Composite transformations, area and invariant lines Problem set A One question at a time

Find the single matrix for each pair of transformations.

1. (a) \(\begin{pmatrix}3 & 0 \\ 1 & 2\end{pmatrix}\) followed by \(\begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}\)

   • \(\text{The first transformation is on the right: } \text{single matrix} = \begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}\begin{pmatrix}3 & 0 \\ 1 & 2\end{pmatrix}.\)
   • \(\begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}\begin{pmatrix}3 & 0 \\ 1 & 2\end{pmatrix} = \begin{pmatrix}5 & 4 \\ 1 & 2\end{pmatrix}\)

2. (b) Rotation 90° anticlockwise about the origin, followed by reflection in the \(x\)-axis

   • \(\text{Rotation: } \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix};\quad \text{reflection in the } x\text{-axis: } \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}.\)
   • \(\text{Reflection} \times \text{rotation} = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} = \begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\)

3. (c) Enlargement with scale factor 2, followed by rotation 90° anticlockwise

   • \(\text{Enlargement: } \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix};\quad \text{rotation: } \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}.\)
   • \(\text{Rotation} \times \text{enlargement} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix} = \begin{pmatrix}0 & -2 \\ 2 & 0\end{pmatrix}\)

Find the area of the image.

1. (a) The unit square under \(\begin{pmatrix}4 & -1 \\ 3 & 8\end{pmatrix}\)

   • \(\text{Area scale factor} = |\det| = |4(8) - (-1)(3)| = 35\)
   • \(\text{Area} = 1 \times 35 = 35\)

2. (b) The triangle with vertices \((0,0),\ (3,0),\ (0,2)\) under \(\begin{pmatrix}2 & 4 \\ 3 & 0\end{pmatrix}\)

   • \(\text{Original area} = \tfrac{1}{2}(3)(2) = 3\)
   • \(\text{Area scale factor} = |2(0) - 4(3)| = 12\)
   • \(\text{Area} = 3 \times 12 = 36\)

3. (c) A rectangle of area 5 under \(\begin{pmatrix}1.6 & 0 \\ 0 & 1\end{pmatrix}\)

   • \(\text{Area scale factor} = |1.6(1) - 0| = 1.6\)
   • \(\text{Area} = 5 \times 1.6 = 8\)

Find the image of each point.

1. (a) \((2, 3)\): enlargement, scale factor 3, then translation by \(\begin{pmatrix}-5 \\ 0\end{pmatrix}\)

   • \(\text{Enlargement: } 3(2,3) = (6,9).\)
   • \(\text{Translation: } (6 - 5,\; 9 + 0) = (1, 9)\)

2. (b) \((2, 3)\): translation by \(\begin{pmatrix}-5 \\ 0\end{pmatrix}\), then rotation 90° anticlockwise about the origin

   • \(\text{Translation: } (2 - 5,\; 3 + 0) = (-3, 3).\)
   • \(\text{Rotation 90° anticlockwise: } (x,y) \to (-y,x). \text{ Matrix: } \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}-3 \\ 3\end{pmatrix} = \begin{pmatrix}-3 \\ -3\end{pmatrix}\)
   • \(\text{Image} = (-3, -3)\)

Find the equations of the invariant lines through the origin for each matrix.

1. (a) \(\begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix}\)

   • \(\text{Invariant lines through O are the eigenvector directions.}\)
   • \(\lambda = 5,\; \mathbf{v} = \begin{pmatrix}1 \\ 1\end{pmatrix} \text{ gives the line } y = x\)
   • \(\lambda = 2,\; \mathbf{v} = \begin{pmatrix}2 \\ -1\end{pmatrix} \text{ gives the line } y = -\tfrac{1}{2}x\)

2. (b) \(\begin{pmatrix}1 & 2 \\ 4 & 3\end{pmatrix}\)

   • \(\text{Invariant lines through O are the eigenvector directions.}\)
   • \(\lambda = 5,\; \mathbf{v} = \begin{pmatrix}1 \\ 2\end{pmatrix} \text{ gives the line } y = 2x\)
   • \(\lambda = -1,\; \mathbf{v} = \begin{pmatrix}1 \\ -1\end{pmatrix} \text{ gives the line } y = -x\)

The transformation \(T\) has matrix \(\begin{pmatrix}1 & 2 \\ 0 & 3\end{pmatrix}\).

1. (a) Find the eigenvalues of the matrix.

   • \(\lambda^2 - 4\lambda + 3 = 0\)
   • \(\lambda = 1 \text{ or } 3\)

2. (b) Find the equation of the line of invariant points.

   • Invariant points satisfy \(M\mathbf{v} = \mathbf{v}\), the eigenvalue 1.
   • Eigenvector: \(\begin{pmatrix}1 \\ 0\end{pmatrix}\), so the line is \(y = 0\).

3. (c) Find the equation of the other invariant line through the origin, and say how far points on it move from the origin.

   • Eigenvector for \(3\): \(\begin{pmatrix}1 \\ 1\end{pmatrix}\), so the line is \(y = x\).
   • The eigenvalue is 3, so each point on it moves 3 times as far from the origin, along the line.

Describe the single transformation equal to a rotation of 90° anticlockwise about the origin followed by a reflection in the line \(y = x\).

• \(\text{Rotation: } \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix};\quad \text{reflection in } y = x: \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}.\)
• \(\text{Single matrix} = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\)
• This sends (x, y) to (x, −y): a reflection in the x-axis.

\(M = \begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}\). Find the area of the image of the unit square under \(M^4\).

• \(\det M = 2(2) - 1(1) = 3\)
• \(\det(M^4) = (\det M)^4 = 3^4 = 81\)
• \(\text{The unit square has area 1, so the image has area } 1 \times 81 = 81.\)

Explain why a rotation about the origin through 90° has no invariant lines through the origin, and a rotation through 180° has infinitely many.

• \(\text{Rotation 90°: } \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix},\; \lambda^2 + 1 = 0 \text{ has no real solutions, so there are no real eigenvectors.}\)
• Every vector is turned through a right angle, so it lands on a different line. No line through the origin maps to itself.
• \(\text{Rotation 180°: } \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix} = -I \text{, so every vector } \mathbf{v} \text{ maps to } -\mathbf{v}.\)
• Each image is on the same line through the origin as the original, so every line through the origin is invariant (eigenvalue −1).

Transformation \(T\) is a reflection in the \(x\)-axis, then an enlargement with scale factor 3, then a rotation of 90° clockwise about the origin.

1. (a) Find the single matrix for \(T\).

   • Matrices in the order done: \(\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\), \(\begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}\), \(\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\).
   • The first is on the right: \(M = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} \begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix} \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} = \begin{pmatrix}0 & -3 \\ -3 & 0\end{pmatrix}\)

2. (b) A shape has area 4. Find the area of its image under \(T\).

   • \(\det M = -9\)
   • Area scale factor \(|\det M| = 9\), so the area is \(9 \times 4 = 36\)

A logo is made from a triangle with vertices \((0,0),\ (4,0)\) and \((0,4)\), and its images under \(T,\ T^2,\ T^3,\ \ldots\), where \(T\) is an enlargement with scale factor \(\frac{1}{2}\) followed by a rotation of 90° anticlockwise about the origin. (a) Find the matrix of \(T\). (b) Find the area of the first triangle and of its image under \(T\). (c) Find the total area of all the triangles.

• \(\text{(a) Enlargement: } \begin{pmatrix}0.5 & 0 \\ 0 & 0.5\end{pmatrix};\quad \text{rotation: } \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}.\)
• \(T = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0.5 & 0 \\ 0 & 0.5\end{pmatrix} = \begin{pmatrix}0 & -0.5 \\ 0.5 & 0\end{pmatrix}\)
• \(\text{(b) First triangle: } \tfrac{1}{2}(4)(4) = 8.\)
• \(\det T = 0(0) - (-0.5)(0.5) = 0.25 \text{, so the image has area } 8 \times 0.25 = 2.\)
• \(\text{(c) The areas } 8,\; 2,\; 0.5,\; \ldots \text{ form a geometric series with } a = 8,\; r = 0.25.\)
• \(S_\infty = \frac{8}{1 - 0.25} = \frac{32}{3}\)