Linear Transformations#
Solution to Exercise 6.1
(a) Let \(\vec{u} = (u_1, u_2), \vec{v} = (v_1, v_2) \in \mathbb{R}^2\) and \(\alpha \in \mathbb{R}\)
\[\begin{split} \begin{align*}
T(\vec{u} + \alpha \vec{v})
&= T\begin{pmatrix} u_1 + \alpha v_1 \\ u_2 + \alpha v_2 \end{pmatrix}
= \begin{pmatrix} 0 \\ u_2 + \alpha v_2 \end{pmatrix}, \\
T(\vec{u}) + \alpha T(\vec{v})
&= T\begin{pmatrix} u_1 \\ u_2 \end{pmatrix} + \alpha T \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}
= \begin{pmatrix} 0 \\ u_2 + \alpha v_2 \end{pmatrix},
\end{align*} \end{split}\]
therefore \(T\) is a linear transformation.
(b) \(T\) is not a linear transformation since
\[\begin{split} \begin{align*}
T\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 2 \\ 5 \end{pmatrix}, \\
T\begin{pmatrix} 1 \\ 0 \end{pmatrix} + T\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 10 \end{pmatrix}.
\end{align*} \end{split}\]
(c) Let \(\vec{u} = (u_1, u_2), \vec{v} = (v_1, v_2) \in \mathbb{R}^2\) and \(\alpha \in \mathbb{R}\)
\[\begin{split} \begin{align*}
T(\vec{u} + \alpha \vec{v})
&= T\begin{pmatrix} u_1 + \alpha v_1 \\ u_2 + \alpha v_2 \end{pmatrix}
= \begin{pmatrix} u_1 + \alpha v_1 \\ u_1 - u_2 + \alpha v_1 - \alpha v_2 \end{pmatrix}, \\
T(\vec{u}) + \alpha T(\vec{v})
&= T\begin{pmatrix} u_1 \\ u_2 \end{pmatrix} + \alpha T\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}
= \begin{pmatrix} u_1 + \alpha v_1 \\ u_1 + \alpha v_1 - u_2 - \alpha v_2\end{pmatrix},
\end{align*} \end{split}\]
therefore \(T\) is a linear transformation.
(d) Let \(\vec{u} = (u_1, u_2, v_3),\vec{v} = (v_1, v_2, v_3)\in \mathbb{R}^3\) and \(\alpha \in \mathbb{R}\)
\[\begin{split} \begin{align*}
T(\vec{u} + \alpha \vec{v})
&= T\begin{pmatrix} u_1 + \alpha v_1 \\ u_2 + \alpha v_2 \\ u_3 + \alpha v_3 \end{pmatrix}
= \begin{pmatrix} u_1 + u_2 + \alpha v_1 + \alpha v_2 \\ u_3 + \alpha v_3 \end{pmatrix} , \\
T(\vec{u}) + \alpha T(\vec{v})
&= T\begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} + \alpha \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}
= \begin{pmatrix} u_1 + \alpha v_1 + u_2 + \alpha v_2 \\ u_3 + \alpha v_3 \end{pmatrix}
\end{align*} \end{split}\]
therefore \(T\) is a linear transformation.
(e) \(T\) is not a linear transformation since
\[\begin{split} \begin{align*}
T\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right)
&= T \begin{pmatrix} 2 \\ 1 \end{pmatrix}
= \begin{pmatrix} 7 \\ 1 \end{pmatrix}, \\
T\begin{pmatrix} 1 \\ 0 \end{pmatrix} + T \begin{pmatrix} 1 \\ 1 \end{pmatrix}
&= \begin{pmatrix} 4 \\ 0 \end{pmatrix} + \begin{pmatrix} 4 \\ 1 \end{pmatrix}
= \begin{pmatrix} 8 \\ 1 \end{pmatrix}.
\end{align*} \end{split}\]
(f) Let \(u = f(x), v = g(x) \in P(\mathbb{R})\) and \(\alpha \in \mathbb{R}\):
\[\begin{split} \begin{align*}
T(u + \alpha v) &= T(f(x) + \alpha g(x)) = \frac{\mathrm{d}}{\mathrm{d} x}(f(x) + \alpha g(x))
= \frac{\mathrm{d}}{\mathrm{d} x}f(x) + \alpha \frac{\mathrm{d}}{\mathrm{d} x} g(x), \\
T(u) + \alpha T(v) &= T(f(x)) + \alpha T(g(x))
= \frac{\mathrm{d}}{\mathrm{d} x}f(x) + \alpha \frac{\mathrm{d}}{\mathrm{d} x}g(x),
\end{align*} \end{split}\]
therefore \(T\) is a linear transformation.
(g) Let \(u = f(x), v = g(x) \in P(\mathbb{R})\) and \(\alpha \in \mathbb{R}\):
\[\begin{split} \begin{align*}
T(u + \alpha v) &= T(f(x) + \alpha g(x)) = x(f(x) + \alpha g(x)) = xf(x) + \alpha x g(x), \\
T(u) + \alpha T(v) &= T(f(x)) + \alpha T(g(x)) = xf(x) + \alpha x g(x),
\end{align*} \end{split}\]
therefore \(T\) is a linear transformation.
Solution to Exercise 6.2
The transformation matrix is
\[\begin{split} A = \begin{pmatrix} -1 & 3 \\ 1 & -4 \end{pmatrix} \end{split}\]
Calculating \(T (2, 5)\)
\[\begin{split} T
\begin{pmatrix} 2 \\ 5 \end{pmatrix} =
\begin{pmatrix} -1 & 3 \\ 1 & -4 \end{pmatrix}
\begin{pmatrix} 2 \\ 5 \end{pmatrix}
= \begin{pmatrix} 13 \\ -18 \end{pmatrix}. \end{split}\]
Solution to Exercise 6.3
The transformation matrix is
\[\begin{split} A = \begin{pmatrix} 1 & -2 \\ 2 & 3 \end{pmatrix}, \end{split}\]
so the inverse is
\[\begin{split} A^{-1} = \frac{1}{7}
\begin{pmatrix}
3 & 2 \\
-2 & 1
\end{pmatrix}. \end{split}\]
Therefore
\[\begin{split} \vec{u} = A^{-1}
\begin{pmatrix} -1 \\ 5 \end{pmatrix} = \frac{1}{7}
\begin{pmatrix}
3 & 2 \\
-2 & 1
\end{pmatrix}
\begin{pmatrix} -1 \\ 5 \end{pmatrix} =
\begin{pmatrix} 1 \\ 1 \end{pmatrix}. \end{split}\]
Solution to Exercise 6.4
The transformation matrix is determined using equation (6.2) which is
\[A = \begin{pmatrix} T(\vec{u}_1) & T(\vec{u}_2) & \cdots & T(\vec{u}_n) \end{pmatrix}
\begin{pmatrix} \vec{u}_1 & \vec{u}_2 & \cdots & \vec{u}_n \end{pmatrix}^{-1}.\]
Using Gauss-Jordan elimination to calculate the inverse of \((\vec{u}_1, \vec{u}_2, \vec{u}_3)^{-1}\)
\[\begin{split} \begin{align*}
& \left( \begin{array}{rrr|rrr}
1 & 0 & -1 & 1 & 0 & 0 \\
-1 & 1 & 1 & 0 & 1 & 0 \\
0 & 2 & 1 & 0 & 0 & 1
\end{array} \right)
\begin{array}{l} \\ R_2 + R_1 \\ \phantom{x} \end{array} \\ \\
\longrightarrow \qquad
& \left( \begin{array}{rrr|rrr}
1 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 0 & 1 & 1 & 0 \\
0 & 2 & 1 & 0 & 0 & 1
\end{array} \right)
\begin{array}{l} \\ \\ R_3 - 2 R_2 \end{array} \\ \\
\longrightarrow \qquad
& \left( \begin{array}{rrr|rrr}
1 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & 0 & 1 & 1 & 0 \\
0 & 0 & 1 & -2 & -2 & 1
\end{array} \right)
\begin{array}{l} R_1 + R_3 \\ \phantom{x} \\ \phantom{x} \end{array} \\ \\
\longrightarrow \qquad
& \left( \begin{array}{rrr|rrr}
1 & 0 & 0 & -1 & -2 & 1 \\
0 & 1 & 0 & 1 & 1 & 0 \\
0 & 0 & 1 & -2 & -2 & 1
\end{array} \right)
\end{align*} \end{split}\]
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
\[\begin{split} \begin{align*}
A &= \begin{pmatrix} 1 & 6 & 2 \\ -2 & 5 & 4 \\ -4 & 10 & 7 \end{pmatrix}
\begin{pmatrix}
-1 & -2 & 1 \\
1 & 1 & 0 \\
-2 & -2 & 1
\end{pmatrix}
= \begin{pmatrix} 1 & 0 & 3 \\ -1 & 1 & 2 \\ 0 & 4 & 3 \end{pmatrix}.
\end{align*} \end{split}\]
Checking \(A\)
\[\begin{split} T \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} =
\begin{pmatrix} 1 & 0 & 3 \\ -1 & 1 & 2 \\ 0 & 4 & 3 \end{pmatrix}
\begin{pmatrix}1 \\ -1 \\ 0 \end{pmatrix} =
\begin{pmatrix} 1 \\ -2 \\ - 4 \end{pmatrix} \quad \checkmark \end{split}\]
Solution to Exercise 6.5
The transformation matrix is
\[\begin{split} \begin{align*}
Rot\left(\pi/6\right) &= \begin{pmatrix}
\cos(\pi/6) & -\sin(\pi/6) \\
\sin(\pi/6) & \cos(\pi/6)
\end{pmatrix} \\
&= \begin{pmatrix}
\sqrt{3}/2 & -1/2 \\
1/2 & \sqrt{3}/2
\end{pmatrix}
\end{align*} \end{split}\]
therefore
\[\begin{split} \begin{align*}
Rot\left(\frac{\pi}{6}\right) \begin{pmatrix} 2 \\ 1 \end{pmatrix}
&= \begin{pmatrix}
\sqrt{3}/2 & -1/2 \\
1/2 & \sqrt{3}/2
\end{pmatrix}
\begin{pmatrix} 2 \\ 1 \end{pmatrix} \\
&= \begin{pmatrix} \sqrt{3} - 1/2 \\ 1 + \sqrt{3}/2 \end{pmatrix}
\approx \begin{pmatrix} 1.2321 \\ 1.8660 \end{pmatrix}
\end{align*} \end{split}\]
Solution to Exercise 6.6
The transformation matrix is
\[\begin{split} Re\!f \left(\pi/3\right) =
\begin{pmatrix}
\cos(2\pi/3) & \sin(2\pi/3) \\
\sin(2\pi/3) & -\cos(2\pi/3)
\end{pmatrix}
\begin{pmatrix}
-1/2 & \sqrt{3}/2 \\
\sqrt{3}/2 & 1/2
\end{pmatrix} \end{split}\]
therefore
\[\begin{split} \begin{align*}
Re\!f \left(\pi/3\right)
\begin{pmatrix} 5 \\ 3 \end{pmatrix}
&= \begin{pmatrix}
-1/2 & \sqrt{3}/2 \\
\sqrt{3}/2 & 1/2
\end{pmatrix}
\begin{pmatrix} 5 \\ 3 \end{pmatrix} \\
&=
\begin{pmatrix} 3\sqrt{3}/2 - 5/2 \\ 3/2 + 5\sqrt{3}/2 \end{pmatrix}
\approx \begin{pmatrix} 0.0981 \\ 5.8301 \end{pmatrix}.
\end{align*} \end{split}\]
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 \((-3, -2)\) so that the centre of the square is at the origin:
\[\begin{split} \begin{align*}
T \begin{pmatrix} -3 \\ -2 \end{pmatrix} =
\begin{pmatrix}
1 & 0 & -3 \\
0 & 1 & -2 \\
0 & 0 & 1
\end{pmatrix}
\end{align*} \end{split}\]
Rotate by \(\pi/3\) clockwise about the origin:
\[\begin{split} \begin{align*}
Rot\left(-\frac{\pi}{3}\right) &=
\begin{pmatrix}
\cos(\pi/3) & \sin(\pi/3) & 0 \\
-\sin(\pi/3) & \cos(\pi/3) & 0 \\
0 & 0 & 1
\end{pmatrix} \\
&= \begin{pmatrix}
1/2 & \sqrt{3}/2 & 0 \\
-\sqrt{3}/2 & 1/2 & 0 \\
0 & 0 & 1
\end{pmatrix} \\
&\approx \begin{pmatrix}
0.5 & 0.8660 & 0 \\
-0.8660 & 0.5 & 0 \\
0 & 0 & 1
\end{pmatrix}
\end{align*} \end{split}\]
Translate by \((3, 2)\) so that the centre of the square is back to \(\vec{c}\)
\[\begin{split} \begin{align*}
T \begin{pmatrix} 3 \\ 2 \end{pmatrix} &=
\begin{pmatrix}
1 & 0 & 3 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{pmatrix}
\end{align*} \end{split}\]
(c) Calculate composite alignment matrix
\[\begin{split} \begin{align*}
A &= T \begin{pmatrix} 3 \\ 2 \end{pmatrix} \cdot Rot\left(-\frac{\pi}{3}\right) \cdot T \begin{pmatrix} -3 \\ -2 \end{pmatrix} \\
&= \begin{pmatrix}
1 & 0 & 3 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1/2 & \sqrt{3}/2 & 0 \\
-\sqrt{3}/2 & 1/2 & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & -3 \\
0 & 1 & -2 \\
0 & 0 & 1
\end{pmatrix} \\
&= \begin{pmatrix}
1/2 & \sqrt{3}/2 & 3/2 - \sqrt{3} \\
-\sqrt{3}/2 & 1/2 & 1 + 3\sqrt{3}/2 \\
0 & 0 & 1
\end{pmatrix} \\
&\approx
\begin{pmatrix}
0.5 & 0.8660 & -0.2321 \\
-0.8660 & 0.5 & 3.5981 \\
0 & 0 & 1
\end{pmatrix}.
\end{align*} \end{split}\]
Apply composite transformation matrix
\[\begin{split} \begin{align*}
AP &= \begin{pmatrix}
1/2 & \sqrt{3}/2 & 3/2 - \sqrt{3} \\
-\sqrt{3}/2 & 1/2 & 1 + 3\sqrt{3}/2 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
2 & 4 & 4 & 2 \\
1 & 1 & 3 & 3 \\
1 & 1 & 1 & 1
\end{pmatrix} \\
&= \begin{pmatrix}
5/2 - \sqrt{3}/2 & 7/2 - \sqrt{3}/2 & \sqrt{3}/2 + 7/2 & \sqrt{3}/2 + 5/2 \\
\sqrt{3}/2 + 3/2 & 3/2 - \sqrt{3}/2 & 5/2 - \sqrt{3}/2 & \sqrt{3}/2 + 5/2 \\
1 & 1 & 1 & 1
\end{pmatrix} \\
&\approx \begin{pmatrix}
1.634 & 2.634 & 4.366 & 3.366 \\
2.366 & 0.634 & 1.634 & 3.366 \\
1 & 1 & 1 & 1
\end{pmatrix}
\end{align*} \end{split}\]
