Linear Transformations Exercises

Exercise 10

A linear transformation \(T: \mathbb{R}^3 \to \mathbb{R}^3\) is defined by \(T(x,y,z) = (x + 2y - z, 3x + z, 2x - y + 3z)\). Determine the transformation matrix for \(T\) and use it to calculate \(T(2, 5, 1)\).

Solution to Exercise 10

\[\begin{split} \begin{align*} T(\mathbf{e}_1) &= T \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}, \\ T(\mathbf{e}_2) &= T \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}, \\ T(\mathbf{e}_3) &= T \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix}, \end{align*} \end{split}\]

therefore the transformation matrix is

\[\begin{split} \begin{align*} A &= \begin{pmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 2 & -1 & 3 \end{pmatrix}. \end{align*} \end{split}\]

Calculating \(T(2, 5, 1)\)

\[\begin{split} \begin{align*} T\begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix} &= \begin{pmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 2 & -1 & 3 \end{pmatrix} \begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix} = \begin{pmatrix} 11 \\ 7 \\ 2 \end{pmatrix}. \end{align*} \end{split}\]

Exercise 11

\(T: \mathbb{R}^3 \to \mathbb{R}^3\) is a linear transformation such that

\[\begin{split} \begin{align*} T\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} &= \begin{pmatrix} 1 \\ -2 \\ -4 \end{pmatrix}, & T\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} &= \begin{pmatrix} 6 \\ 5 \\ 10 \end{pmatrix}, & T\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} &= \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix}. \end{align*} \end{split}\]

Find the transformation matrix for \(T\).

Solution to Exercise 11

The transformation matrix is determined using equation (20) which is

\[\begin{split} \begin{align*} A &= (T(\mathbf{u}_1), T(\mathbf{u}_2), \ldots, T(\mathbf{u}_n)) \cdot (\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_n)^{-1}, \\ &= \begin{pmatrix} 1 & 6 & 2 \\ -2 & 5 & 4 \\ -4 & 10 & 7 \end{pmatrix} \begin{pmatrix} 1 & 0 & -1 \\ -1 & 1 & 1 \\ 0 & 2 & 1 \end{pmatrix}^{-1} \end{align*} \end{split}\]

The inverse of the right-hand matrix is

\[\begin{split} \begin{align*} \begin{pmatrix} 1 & 0 & -1 \\ -1 & 1 & 1 \\ 0 & 2 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} -1 & -2 & 1 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix} \end{align*} \end{split}\]

so

\[\begin{split} \begin{align*} A &= \begin{pmatrix} 1 & 6 & 2 \\ -2 & 5 & 4 \\ -4 & 10 & 7 \end{pmatrix} \begin{pmatrix} -1 & -2 & 1 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 3 \\ -1 & 1 & 2 \\ 0 & 4 & 3 \end{pmatrix}. \end{align*} \end{split}\]

Checking \(A\)

\[\begin{split} \begin{align*} T\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} &= \begin{pmatrix} 1 & 0 & 3 \\ -1 & 1 & 2 \\ 0 & 4 & 3 \end{pmatrix} \begin{pmatrix}1 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ - 4 \end{pmatrix} \quad \checkmark \end{align*} \end{split}\]

Exercise 12

An equilateral triangle is defined in \(\mathbb{R}^2\) is centered \((3, 2)\) at the origin with side lengths 2 and with one side parallel to the \(x\)-axis. The triangle is to be translated by the translation vector \(\mathbf{t} = (3, 1)\).

(a) determine the co-ordinates off the three vertices of the triangle before the translation has been applied;

(b) determine the transformation matrix for the translation;

(c) using MATLAB, calculate the vertex co-ordinates of the translated triangle;

(d) plot the pre and post-translated triangle on the same axes.

Solution to Exercise 12

(a) Since the side lengths are 2 then the height of the triangle is \(h = \sqrt{3}\). Let \((x_1,y_1)\), \((x_2,y_2)\) and \((x_3,y_3)\) be the vertex co-ordinates then \(x_1 = 2\), \(x_2 = 4\), \(x_3 = 3\), \(y_2 = y_1\) and \(y_3 = y_1 + h\). The triangle is centred at \((3, 2)\) with a side parallel to the \(x\)-axis so

\[\begin{split} \begin{align*} 2 &= \frac{y_1 + y_2 + y_3}{3} = \frac{y_1 + y_1 + y_1 + h}{3} = \frac{3y_1 + \sqrt{3}}{3} \\ \therefore y_1 &= 2 - \frac{\sqrt{3}}{3} \approx 1.4226, \end{align*} \end{split}\]

so the vertex co-ordinates are

\[\begin{split} \begin{align*} &\begin{pmatrix} 2 \\ 2 - \sqrt{3}/3 \end{pmatrix}, & &\begin{pmatrix} 4 \\ 2 - \sqrt{3}/3 \end{pmatrix}, & &\begin{pmatrix} 3 \\ 2 + 2\sqrt{3}/3 \end{pmatrix}. \end{align*} \end{split}\]

(b) \( T \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\)

(c)

clear

% Define vertex array
P = [ 2, 4, 3 ;
      2 - sqrt(3)/3, 2 - sqrt(3)/3, 2 + 2 * sqrt(3)/3 ;
      1, 1, 1 ];

% Define translation matrix
T = @(t) [1, 0, t(1) ; 
          0, 1, t(2) ;
          0, 0, 1 ];

% Apply translation
t = [3 ; 1];
P1 = T(t) * P

\[\begin{split}P_1 = \begin{pmatrix} 5 & 7 & 6 \\ 2.4226 & 2.4226 & 4.1547 \\ 1 & 1 & 1 \end{pmatrix} \end{split}\]

(d)

figure
patch(P(1,:), P(2,:), 'b', FaceAlpha=0.5)
patch(P1(1,:), P1(2,:), 'r', FaceAlpha=0.5)
axis equal
axis([0, 8, 0, 5])
xlabel('$x$', FontSize=12, Interpreter='latex')
ylabel('$y$', FontSize=12, Interpreter='latex')

Exercise 13

The translated triangle from Exercise 12 is scaled by a factor of 0.5 in both directions. In MATLAB, calculate the vertex co-ordinates of the scaled triangle and produce a plot showing the pre and post scaled triangle on the same axes.

Solution to Exercise 13

% Calculate centre of the triangle
c = mean(P1, 2);

% Define scaling matrix
S = @(s) [ s(1), 0, 0 ;
           0, s(2), 0 ;
           0, 0, 1 ];

% Apply transformations (remembering to translate to and from origin)
s = [0.5 ; 0.5];
P2 = T(c) * S(s) * T(-c) * P1;

% Plot pre and post scaled triangle
figure
patch(P1(1,:), P1(2,:), 'b', FaceAlpha=0.5)
patch(P2(1,:), P2(2,:), 'r', FaceAlpha=0.5)
axis equal
axis([0, 8, 0, 5])
xlabel('$x$', FontSize=12, Interpreter='latex')
ylabel('$y$', FontSize=12, Interpreter='latex')

Exercise 14

The translated triangle from Exercise 12 is rotated by \(\theta=\pi/4\) anti-clockwise about its centre. In MATLAB, calculate the vertex co-ordinates of the rotated triangle and produce a plot showing the pre and post rotated triangle on the same axes.

Solution to Exercise 14

% Calculate centre of the triangle
c = mean(P1, 2);

% Define rotation matrix
R = @(th) [ cos(theta), -sin(theta), 0 ;
            sin(theta), cos(theta), 0 ;
            0, 0, 1 ];

% Apply transformations (remembering to translate to and from origin)
theta = pi / 4;
P3 = T(c) * R(theta) * T(-c) * P1;

% Plot pre and post scaled triangle
figure
patch(P1(1,:), P1(2,:), 'b', FaceAlpha=0.5)
patch(P3(1,:), P3(2,:), 'r', FaceAlpha=0.5)
axis equal
axis([0, 8, 0, 5])
xlabel('$x$', FontSize=12, Interpreter='latex')
ylabel('$y$', FontSize=12, Interpreter='latex')

Exercise 15

The MATLAB code below produces a 5 second animation of a point \(\mathbf{p}\) travelling in a circular arc centred at \((10,10)\) with radius 8. The point performs one full rotation per second.

% Define arc parameters
c = [10 ; 10];
r = 8;

% Loop through animation frames
fps = 60;
for t = 0 : 1/fps : 5

    % Calculate point position
    phi = t * 2 * pi;
    p = c + r * [cos(phi) ; sin(phi)];

    % Plot point
    clf
    plot(p(1), p(2), 'bo', MarkerFaceColor='b')
    axis equal
    axis([0, 20, 0, 20])
    title(sprintf('time = %1.2f s', t))
    xlabel('$x$', FontSize=12, Interpreter='latex')
    ylabel('$y$', FontSize=12, Interpreter='latex')
    box on
    drawnow
end

Copy this code into a MATLAB live script and edit it so that it does the following:

(a) animate a square with side lengths 1 travelling with a centre at \(\mathbf{p}\). (Hint: define the square vertices so it is centred at the origin);

(b) the square from part (a) is scaled by the scaling vector \(\mathbf{s} = (2+\cos(5t), 2+\cos(5t))\) about its centre;

(c) the square rotates in an clockwise direction about its centre so that the square performs 8 rotations per second.

Solution to Exercise 15

clear

% Define arc parameters
c = [10 ; 10];
r = 8;

% Define square centred at the origin
P = [ -0.5,  0.5, 0.5, -0.5 ;
      -0.5, -0.5, 0.5,  0.5 ;
       1,    1,   1,    1 ];

% Define transformation matrices
T = @(t) [1, 0, t(1) ; 
          0, 1, t(2) ;
          0, 0, 1 ];
S = @(s) [ s(1), 0, 0 ;
           0, s(2), 0 ;
           0, 0, 1 ];
R = @(th) [ cos(th), -sin(th), 0 ;
            sin(th), cos(th), 0 ;
            0, 0, 1 ];

fps = 60;
for t = 0 : 1/fps : 5

    % Calculate square centre position
    phi = t * 2 * pi;
    p = c + r * [cos(phi) ; sin(phi)];

    % Transform square
    P1 = T(p) * R(-8*t) * S([2+cos(5*t), 2+cos(5*t)]) * P;
    
    % Plot square
    clf
    patch(P1(1,:), P1(2,:), 'b', FaceAlpha=0.5)
    axis equal
    axis([0, 20, 0, 20])
    title(sprintf('time = %1.2f s', t))
    xlabel('$x$', FontSize=12, Interpreter='latex')
    ylabel('$y$', FontSize=12, Interpreter='latex')
    box on
    drawnow
end

Exercise 16

A set of co-ordinates is to be rotated by \(\theta=\pi/6\) anti-clockwise about the line that passes through the point \((2, -4, -3)\) and has the direction vector \((-2, 3, 4)\). Determine the individual transformation matrices used to perform this rotation and write down an expression for calculating a single composite transformation matrix.

Solution to Exercise 16

First translate by \(\mathbf{-p}\) so the translation matrix is

\[\begin{split} \begin{align*} T(-\mathbf{p}) &= \begin{pmatrix} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

Rotate the direction vector \(\mathbf{d}=(-2,3,4)\) anti-clockwise about the \(z\)-axis by angle \(\phi\) so that it is in the \(xz\) plane. The sine and cosine of \(\phi\) are

\[\begin{split} \begin{align*} \cos(\phi) &= \frac{|d_x|}{d_1} = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}, \\ \sin(\phi) &= \frac{|d_y|}{d_1} = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}, \end{align*} \end{split}\]

so the rotation matrix is

\[\begin{split} \begin{align*} R_z(\phi) &= \begin{pmatrix} 2\sqrt{13}/13 & -3\sqrt{13}/13 & 0 & 0 \\ 3\sqrt{13}/13 & 2\sqrt{13}/13 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

Rotate the direction vector anti-clockwise about the \(y\) axis by angle \(\psi\) so that it points along the \(z\)-axis. The sine and cosine of \(\psi\) are

\[\begin{split} \begin{align*} \cos(\psi) &= \frac{|d_z|}{|\mathbf{d}|} = \frac{4}{\sqrt{29}} = \frac{4\sqrt{29}}{29}, \\ \sin(\psi) &= \frac{d_1}{|\mathbf{d}|} = \frac{\sqrt{13}}{\sqrt{29}} = \frac{\sqrt{13}\sqrt{29}}{29}, \end{align*} \end{split}\]

so the rotation matrix is

\[\begin{split} \begin{align*} R_y(\psi) &= \begin{pmatrix} 4\sqrt{29}/29 & 0 & \sqrt{13}\sqrt{29}/29 & 0 \\ 0 & 1 & 0 & 0 \\ -\sqrt{13}\sqrt{29}/29 & 0 & 4\sqrt{29}/29 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

Now that the direction vector points along the \(z\)-axis, we perform the rotation by \(\theta=\pi/6\) so the rotation matrix is

\[\begin{split} \begin{align*} R\left(\frac{\pi}{6}\right) &= \begin{pmatrix} \cos(\pi/6) & \sin(\pi/6) & 0 & 0 \\ -\sin(\pi/6) & \cos\pi/6) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} \sqrt{3}/2 & -1/2 & 0 & 0 \\ 1/2 & \sqrt{3}/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

The inverse transformations are

\[\begin{split} \begin{align*} R_y(-\psi) &= \begin{pmatrix} 4\sqrt{29}/29 & 0 & -\sqrt{13}\sqrt{29}/29 & 0 \\ 0 & 1 & 0 & 0 \\ \sqrt{13}\sqrt{29}/29 & 0 & 4\sqrt{29}/29 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \\ R_z(-\phi) &= \begin{pmatrix} 2\sqrt{13}/13 & -3\sqrt{13}/13 & 0 & 0 \\ 3\sqrt{13}/13 & 2\sqrt{13}/13 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \\ T(\mathbf{p}) &= \begin{pmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 1 \end{pmatrix} \end{align*} \end{split}\]

and the composite transformation matrix is

\[ \begin{align*} A &= T(\mathbf{p}) \cdot R_z(-\phi) \cdot R_y(-\psi) \cdot R_z\left(\frac{\pi}{6}\right) \cdot R_y(\psi) \cdot R_z(\phi) \cdot T(-\mathbf{p}). \end{align*} \]