Answers to exercises

Exercise 1

If $X$ is on $L$ then $X' = X$, and the closest point on $L$ to $X$ is $X$ itself. So assume $L$ is on $X$. For any point $Y$ on $L$ other than $X'$, $XX'Y$ forms a triangle with a right angle at $X'$, so $XY > XX'$. Hence $X'$ is the closest point to $X$ on $L$.

Exercise 2

Take any vector ${\bf v}$ and $c=0$ in $F(c{\bf v}) = cF({\bf v})$ to obtain $F({\bf 0}) = {\bf 0}$.

Exercise 3

Rotation $\pi/3$ anticlockwise about the origin.
Dilation by factor $7$ in all directions from the origin.
Rotation by $\pi$ about the origin.
Dilation from the $x$-axis with factor $3$, and dilation from the $y$-axis with factor $2$.
Reflection in the line $y=x$.

Exercise 4

We compute $f \circ f (x,y) = f(y,-x+y) = (-x+y, -x)$ and then $f^3 (x,y) = f \circ f \circ f (x,y) = f(-x+y,-x) = (-x, -y)$. Now $f^6$ is obtained by composing $f^3$ with itself twice, so $f^6 (x,y) = f^3 (f^3 (x,y)) = f^3 (-x,-y) = (x,y)$. Thus $f^6$ is the identity.

Exercise 5

\[ MM^{-1} = \frac{1}{ad-bc} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{ad-bc} \begin{bmatrix} ad-bc & 0 \\ 0 & ad-bc \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \text{Id} \]

The computation for $M^{-1} M$ is similar.

Exercise 6

Note that $x' = x+2y$, $y'=2x+4y$ implies $y'=2x'$, so the image of $F$ lies on the line $y'=2x'$. So $F$ is not surjective: for instance, there is no $(x,y)$ such that $F(x,y) = (0,1)$; if there were, then $x+2y = 0$ and $2x+4y = 1$, but then $2x+4y = 2(x+2y) = 0$, a contradiction. Alternatively, $F$ is not injective as, for instance, $F(0,0) = F(-2,1) = (0,0)$.

Exercise 7

Let $M$ be the desired matrix, so $M \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & -3 \end{bmatrix}$. Thus $M = \begin{bmatrix} 1 & 2 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix}^{-1} = \frac{1}{7} \begin{bmatrix} 1 & 2 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 0 & 7 \\ 7 & -7 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}$. Hence $T_M(x,y) = (y,x-y)$.

Exercise 8

The $x$ and $y$-axes are sent to the $y$-axis and negative $x$-axis respectively, so a point $(x,y)$ is mapped to $(-y,x)$.

Exercise 9

The point $(0,1)$ is on the unit circle at an angle of $\pi/2$ from the positive $x$-axis. Rotating by $\phi$ gives the point $(\cos (\phi + \frac{\pi}{2}), \; \sin (\phi + \frac{\pi}{2} ) ) = (-\sin \phi, \; \cos \phi)$.

Exercise 10

We verify

\[ \begin{bmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{bmatrix} = \begin{bmatrix} \cos \theta \cos \phi - \sin \theta \sin \phi & - \cos \theta \sin \phi - \sin \theta \cos \phi \\ \sin \theta \cos \phi + \cos \theta \sin \phi & - \sin \theta \sin \phi + \cos \theta \cos \phi \end{bmatrix} \]

which by sine and cosine addition formulas is the rotation matrix for angle $\theta + \phi$. Hence the composition of rotations by $\theta$ and $\phi$ is rotation by $\theta + \phi$.

Exercise 11

If $\phi$ is not a multiple of $2\pi$, then a rotation of $\phi$ about $P$ sends no point to itself except $P$. So if $P \neq {\bf 0}$ then $\text{Rot}_{P, \; \phi} ({\bf 0}) \neq {\bf 0}$. By exercise \ref{ex:origin} a linear transformation must send the origin to the origin, so $\text{Rot}_{P, \; \phi}$ cannot be linear.

Exercise 12

Translations by ${\bf v}$ and $-{\bf v}$ undo each other; they are inverse transformations. Thus $\text{Trans}_{{\bf v}}$ is bijective, with inverse $\text{Trans}_{-{\bf v}}$.

Exercise 13

Let ${\bf x}$ be a point in the plane. Translation by $-{\bf v}$ takes $P$ to the origin and carries ${\bf x}$ along to ${\bf x} - {\bf v}$. Then rotating by $\phi$ about ${\bf 0}$ preserves ${\bf 0}$ and rotates ${\bf x} - {\bf v}$ around it. Then translating by ${\bf v}$ moves the origin back to $P$ and takes our point to the point rotated by $\phi$ about $P$. Thus

\begin{align*} \text{Rot}_{P, \; \phi} \begin{bmatrix} x \\ y \end{bmatrix} &= \text{Trans}_{{\bf v}} \circ \text{Rot}_{{\bf 0}, \; \phi} \circ \text{Trans}_{-{\bf v}} \begin{bmatrix} x \\ y \end{bmatrix} = \text{Trans}_{{\bf v}} \begin{bmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{bmatrix} \begin{bmatrix} x - a \\ y - b \end{bmatrix} \\ &= \text{Trans}_{{\bf v}} \begin{bmatrix} (x-a) \cos \phi - (y-b) \sin \phi \\ (x-a) \sin \phi + (y-b) \cos \phi \end{bmatrix} = \begin{bmatrix} (x-a) \cos \phi - (y-b) \sin \phi + a \\ (x-a) \sin \phi + (y-b) \cos \phi + b \end{bmatrix}. \end{align*}

Exercise 14

Let $Y$ be a point not on $L$. Then there is no $X$ such that $\text{Proj}_L (X) = Y$, so $\text{Proj}_L$ is not surjective.
Let $L'$ be a line perpendicular to $L$, intersecting $L$ at a point $Y$. Then for any point $X$ on $L'$, $\text{Proj}_L (X) = Y$. Hence $\text{Proj}_L$ is not injective.
If $P$ is on $L$, then the perpendicular to $L$ through $P$ intersects $L$ at $P$, so $\text{Proj}_L (P) = P$.
For any point $X$, $\text{Proj}_L (X)$ lies on $L$, so using the previous part with $P = \text{Proj}_L (X)$ we have $\text{Proj}_L \left( \text{Proj}_L (X) \right) = \text{Proj}_L (X)$.

Exercise 15

The line through $(x_0, y_0)$ with gradient $-1$ has equation $y-y_0 = -(x-x_0)$, or equivalently $y = -x + x_0 + y_0$. At the intersection with the line $y=x$ we have $y = -x + x_0 + y_0 = x$ so $x = \frac{1}{2}(x_0 + y_0)$ and then $y = x = \frac{1}{2}(x_0 + y_0)$ as desired.

Exercise 16

The line $L$ given by $y=2x$ makes an angle $\phi$ with the positive $x$-axis such that $\tan \phi = 2$, hence $\sin \phi = \frac{2}{\sqrt{5}}$ and $\cos \phi = \frac{1}{\sqrt{5}}$. Thus

\[ \text{Proj}_L \begin{bmatrix} 8 \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{4}{5} \end{bmatrix} \begin{bmatrix} 8 \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{14}{5} \\ \frac{28}{5} \end{bmatrix}. \]

Exercise 17

A proof may be given by explaining the following pictures.
Dropping a perpendicular from $(1,0)$ to $L$ as shown we obtain a right-angled triangle with hypotenuse $1$ and angle $\phi$; so the projection of $(1,0)$ to $L$ is a distance of $\cos \phi$ along $L$ from the origin. Lying on a ray at angle $\phi$ at a distance $\cos \phi$ from the origin, then, $\text{Proj}_L (1,0) = \cos \phi (\cos \phi, \sin \phi) = (\cos^2 \phi, \sin \phi \cos \phi)$. Similarly $\text{Proj}_L (0,1)$ lies at distance $\sin \phi$ from the origin on $L$, so equals $(\sin \phi \cos \phi, \sin^2 \phi)$.
The first part shows $\text{Proj}_L$ is linear, hence given by a matrix. The second part finds the columns of this matrix. So $\text{Proj}_l$ is as claimed.

Exercise 18

If $P = (x,y)$ is on $L$, then $(x,y) = (s \cos \phi, s \sin \phi)$ for some $s$. Hence
We compute

As this is the matrix for $\text{Proj}_L$, we conclude $\text{Proj}_L \circ \text{Proj}_L = \text{Proj}_L$ as desired.

Exercise 19

Under $\text{Proj}_L$, points on $L$ are sent to themselves, but any point not on $L$ is sent to a different point. So if $L$ does not pass through the origin ${\bf 0}$, then $\text{Proj}_L ({\bf 0}) \neq {\bf 0}$. By exercise \ref{ex:origin}, a linear transformation must send ${\bf 0}$ to ${\bf 0}$. Hence $\text{Proj}_L$ is not linear.

Exercise 20

Rotating by $-\phi$ about the origin brings $L$ to the $x$-axis, carrying along a point $P$. Then reflecting in the $x$-axis flips the point over the $x$-axis; rotating back by $\phi$ sends the $x$-axis back to $L$, and carries our point to the reflection of $P$ in $L$. Hence $\text{Ref}_L$ is given by the desired composition of transformations.
As $\text{Rot}_{{\bf 0}, \phi}$, $\text{Ref}_{\text{$x$-axis}}$ and $\text{Rot}_{{\bf 0}, -\phi}$ are all linear, and we know their matrices, we can multiply them together to find a matrix for $\text{Ref}_L$, which must also be linear. The matrix is \begin{align*} \begin{bmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{bmatrix} &\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \cos \phi & \sin \phi \\ -\sin \phi & \cos \phi \end{bmatrix} = \begin{bmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{bmatrix} \begin{bmatrix} \cos \phi & \sin \phi \\ \sin \phi & - \cos \phi \end{bmatrix} \\ &= \begin{bmatrix} \cos^2 \phi - \sin^2 \phi & 2 \sin \phi \cos \phi \\ 2 \sin \phi \cos \phi & \sin^2 \phi - \cos^2 \phi \end{bmatrix} = \begin{bmatrix} \cos (2\phi) & \sin (2\phi) \\ \sin (2\phi) & - \cos (2\phi) \end{bmatrix}. \end{align*}

Exercise 21

A proof may be given by explaining the following diagrams.
Since $(1,0)$ is distance $1$ from the origin on an angle of $0$, reflection in a line at angle $\phi$ sends $(1,0)$ to a point distance $1$ from the origin on an angle of $2\phi$. Thus $\text{Ref}_L (1,0) = (\cos(2\phi), \sin(2\phi))$. Similarly one can show $\text{Ref}_L (0,1) = (\sin(2\phi), -\cos(2\phi))$.
The first part shows $\text{Ref}_L$ is linear; the second part calculates the columns of the corresponding matrix, which is as claimed.

Exercise 22

The line $L$ given by $y=2x$ passes through the origin on an angle $\phi$ such that $\tan \phi = 2$, hence $\sin \phi = \frac{2}{\sqrt{5}}$ and $\cos \phi = \frac{1}{\sqrt{5}}$. Hence $\cos (2\phi) = \cos^2 \phi - \sin^2 \phi = \frac{-3}{5}$ and $\sin(2\phi) = 2 \sin \phi \cos \phi = \frac{4}{5}$. So reflection in $L$ is given by

\[ \text{Ref}_L \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{-3}{5} & \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \]

Exercise 23

The vector from $\text{Proj}_L ({\bf x})$ to ${\bf x}$, and the vector from $\text{Proj}_L ({\bf x})$ to $\text{Ref}_L ({\bf x})$, have the same length and point in opposite directions. Hence

\[ {\bf x} - \text{Proj}_L ({\bf x}) = - \left( \text{Ref}_L ({\bf x}) - \text{Proj}_L ({\bf x}) \right), \]

which rearranges to $\text{Ref}_L ({\bf x}) = 2 \text{Proj}_L ({\bf x}) - {\bf x}$, as desired.

Exercise 24

From the previous exercise, the matrix for $\text{Ref}_L$ is given by twice the matrix for $\text{Proj}_L$, minus the identity matrix, which is

\[ 2 \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \sin \phi \cos \phi & \sin^2 \phi \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 \cos^2 \phi - 1 & 2 \sin \phi \cos \phi \\ 2 \sin \phi \cos \phi & 2 \sin^2 \phi - 1 \end{bmatrix} = \begin{bmatrix} \cos(2\phi) & \sin(2\phi) \\ \sin(2\phi) & -\cos(2\phi) \end{bmatrix}. \]

Exercise 25

If $P = (x,y)$ lies on $L$ then $(x,y) = (s \cos \phi, s \sin \phi)$ for some $s$, so \begin{align*} \begin{bmatrix} \cos(2\phi) & \sin(2\phi) \\ \sin(2\phi) & -\cos(2\phi) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix} \cos(2\phi) & \sin(2\phi) \\ \sin(2\phi) & -\cos(2\phi) \end{bmatrix} \begin{bmatrix} s \cos \phi \\ s \sin \phi \end{bmatrix} \\ &= \begin{bmatrix} s \cos (2\phi - \phi) \\ s \sin (2\phi - \phi) \end{bmatrix} = \begin{bmatrix} s \cos \phi \\ s \sin \phi \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \end{align*}
We verify

is diagonal, with both diagonal entries given by $\cos^2 (2\phi) + \sin^2 (2\phi) = 1$, hence is the identity matrix.

Exercise 26

We note that projection onto $L$ is given by $\text{Proj}_L (x,y) = (d,y)$. Using exercise \ref{ex:ref_proj} then $\text{Ref}_L (x,y) = 2 \text{Proj}_L (x,y) - (x,y) = 2(d,y) - (x,y) = (2d-x,y)$.

Exercise 27

Rotating by $-\phi$ about the origin brings $L$ to the $x$-axis, and brings a point $P$ along with it; then dilating from the $x$-axis and rotating back by $\phi$ brings our line back to $L$ and sends our point to its dilation from $L$.
The three transformations in the previous part are all linear; multiplying them will give a matrix for $\text{Dil}_{L,k}$, which must be linear.

Exercise 28

The line $L$ given by $y=2x$ makes angle $\phi$ with the $x$-axis, where $\tan \phi = 2$, so $\sin \phi = \frac{2}{\sqrt{5}}$ and $\cos \phi = \frac{1}{\sqrt{5}}$. Hence

\[ \text{Dil}_{L,5} \begin{bmatrix} 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 + 4 \cdot \frac{4}{5} & -4 \cdot \frac{2}{5} \\ -4 \cdot \frac{2}{5} & 1 + 4 \cdot \frac{1}{5} \end{bmatrix} \begin{bmatrix} 3 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{21}{5} & \frac{-8}{5} \\ \frac{-8}{5} & \frac{9}{5} \end{bmatrix} \begin{bmatrix} 3 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{63}{5} \\ \frac{-24}{5} \end{bmatrix}. \]

Exercise 29

We first note that $\text{Dil}_{L,k}$ has determinant

\[ \left( 1 + (k-1) \sin^2 \phi \right) \left( 1 + (k-1) \cos^2 \phi \right) - (1-k)^2 \sin^2 \phi \cos^ \phi = 1 + (k-1) \sin^2 \phi + (k-1) \cos^2 \phi = k. \]

Hence the inverse of the matrix for $\text{Dil}_{L,k}$ is

\[ \frac{1}{k} \begin{bmatrix} 1 + (k-1) \cos^2 \phi & (k-1) \sin \phi \cos \phi \\ (k-1) \sin \phi \cos \phi & 1 + (k-1) \sin^2 \phi \end{bmatrix} = \begin{bmatrix} \frac{1}{k} + \left( 1 - \frac{1}{k} \right) \cos^2 \phi & \left( 1 - \frac{1}{k} \right) \sin \phi \cos \phi \\ \left( 1 - \frac{1}{k} \right) \sin \phi \cos \phi & \frac{1}{k} + \left( 1 - \frac{1}{k} \right) \sin^2 \phi \end{bmatrix}. \]

Now $\frac{1}{k} + (1- \frac{1}{k}) \cos^2 \phi = \frac{1}{k} (1 - \cos^2 \phi) + \cos^2 \phi = \frac{1}{k} \sin^2 \phi + 1 - \sin^2 \phi = 1 + \left( \frac{1}{k} - 1 \right) \sin^2 \phi$ and, similarly, $\frac{1}{k} + \left( 1 - \frac{1}{k} \right) \sin^2 \phi = 1 + \left( \frac{1}{k} - 1 \right) \cos^2 \phi$. Hence this matrix is the matrix for dilation by factor $1/k$ from $L$.

Exercise 30

When $k=0$, the matrix for $\text{Dil}_{L,k}$ becomes \[ \begin{bmatrix} 1 - \sin^2 \phi & \sin \phi \cos \phi \\ \sin \phi \cos \phi & 1 - \sin^2 \phi \end{bmatrix} = \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \sin \phi \cos \phi & \cos^2 \phi \end{bmatrix}, \]
which is the matrix for projection onto $L$.
When $k=-1$ we have \[ \begin{bmatrix} 1 - 2 \sin^2 \phi & 2 \sin \phi \cos \phi \\ 2 \sin \phi \cos \phi & 1 - 2 \cos^2 \phi \end{bmatrix} = \begin{bmatrix} \cos(2\phi) & \sin (2\phi) \\ \sin (2\phi) & - \cos (2\phi) \end{bmatrix} \] which is the matrix for reflection in $L$.

Exercise 31

A reflection is a dilation by factor $-1$, so applying it twice gives a dilation with factor $(-1)^2 = 1$, i.e. the identity.
A projection is a dilation by factor $0$, so applying it twice gives a dilation with factor $0^2 = 0$, i.e. the same projection.

Exercise 32

A projection matrix clearly has the desired properties. For the converse, suppose $M$ is symmetric ($b=c$), $ad = b^2$ and $a+d=1$. Since $a$ and $d$ have sum and product $\geq 0$, both are $\geq 0$; since they sum to $1$, both lie in $[0,1]$. Hence we have $a = \cos^2 \phi$ and $b = \sin^2 \phi$ for some $\phi$. Then $b^2 = \cos^2 \phi \sin^2 \phi$ so $b = \sin \phi \cos \phi$ or $- \sin \phi \cos \phi$. In the first case we have the matrix for projection onto a line at angle $\phi$; in the second case for projection onto a line at angle $-\phi$.
A reflection matrix clearly has the desired properties. For the converse, suppose $M$ is symmetric ($b=c$), $a=-d$ and $ad - b^2 =-1$. Substituting $a=-d$ then gives $-a^2 - b^2 = -1$, so $(a,b)$ lies on he unit circle and hence $a = \cos(2\phi)$, $b = \sin(2\phi)$ for some $\phi$. We then have a reflection matrix.
A dilation matrix can be verified to have the desired properties. For the converse, suppose $b=c$ and $a+d = 1 + ad - b^2$. Let $ad-b^2 = k$, so $a+d=k+1$. Since $(a-1)+(d-1) = k-1$ and $(a-1)(d-1) = ad-a-d+1 = b^2$, both $a-1$ and $d-1$ have the same sign as $k-1$. Since they sum to $k-1$, they lie between $0$ and $k-1$. Thus we may write $a-1 = (k-1) \sin^2 \phi$ and $d-1 = (k-1) \cos^2 \phi$ for some $\phi$. We then have $b^2 = ad-k = [1+(k-1)\sin^2 \phi][1+(k-1) \cos^2 \phi]-k = (k-1)^2 \sin^2 \phi \cos^2 \phi$. Thus $b = \pm (1-k) \sin \phi \cos \phi$. Replacing $\phi$ with $-\phi$ if necessary, we have a dilation matrix.

Exercise 33

Applying the translation we have $(x,y) \mapsto (x+3,y)$, then applying the dilation we have $(x+3,y) \mapsto (x+3,2y) = (x',y')$. So the transformation is the same, and the answer is the same.

Exercise 34

Applying the translation we have $(x,y) \mapsto (x+1,y+3)$; then applying the reflection we have $(x+1,y+3) \mapsto (-x-1,y+3)$; then applying the dilation we have $(-x-1,y+2) \mapsto (-3x-3,y+2)$. The result is a different transformation, giving a different answer.

Exercise 35

Translating $k$ units to the left gives $y=e^{x+k}$. Dilating by $e^k$ form the $x$-axis gives $y = e^k e^x$. These are the same graph since $e^{x+k} = e^k e^x$.

Exercise 36

From $(x,y)$, dilating and then rotating gives an ellipse; the transformation is

\[ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\sqrt{2} \\ \frac{1}{\sqrt{2}} & \sqrt{2} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \]

and hence, inverting,

\[ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & - \sqrt{2} \\ \frac{1}{\sqrt{2}} & \sqrt{2} \end{bmatrix}^{-1} \begin{bmatrix} x' \\ y' \end{bmatrix} = \frac{1}{2} \begin{bmatrix} \sqrt{2} & \sqrt{2} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} x' + y' \\ \frac{-1}{2} x' + \frac{1}{2} y' \end{bmatrix}. \]

Thus the expression $x^2 + y^2$ becomes

\[ \frac{1}{2} (x' + y')^2 + \frac{1}{2} \left( -\frac{1}{2} x' + \frac{1}{2} y' \right)^2 = \frac{5}{8} x'^2 + \frac{5}{8} y'^2 + \frac{3}{4} x'y'. \]

The equation of the transformed circle is thus $\frac{5}{8} x^2 + \frac{5}{8} y^2 + \frac{3}{4} xy = 1$, which is an ellipse.

Exercise 37

Reflection in the line $y=x$ takes $(x,y)$ to $(x',y') = (y,x)$. The equation $y=f(x)$ becomes $x' = f(y')$, and hence the transformed graph is given by $x = f(y)$, or equivalently, $y = f^{-1}(x)$.

Exercise 38

The parallelograms spanned by ${\bf v}, {\bf w}$ and ${\bf v}, {\bf w} + k {\bf v}$ both have base ${\bf v}$ the same height above this base, and the same orientation, so have the same signed area. The same applies to ${\bf v}, {\bf w}$ and ${\bf v} + k{\bf w}, {\bf w}$ above a base of ${\bf w}$.
If we reverse the order of ${\bf v}$ and ${\bf w}$ the parallelogram is unchanged but has its orientation reversed, hence its signed area changes sign.
The vectors $(a,0)$ and $(b,d)$ span a parallelogram with base $a$ and perpendicular height $d$, hence area $|ad|$. The parallelogram is positively oriented if $a$ and $d$ have the same sign, and negatively oriented otherwise, so the signed area is $ad$. A similar argument applies to $(a,c)$ and $(b,0)$.
Applying the first part to $(a,c)$ and $(b,d)$, the signed area they span is equal to the signed area spanned by $(a,c) - \frac{c}{d} (b,d)$ and $(b,d)$.
For any vectors ${\bf v} = (a,c)$ and ${\bf w} = (b,d)$, if $d \neq 0$ then the previous part applies and gives the signed area as $\det M$. If $d = 0$ then part c applies and gives the signed area as $-bc = ad-bc = \det M$.

Exercise 39

$\text{We compute}\det \begin{bmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{bmatrix} = \cos^2 \phi + \sin^2 \phi = 1$, $\det \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \sin \phi \cos \phi & \sin^2 \phi \end{bmatrix} = \cos^2 \phi \sin^2 \phi - \cos^2 \phi \sin^2 \phi = 0$ and $\det \begin{bmatrix} \cos(2\phi) & \sin(2\phi) \\ \sin(2\phi) & -\cos(2\phi) \end{bmatrix} = -\cos^2 (2\phi) - \sin^2 (2\phi) = -1$.

Exercise 40

$\text{Compute} \det \begin{bmatrix} 1 + (k-1) \sin^2 \phi & (1-k) \sin \phi \cos \phi \\ (1-k) \sin \phi \cos \phi & 1 + (k-1) \cos^2 \phi \end{bmatrix} = \left[ 1 + (k-1) \sin^2 \phi \right] \left[ 1 + (k-1) \cos^2 \phi \right] - (1-k)^2 \sin^2 \phi \cos^2 \phi$, which simplifies to $1 + (k-1) (\sin^2 \phi + \cos^2 \phi) + (k-1)^2 \sin^2 \phi \cos^2 \phi - (1-k)^2 \sin^2 \phi \cos^2 \phi = 1 + (k-1) = k$.

Exercise 41

Let $A,B$ be points in the plane and let $A'B'$ be their images under transformation; to demonstrate an isometry we show $AB=A'B'$.

Under a translation, $ABB'A'$ forms a parallelogram so $AB = A'B'$.

For a reflection, let $X$ be a point on the line of reflection. Then triangles $XAB$ and $XA'B'$ are congruent, so $AB = A'B'$.

For a projection, take $A,B$ to be distinct points projecting onto the same point. Then $A'=B'$ so $A'B' = 0 < AB$. As $AB \neq A'B'$, a projection is not an isometry.

For a dilation, take $A$ to be a point on the line of dilation and $B$ a point which projects onto $A$. Then $A'B' = |k| AB \neq AB$. Hence this transformation is not an isometry.

Exercise 42

Since $\det M = 1$, $T_M$ preserves areas. But, for instance, let $A = (0,0)$ and $B = (1,0)$, so $A' = (0,0)$ and $B' = (2,0)$. Then $AB = 1$ but $A'B' = 2$ so $T_M$ is not an isometry.

Exercise 43

We compute the matrix for the composition $\text{Ref}_{L_2} \circ \text{Ref}_{L_1}$ as

\[ \begin{bmatrix} \cos(2\phi) & \sin(2\phi) \\ \sin(2\phi) & - \cos(2\phi)\end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} \cos(2\phi) & - \sin(2\phi) \\ \sin(2\phi) & \cos(2\phi) \end{bmatrix}, \]

which is the matrix for rotation by $2\phi$ about the origin.