6.7. Transformations exercises#
Exercise 6.1
Which of the following transformations are linear transformations?
(a) \(T: (x, y)^\mathsf{T} \mapsto (0, y)^\mathsf{T}\)
(b) \(T: (x, y)^\mathsf{T} \mapsto (x, 5)^\mathsf{T}\)
(c) \(T: (x, y, z)^\mathsf{T} \mapsto (x, x - y)^\mathsf{T}\)
(d) \(T: (x, y, z)^\mathsf{T} \mapsto \begin{pmatrix} x + y \\ z \end{pmatrix}\)
(e) \(T: (x, y)^\mathsf{T} \mapsto (3x + 1, y)^\mathsf{T}\)
(f) \(T: f(x) \mapsto \dfrac{\mathrm{d}}{\mathrm{d}x} f(x)\)
(g) \(T: f(x) \mapsto xf(x)\)
(h) \(T: \mathbb{C}^2 \to \mathbb{C}\) where \(T: (x, y)^\mathsf{T} \mapsto x + y\)
(i) \(T: \mathbb{C}^2 \to \mathbb{C}\) where \(T: (x, y)^\mathsf{T} \mapsto x y\)
(j) \(T: \mathbb{C}^2 \to \mathbb{C}\) where \(T: (x, y)^\mathsf{T} \mapsto \bar{y}\)
(\(\bar{x}\) is the complex conjugate of \(x = a + bi\) defined by \(\bar{x} = a - bi\).)
Solution to Exercise 6.1
(a) Let \(\vec{u} = (u_1, u_2)^\mathsf{T}, \vec{v} = (v_1, v_2)^\mathsf{T} \in \mathbb{R}^2\) and \(\alpha \in \mathbb{R}\)
therefore \(T\) is a linear transformation.
(b) \(T\) is not a linear transformation since
(c) Let \(\vec{u} = (u_1, u_2)^\mathsf{T}, \vec{v} = (v_1, v_2)^\mathsf{T} \in \mathbb{R}^2\) and \(\alpha \in \mathbb{R}\)
therefore \(T\) is a linear transformation.
(d) Let \(\vec{u} = (u_1, u_2, v_3)^\mathsf{T},\vec{v} = (v_1, v_2, v_3)^\mathsf{T}\in \mathbb{R}^3\) and \(\alpha \in \mathbb{R}\)
therefore \(T\) is a linear transformation.
(e) \(T\) is not a linear transformation since
(f) Let \(u = f(x), v = g(x) \in P(\mathbb{R})\) and \(\alpha \in \mathbb{R}\):
therefore \(T\) is a linear transformation.
(g) Let \(u = f(x), v = g(x) \in P(\mathbb{R})\) and \(\alpha \in \mathbb{R}\):
therefore \(T\) is a linear transformation.
Exercise 6.2
A linear transformation \(T: \mathbb{R}^2 \to \mathbb{R}^2\) is defined by \(T: (x, y)^\mathsf{T} \mapsto (-x + 3y, x - 4y)^\mathsf{T}\). Determine the transformation matrix for \(T\) and hence calculate \(T (2, 5)^\mathsf{T}\).
Solution to Exercise 6.2
The transformation matrix is
Calculating \(T \begin{pmatrix} 2 \\ 5 \end{pmatrix}\)
therefore \(T\begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 13 \\ -18 \end{pmatrix}\).
Exercise 6.3
A linear transformation \(T: \mathbb{R}^2 \to \mathbb{R}^2\) is defined by \(T: (x, y)^\mathsf{T} \mapsto (x - 2y, 2x + 3y)^\mathsf{T}\). Given \(T(\vec{u}) = (-1, 5)^\mathsf{T}\) determine \(\vec{u}\).
Solution to Exercise 6.3
The transformation matrix is
so the inverse is
Therefore
Exercise 6.4
\(T: \mathbb{R}^3 \to \mathbb{R}^3\) is a linear transformation such that
Find the transformation matrix for \(T\).
Solution to Exercise 6.4
The transformation matrix is determined using equation (6.2) which is
Using Gauss-Jordan elimination to calculate the inverse of \((\vec{u}_1, \vec{u}_2, \vec{u}_3)^{-1}\)
So \((\vec{u}_1, \vec{u}_2, \vec{u}_3)^{-1} = \begin{pmatrix} -1 & -2 & 1 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix}\) and
Checking \(A\)
Exercise 6.5
Rotate the position vector \((2, 1)^\mathsf{T} \in \mathbb{R}^2\) by angle \(\pi/6\) anti-clockwise about the origin.
Solution to Exercise 6.5
The transformation matrix is
therefore
Exercise 6.6
Reflect the position vector \((5, 3)^\mathsf{T} \in \mathbb{R}^2\) about the line that passes through \((0, 0)\) and makes an angle \(\pi/3\) with the \(x\)-axis.
Solution to Exercise 6.6
The transformation matrix is
therefore
Exercise 6.7
A square with side lengths 2 is centred at the co-ordinates \((3, 2)\). It is to be translated so the centre is at the origin, rotated by an angle \(\pi/3\) clockwise about the origin and then translated back to its initial position.
(a) Write down a matrix containing the homogeneous co-ordinates for the vertices of the square.
(b) Determine the transformation matrices that perform the three transformations.
(c) Calculate the composite transformation matrix and apply with to the co-ordinate matrix from part (a).
Solution to Exercise 6.7
(a) \(\begin{pmatrix} 2 & 4 & 4 & 2 \\ 1 & 1 & 3 & 3 \\ 1 & 1 & 1 & 1 \end{pmatrix} \)
(b) Translate by \((-3, -2)^\mathsf{T}\) so that the centre of the square is at the origin:
Rotate by \(\pi/3\) clockwise about the origin:
Translate by \((3, 2)^\mathsf{T}\) so that the centre of the square is back to \(\vec{c}\)
(c) Calculate composite alignment matrix
Apply composite transformation matrix
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