Operations on linear transformations in 2D

Unit 11 Operations on linear transformations in 2D

Key unit competence: Apply linear and geometric transformations in two dimensions in solving mathematical problems.

Key ideas

• Matrices of common transformations: reflection in the x-axis $\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$; in the y-axis $\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$; in $y=x$: $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$; central symmetry (centre $O$): $\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$.
• Rotation about $O$ through $\theta$: $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. Reflection in the line through $O$ making angle $\alpha$ with the x-axis: $\left(\begin{array}{cc}\cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{array}\right)$.
• Orthogonal transformations keep lengths and angles; their matrices satisfy $A^{T}A=I$ and $\det A=\pm 1$ (rotations $+1$, reflections $-1$).
• Composite: the matrix of $g\circ f$ is $M_g{M}_f$ (apply $f$ first: its matrix is on the right).
• Kernel $\ker T=\{\vec{v}:T(\vec{v})=\vec{0}\}$; image (range) $\mathrm{Im}T=\{T(\vec{v})\}$. Rank–nullity: $\dim \ker T+\dim \mathrm{Im}T=2$.
• The area of an image figure = $|\det A|\times$ the area of the object.

Formulas to know

Common misconceptions

• "The matrix of '$f$ then $g$' is $M_f{M}_g$." No: it is $M_g{M}_f$.
• "$\ker T$ is always $\{\vec{0}\}$." Only when $\det A\ne 0$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find the matrix of the composite "rotate $90^{\circ }$ about $O$ then reflect in the y-axis" and describe it.

   Show answer

   $\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$: reflection in $y=x$.

   Common mistake: Multiplying in the wrong order.

2. Find the matrix of the reflection in the line $y=x\tan \alpha$ for $\alpha =45^{\circ }$ using the general formula, and check with the known matrix.

   Show answer

   $\left(\begin{array}{cc}\cos 90^{\circ }& \sin 90^{\circ }\\ \sin 90^{\circ }& -\cos 90^{\circ }\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, reflection in $y=x$. ✓

   Common mistake: Using $\alpha$ instead of $2\alpha$.

3. Show that $A=\left(\begin{array}{cc}\frac{3}{5}& -\frac{4}{5}\\ \frac{4}{5}& \frac{3}{5}\end{array}\right)$ is orthogonal and identify it.

   Show answer

   $A^{T}A=I$ (check the entries: $\frac{9+16}{25}=1$, off-diagonal 0). $\det =1$: a rotation with $\cos \theta =\frac{3}{5}$, $\theta \approx 53.1^{\circ }$.

   Common mistake: Only checking the determinant.

4. Verify rank–nullity for $T(x,y)=(3x+6y,x+2y)$.

   Show answer

   Kernel: $x+2y=0$, dimension 1 (spanned by $(2,-1)$). Image: multiples of $(3,1)$, dimension 1. Total 2. ✓

   Common mistake: Taking the dimension of the image as 2.

5. Find the kernel and image of $T(x,y)=(x-y,y-x)$.

   Show answer

   $\ker T=\{(t,t)\}$ (line $y=x$); $\mathrm{Im}T=\{(s,-s)\}$ (line $y=-x$); dimensions 1 and 1.

   Common mistake: Saying the kernel is $\{\vec{0}\}$ without solving.

6. Show that reflecting in the x-axis and then in the y-axis is the same as a rotation of $180^{\circ }$.

   Show answer

   $\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=R_{180^{\circ }}$.

   Common mistake: Adding the matrices.

7. The square with vertices $(0,0),(1,0),(1,1),(0,1)$ is transformed by $\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$. Find the image vertices and area.

   Show answer

   $(0,0),(2,1),(3,4),(1,3)$; area $|\det |=5$.

   Common mistake: Multiplying row vectors on the wrong side.

8. Write the matrix of rotation by $45^{\circ }$ about $O$ and find the image of $(2,0)$.

   Show answer

   $\frac{\sqrt{2}}{2}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$; image $(\sqrt{2},\sqrt{2})$.

   Common mistake: Using $\sin$ and $\cos$ in the wrong places.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. $T$ maps $(1,0)$ to $(2,1)$ and $(0,1)$ to $(1,1)$. Find $T(1,-1)$.	a. $(1,0)$
   2. Find the inverse of the rotation through $45^{\circ }$.	b. $\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$
   3. Rotate $(2,0)$ through $60^{\circ }$ about $O$.	c. $(1,\sqrt{3})$
   4. Which matrix represents the rotation of $180^{\circ }$ about $O$?	d. $\left(\begin{array}{cc}\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\end{array}\right)$

   Show answer

   1 → a ($(1,0)$) · 2 → d ($\left(\begin{array}{cc}\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\end{array}\right)$) · 3 → c ($(1,\sqrt{3})$) · 4 → b ($\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$)

   • Step 1. $T(1,-1)=T(1,0)-T(0,1)$.
     Step 2. $(2,1)-(1,1)=(1,0)$.
   • Step 1. The inverse rotates through $-45^{\circ }$.
     Step 2. $\cos (-45^{\circ })=\frac{\sqrt{2}}{2}$, $\sin (-45^{\circ })=-\frac{\sqrt{2}}{2}$.
   • Step 1. $(2\cos 60^{\circ },2\sin 60^{\circ })$.
     Step 2. $(1,\sqrt{3})$.
   • Step 1. $\cos 180^{\circ }=-1$, $\sin 180^{\circ }=0$.
     Step 2. $R_{180^{\circ }}=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\{(0,0)\}$  ·  $\{(1,2)\}$  ·  $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$  ·  $\mathrm{span}\{(1,-1)\}$
   1. $T(x,y)=(2x-y,-2x+y)$. A basis of $\ker T$ is .
   2. Two rotations about $O$ through $30^{\circ }$ and $60^{\circ }$ give the matrix .
   3. For $T(x,y)=(2x-y,-2x+y)$, $\mathrm{Im}T$ is .
   Show answer

   1: $\{(1,2)\}$ · 2: $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ · 3: $\mathrm{span}\{(1,-1)\}$

   • Step 1. $2x-y=0$ (the second equation is the same up to sign).
     Step 2. $y=2x$: $(x,y)=x(1,2)$.
   • Step 1. Angles add: $30^{\circ }+60^{\circ }=90^{\circ }$.
     Step 2. $R_{90^{\circ }}=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$.
   • Step 1. $T(x,y)=(2x-y)(1,-1)$.
     Step 2. So every image is a multiple of $(1,-1)$; check $1+1=2$ (rank–nullity).

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find a basis of the kernel of the transformation with matrix $\left(\begin{array}{cc}1& -2\\ 2& -4\end{array}\right)$.	a. a reflection (orthogonal, $\det =-1$)
   2. Write the matrix of the reflection in the line through $O$ at $30^{\circ }$ to the $x$-axis.	b. $0$
   3. An invertible transformation $T$ of ${\mathbb{R}}^2$ has $\dim \ker T=$	c. $\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right)$
   4. The matrix $\left(\begin{array}{cc}0.6& 0.8\\ 0.8& -0.6\end{array}\right)$ represents:	d. $\{(2,1)\}$

   Show answer

   1 → d ($\{(2,1)\}$) · 2 → c ($\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right)$) · 3 → b ($0$) · 4 → a (a reflection (orthogonal, $\det =-1$))

   • Step 1. $x-2y=0$ (the second equation is twice the first).
     Step 2. $(x,y)=y(2,1)$.
   • Step 1. Use $\left(\begin{array}{cc}\cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{array}\right)$ with $2\alpha =60^{\circ }$.
     Step 2. $\cos 60^{\circ }=\frac{1}{2}$, $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$.
   • Step 1. Invertible means only $\vec{0}$ is sent to $\vec{0}$: $\ker T=\{\vec{0}\}$.
     Step 2. Then $\dim \ker T=0$ (and $\dim \mathrm{Im}T=2$).
   • Step 1. Columns have length 1 and are perpendicular: $A^{T}A=I$.
     Step 2. $\det =-0.36-0.64=-1$: a reflection.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\left(\begin{array}{cc}-3& 0\\ 0& 3\end{array}\right)$  ·  the reflection in the $x$-axis  ·  $1$  ·  $\left(\begin{array}{cc}3& 0\\ 0& -3\end{array}\right)$
   1. The determinant of every rotation matrix is .
   2. A rotation of $90^{\circ }$ is followed by the reflection in $y=x$. The result is .
   3. An enlargement with centre $O$ and factor 3 is followed by the reflection in the $x$-axis. The single matrix is .
   Show answer

   1: $1$ · 2: the reflection in the $x$-axis · 3: $\left(\begin{array}{cc}3& 0\\ 0& -3\end{array}\right)$

   • Step 1. $\det \left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)={\cos }^2\theta +{\sin }^2\theta$.
     Step 2. $=1$.
   • Step 1. $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.
     Step 2. This matrix is the reflection in the $x$-axis.
   • Step 1. $M=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}3& 0\\ 0& 3\end{array}\right)$ (the first transformation on the right).
     Step 2. $=\left(\begin{array}{cc}3& 0\\ 0& -3\end{array}\right)$.

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