of reflection is on the \(x\)-axis (Fig. 6.5). This means we can then use equation (6.9) to reflect the rotated \(\vec{u}\) vector to give the vector \(\vec{v}\) (Fig. 6.6) before rotating by \(\theta\) so that the line of reflection is back to its original position (Fig. 6.7).
Using composite transformations we can calculate the transformation matrix for reflection about a line
therefore
Using double angle formulae
then
\(\mathbb{R}^2\))
(Reflection inThe reflection of a vector \(\vec{u} = (u_x, u_y)^\mathsf{T} \in \mathbb{R}^2\) about a line which passes through the origin and makes an angle \(\theta\) with the \(x\)-axis (i.e., \(y = \tan(\theta)x\)) is the linear transformation \(Re\!f(\theta) : \mathbb{R}^2 \to \mathbb{R}^2\) defined by
The transformation matrix for reflection in \(\mathbb{R}^2\) is
6.3.2.1. Inverse reflection transformation#
The inverse of reflecting about a line is to simply perform the reflection again, i.e.,
Reflect the vector \(\vec{u} = (3, -1)^\mathsf{T}\) about the line that passes through the origin and has gradient of 1.
Solution
Since the line of reflection has gradient 1 then \(\theta = \tan^{-1}(1) = \pi/4\) and the reflection matrix is
Applying the transformation to \(\vec{u} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}\)
6.3.3. Scaling#
(Scaling transformation)
The linear transformation \(S(\vec{s}) : \mathbb{R}^n \to \mathbb{R}^n\) where \(\vec{s} \in \mathbb{R}^n\) applied to the position vector \(\vec{u} \in \mathbb{R}^n\) scales \(\vec{u}\) so that the head is moved closer or further away from the origin.
The scaling of a vector \(\vec{u} = (u_1, u_2, \ldots, u_n)^\mathsf{T} \in \mathbb{R}^n\) by a scaling vector \(\vec{s} = (s_1, s_2, \ldots, s_n)^\mathsf{T}\) is achieved by multiplying the corresponding elements in \(\vec{u}\) and \(\vec{s}\).
(Scaling transformation)
The scaling of a vector \(\vec{u} \in \mathbb{R}^n\) by the scaling vector \(\vec{s}\) is the linear transformation \(S(\vec{s}): \vec{u} \mapsto \vec{v}\) calculated using
The transformation matrix for scaling is
6.3.3.1. Inverse scaling transformation#
Since the inverse operation of scalar multiplication is scalar division then the inverse scaling matrix is
Scale the point with position vector \(\vec{u} = (2, 1)^\mathsf{T}\) by scaling scaling vector \(\vec{s} = (2, 3)^\mathsf{T}\).
Solution
The transformation matrix is
Applying the transformation to \(\vec{u} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\)