Exercise 0.1: Computing the image of a point under a transformation: ten direct calculations

For each line, write the matrix of the transformation, always centred at the origin, then compute what is asked: the two coordinates of the image (to two decimal places when they are irrational), the area scale factor, or the preimage. Rotations are counterclockwise unless stated otherwise.

One method for all ten lines: the columns of the matrix are the images of $(1, 0)$ and $(0, 1)$ , which you write down first. Then multiply the matrix by the point written as a column, row by column. For a composition, the matrix of the transformation applied FIRST goes on the RIGHT of the product; for a preimage, apply the inverse matrix; for an area factor, take the absolute value of the determinant.

a) The image of the point $(-4, 6)$ under the dilation with factor $\frac{5}{2}$ .
b) The image of the point $(-7, 4)$ under the reflection about the $y$ -axis.
c) The image of the point $(6, -1)$ under the rotation through $180^{\circ}$ .
d) The image of the point $(5, 2)$ under the reflection about the line $y=-x$ .
e) The image of the point $(-3, 8)$ under the orthogonal projection onto the $y$ -axis.
f) The image of the point $(2, -5)$ under the vertical shear with factor $3$ , which sends $(1, 0)$ to $(1, 3)$ and leaves $(0, 1)$ fixed.
g) The image of the point $(4, 2)$ under the rotation through $30^{\circ}$ , exactly and then to two decimal places.
h) The image of the point $(2, 1)$ under the transformation that first applies the rotation through $90^{\circ}$ CLOCKWISE, then the shear with matrix $\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ .
i) The factor by which the transformation with matrix $\begin{pmatrix} 3 & -2 \\ 4 & -5 \end{pmatrix}$ multiplies areas.
j) The preimage of the point $(4, 7)$ under the transformation with matrix $\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$ .

Type your answers, the page tells you right or wrong 0/19

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Answers

a) $\begin{pmatrix} \frac{5}{2} & 0 \\ 0 & \frac{5}{2} \end{pmatrix}$ ; image $(-10, 15)$
b) $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ ; image $(7, 4)$
c) $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ ; image $(-6, 1)$
d) $\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$ ; image $(-2, -5)$
e) $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ ; image $(0, 8)$
f) $\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$ ; image $(2, 1)$
g) image $(2\sqrt{3}-1, 2+\sqrt{3})\approx(2.46, 3.73)$
h) $T=\begin{pmatrix} 1 & 1 \\ -1 & 0 \end{pmatrix}$ ; image $(3, -2)$
i) $\det=-7$ ; area factor $7$
j) $A^{-1}=\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$ ; preimage $(5, -6)$

The move that decides everything: write the matrix by its COLUMNS, that is, find where $(1, 0)$ and $(0, 1)$ go, instead of reciting it from memory. Every sign error in this list comes from a recited matrix; none survives a sketch of the two basis vectors.

a)

$H=\begin{pmatrix} \frac{5}{2} & 0 \\ 0 & \frac{5}{2} \end{pmatrix}$ , so the image is $\frac{5}{2}(-4, 6)=(-10, 15)$ . Each coordinate is multiplied by the factor.

b)

$(1, 0)$ goes to $(-1, 0)$ and $(0, 1)$ stays fixed: $S_{y}=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ , and the image is $(7, 4)$ . Pitfall: changing the sign of $y$ , which is the reflection about the $x$ -axis and gives $(-7, -4)$ .

c)

$R_{180}=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ , since $\cos 180^{\circ}=-1$ and $\sin 180^{\circ}=0$ : the image is $(-6, 1)$ , the point symmetric to $(6, -1)$ through the origin.

d)

$(1, 0)$ goes to $(0, -1)$ and $(0, 1)$ to $(-1, 0)$ : $\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$ . The image of $(5, 2)$ is $(-2, -5)$ : swap the coordinates AND change both signs. Pitfall: swapping without changing the signs, which gives $(2, 5)$ , the image under the reflection about $y=x$ .

e)

$(1, 0)$ goes to $(0, 0)$ and $(0, 1)$ stays fixed: $Q_{y}=\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ , so the image is $(0, 8)$ . Pitfall: keeping the $x$ -coordinate, $(-3, 0)$ , which is the projection onto the $x$ -axis.

f)

The columns are $(1, 3)$ and $(0, 1)$ : $\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$ . The image is $(2, 3\times 2+(-5))=(2, 1)$ : the $x$ -coordinate does not move, the $y$ -coordinate slides by $3x$ . Pitfall: writing the horizontal shear $\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$ , which gives $(-13, -5)$ .

g)

$R_{30}=\begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}$ , so $x'=4\cdot\frac{\sqrt{3}}{2}-2\cdot\frac{1}{2}=2\sqrt{3}-1\approx 2.46$ and $y'=4\cdot\frac{1}{2}+2\cdot\frac{\sqrt{3}}{2}=2+\sqrt{3}\approx 3.73$ . Pitfall: putting the $-\sin$ at the bottom left, which rotates through $30^{\circ}$ clockwise.

h)

The clockwise rotation through $90^{\circ}$ sends $(1, 0)$ to $(0, -1)$ and $(0, 1)$ to $(1, 0)$ : $R=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ . Applied first, it goes on the right: $T=CR=\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 1 \\ -1 & 0 \end{pmatrix}$ . The image of $(2, 1)$ is $(3, -2)$ ; step by step, $R(2, 1)=(1, -2)$ then $C(1, -2)=(3, -2)$ . Pitfall: the product $RC$ , which applies the shear first and gives $(1, -1)$ .

i)

$\det\begin{pmatrix} 3 & -2 \\ 4 & -5 \end{pmatrix}=3\times(-5)-(-2)\times 4=-15+8=-7$ . Areas are multiplied by $|\det|=7$ ; the minus sign only says that the orientation is reversed. Pitfall: answering $-7$ ; an area is never negative.

j)

$\det A=2\times 3-1\times 5=1$ , so $A^{-1}=\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$ : swap the diagonal entries, change the sign of the other two, divide by the determinant. The preimage is $A^{-1}(4, 7)=(12-7, -20+14)=(5, -6)$ , and we check $A(5, -6)=(10-6, 25-18)=(4, 7)$ . Pitfall: applying $A$ instead of $A^{-1}$ , which gives the image $(15, 41)$ and not the preimage.

Check: for each line, redo the calculation on a sketch. A reflection keeps the distance to the origin, and so does a rotation ( $\sqrt{4^{2}+2^{2}}=\sqrt{20}$ and $(2\sqrt{3}-1)^{2}+(2+\sqrt{3})^{2}=20$ in g); a projection flattens the point onto its axis; and a preimage, once transformed, gives back the starting point.

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Exercise 1: Matrices of the standard geometric transformations

All the transformations in this exercise are centred at the origin. Let $P(3, 2)$ be the point whose image we follow throughout.

a) Give the matrix $R$ of the counterclockwise rotation through $90^{\circ}$ , then the image of $P$ .
b) Give the matrix $S$ of the reflection about the line $y=x$ , then the image of $P$ .
c) Give the matrix $H$ of the dilation with factor $3$ , then the image of $P$ .
d) Give the matrix $Q$ of the orthogonal projection onto the $x$ -axis, then the image of $P$ . Compute the four determinants and say what each one reveals.

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Answers

a) $R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ ; $(-2, 3)$
b) $S=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ ; $(2, 3)$
c) $H=3I$ ; $(9, 6)$
d) $Q$ : $(3, 0)$ ; determinants $1$ , $-1$ , $9$ , $0$

a)

Apply the formula for $R_{\theta}$ with $\theta=90^{\circ}$ , so $\cos 90^{\circ}=0$ and $\sin 90^{\circ}=1$ : $R=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ . Image of $P$ : $R\begin{pmatrix} 3 \\ 2 \end{pmatrix}=\begin{pmatrix} (0)(3)+(-1)(2) \\ (1)(3)+(0)(2) \end{pmatrix}=\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ . Visual check: the point was in the first quadrant and moves to the second, which is exactly what a counterclockwise quarter turn does.

b)

The reflection about $y=x$ swaps the two coordinates, so $S=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ . Image of $P$ : $(2, 3)$ . The matrix can also be recovered from its columns: $(1, 0)$ reflects to $(0, 1)$ , which gives the first column, and $(0, 1)$ reflects to $(1, 0)$ , which gives the second.

c)

$H=\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ , and the image of $P$ is $(9, 6)$ . A dilation centred at the origin reads off directly: each coordinate is multiplied by the factor.

d)

The projection onto the $x$ -axis keeps the $x$ -coordinate and sets the $y$ -coordinate to zero, so $Q=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ , and the image of $P$ is $(3, 0)$ .

The determinants: $\det(R)=(0)(0)-(-1)(1)=1$ , $\det(S)=(0)(0)-(1)(1)=-1$ , $\det(H)=(3)(3)-0=9$ and $\det(Q)=(1)(0)-0=0$ .

What each one says. $\det(R)=1$ : the rotation preserves areas and orientation, as expected of a rigid motion. $\det(S)=-1$ : the reflection preserves areas in absolute value but reverses orientation; a triangle traced counterclockwise is traced clockwise after the reflection. $\det(H)=9=3^{2}$ : areas are multiplied by $9$ , not by $3$ , because the enlargement acts in both directions at once. $\det(Q)=0$ : the projection flattens the whole plane onto a line, every area becomes zero, and the transformation is therefore not invertible, since there is no way back from a point of the axis to its preimage.

Exercise 2: Rotation through 60 degrees and powers of a matrix

Let $R_{60}$ be the matrix of the counterclockwise rotation through $60^{\circ}$ about the origin. Recall that $\cos 60^{\circ}=\dfrac{1}{2}$ and $\sin 60^{\circ}=\dfrac{\sqrt{3}}{2}$ .

The blue arrow shows the direction of the rotation.

a) Write the matrix $R_{60}$ .
b) Find the image of the point $A(2, 0)$ , and check that the distance to the origin is preserved.
c) Compute $R_{60}^{2}$ and identify the resulting transformation geometrically.
d) Without computing the product, what is $R_{60}^{6}$ ? Justify geometrically, then give $\det(R_{60}^{6})$ .

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Answers

a) $R_{60}=\begin{pmatrix}\frac12&-\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&\frac12\end{pmatrix}$
b) $(1, \sqrt3)$ , distance 2
c) Rotation through $120^\circ$
d) $R_{60}^6=I$ , determinant 1

a)

$R_{60}=\begin{pmatrix} \cos 60^{\circ} & -\sin 60^{\circ} \\ \sin 60^{\circ} & \cos 60^{\circ} \end{pmatrix}=\begin{pmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\[4pt] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix}$ .

b)

$R_{60}\begin{pmatrix} 2 \\ 0 \end{pmatrix}=\begin{pmatrix} \dfrac{1}{2}(2)-\dfrac{\sqrt{3}}{2}(0) \\[4pt] \dfrac{\sqrt{3}}{2}(2)+\dfrac{1}{2}(0) \end{pmatrix}=\begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix}$ . The image is $A'(1, \sqrt{3})$ , whose distance to the origin is $\sqrt{1+3}=2$ , exactly that of $A$ . A rotation preserves lengths, and this check should become a reflex.

c)

$R_{60}^{2}=\begin{pmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\[4pt] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix}\begin{pmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\[4pt] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix}$ . Top-left entry: $\dfrac{1}{4}-\dfrac{3}{4}=-\dfrac{1}{2}$ . Top right: $-\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}=-\dfrac{\sqrt{3}}{2}$ . Bottom left: $\dfrac{\sqrt{3}}{4}+\dfrac{\sqrt{3}}{4}=\dfrac{\sqrt{3}}{2}$ . Bottom right: $-\dfrac{3}{4}+\dfrac{1}{4}=-\dfrac{1}{2}$ .

So $R_{60}^{2}=\begin{pmatrix} -\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\[4pt] \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \end{pmatrix}$ . This is $R_{\theta}$ with $\cos\theta=-\dfrac{1}{2}$ and $\sin\theta=\dfrac{\sqrt{3}}{2}$ , that is $\theta=120^{\circ}$ . Applying a rotation through $60^{\circ}$ twice gives a rotation through $120^{\circ}$ : the angles add when the matrices multiply.

d)

Six rotations through $60^{\circ}$ make $360^{\circ}$ , one full turn, which brings every point back to where it started. So $R_{60}^{6}=I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ , with no matrix computation at all. Consequently $\det(R_{60}^{6})=\det(I)=1$ , consistent with $\big(\det R_{60}\big)^{6}=1^{6}=1$ .

Remember the mechanism: $R_{\alpha}R_{\beta}=R_{\alpha+\beta}$ . It turns the power of a matrix into a simple sum of angles, and it is exactly what De Moivre's formula does with complex numbers. The two chapters describe the same phenomenon in two languages.

Exercise 3: Composition of transformations and non-commutativity

Let $S$ be the reflection about the $x$ -axis, with matrix $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ , and $R$ the counterclockwise rotation through $90^{\circ}$ , with matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ .

a) Compute the matrix of $R\circ S$ , that is, the transformation that applies $S$ first and then $R$ . Identify the result geometrically.
b) Compute the matrix of $S\circ R$ and identify it as well.
c) Compare a) and b). What can you conclude? Illustrate with the point $P(1, 0)$ .
d) Compute the determinants of $R$ , $S$ , $R\circ S$ and $S\circ R$ , and check that they agree with the property $\det(AB)=\det(A)\det(B)$ .

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Answers

a) $RS$ : reflection about $y=x$
b) $SR$ : reflection about $y=-x$
c) $RS\neq SR$ ; $(0, 1)$ versus $(0, -1)$
d) All the determinants equal $\pm 1$ , as expected

a)

Applying $S$ then $R$ corresponds to the product $RS$ , since the matrix of the transformation applied first goes on the RIGHT. $RS=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}=\begin{pmatrix} (0)(1)+(-1)(0) & (0)(0)+(-1)(-1) \\ (1)(1)+(0)(0) & (1)(0)+(0)(-1) \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ . This is the matrix of the reflection about the line $y=x$ .

b)

Applying $R$ then $S$ corresponds to the product $SR=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix} (1)(0)+(0)(1) & (1)(-1)+(0)(0) \\ (0)(0)+(-1)(1) & (0)(-1)+(-1)(0) \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$ . This is the matrix of the reflection about the line $y=-x$ .

c)

The two results differ: $RS\neq SR$ . Composition of transformations is therefore NOT commutative, which mirrors the non-commutativity of matrix multiplication. The order in which the transformations are chained changes the geometric result.

Illustration with $P(1, 0)$ . Under $R\circ S$ : the reflection leaves $P$ unchanged since it lies on the $x$ -axis, then the rotation sends it to $(0, 1)$ . Under $S\circ R$ : the rotation sends $P$ to $(0, 1)$ , then the reflection about the $x$ -axis sends it to $(0, -1)$ . The images are $(0, 1)$ and $(0, -1)$ , two different points, from the same starting point and the same two transformations. This is the concrete proof that order matters.

d)

$\det(R)=1$ , $\det(S)=-1$ , $\det(RS)=(0)(0)-(1)(1)=-1$ and $\det(SR)=(0)(0)-(-1)(-1)=-1$ . The property holds both ways: $\det(R)\det(S)=(1)(-1)=-1=\det(RS)$ , and likewise for $SR$ .

Notice what this reveals: the determinant of the product does not depend on the order of the factors, whereas the product itself does. The determinant only sees the effect on areas and orientation; it cannot tell apart two reflections about different axes. Two transformations can have the same determinant without being the same transformation, and that is precisely the case here.

Exercise 4: Recovering the matrix, image of a figure and area factor

A linear transformation $T$ of the plane is defined by what it does to the vectors of the standard basis: $T(1, 0)=(3, 1)$ and $T(0, 1)=(-1, 2)$ .

a) Find the matrix $A$ of $T$ .
b) Find the image of the vector $(4, -2)$ .
c) The triangle with vertices $(0, 0)$ , $(1, 0)$ and $(0, 1)$ has area $\dfrac{1}{2}$ . Find the area of its image under $T$ .
d) Is the transformation $T$ invertible? If so, find $A^{-1}$ and the preimage of the point $(7, 7)$ .

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Answers

a) $A=\begin{pmatrix}3&-1\\1&2\end{pmatrix}$
b) $(14, 0)$
c) $\det A=7$ : area $3.5$
d) $A^{-1}=\frac17\begin{pmatrix}2&1\\-1&3\end{pmatrix}$ ; $(3, 2)$

a)

The columns of the matrix are the images of the standard basis vectors, in order. The first column is therefore $T(1, 0)=(3, 1)$ and the second $T(0, 1)=(-1, 2)$ , which gives $A=\begin{pmatrix} 3 & -1 \\ 1 & 2 \end{pmatrix}$ . Be careful not to write the images as ROWS: it is the most common mistake, and it produces the transpose, hence a different transformation.

b)

$A\begin{pmatrix} 4 \\ -2 \end{pmatrix}=\begin{pmatrix} (3)(4)+(-1)(-2) \\ (1)(4)+(2)(-2) \end{pmatrix}=\begin{pmatrix} 12+2 \\ 4-4 \end{pmatrix}=\begin{pmatrix} 14 \\ 0 \end{pmatrix}$ . There is another way to check, by linearity: $T(4, -2)=4\,T(1, 0)-2\,T(0, 1)=4(3, 1)-2(-1, 2)=(12+2, 4-4)=(14, 0)$ . Both routes give the same result, and the second one shows directly what linear means.

c)

$\det(A)=(3)(2)-(-1)(1)=6+1=7$ . The area of the image is $|\det(A)|$ times the original area, that is $7\times\dfrac{1}{2}=\dfrac{7}{2}=3.5$ square units. Since the determinant is positive, the orientation of the triangle is preserved.

d)

$\det(A)=7\neq 0$ , so $T$ is invertible. For a $2\times 2$ matrix, $A^{-1}=\dfrac{1}{\det(A)}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}=\dfrac{1}{7}\begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}$ .

Preimage of $(7, 7)$ : $A^{-1}\begin{pmatrix} 7 \\ 7 \end{pmatrix}=\dfrac{1}{7}\begin{pmatrix} (2)(7)+(1)(7) \\ (-1)(7)+(3)(7) \end{pmatrix}=\dfrac{1}{7}\begin{pmatrix} 21 \\ 14 \end{pmatrix}=\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ . Direct check: $T(3, 2)=(9-2, 3+4)=(7, 7)$ , as required.

A remark that holds for the whole chapter: the area of the image of the triangle is also the area of the triangle built on the two column vectors $(3, 1)$ and $(-1, 2)$ , namely $\dfrac{1}{2}|\det(A)|=\dfrac{7}{2}$ . The determinant is not just an abstract multiplying factor: it IS the area of the parallelogram onto which the unit square is mapped.

Exercise 5: The shear: a transformation that preserves area without being a rotation

Consider the transformation $T$ with matrix $A=\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ , called a horizontal shear. It slides points further the higher they are.

a) Find the images of the four vertices of the unit square $(0, 0)$ , $(1, 0)$ , $(1, 1)$ , $(0, 1)$ , and describe the resulting figure.
b) Compute $\det(A)$ and deduce the area of the image of the unit square.
c) Does the shear preserve lengths? Compare $\|(0, 1)\|$ with the norm of its image, and conclude.
d) Find all the points left fixed by $T$ . What does this set represent geometrically?

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Answers

a) Parallelogram $(0, 0)$ , $(1, 0)$ , $(3, 1)$ , $(2, 1)$
b) $\det=1$ : area 1
c) $\sqrt5\neq 1$ : lengths not preserved
d) The $x$ -axis

a)

$T(0, 0)=(0, 0)$ ; $T(1, 0)=(1, 0)$ ; $T(1, 1)=(1+2, 1)=(3, 1)$ ; $T(0, 1)=(2, 1)$ . The image of the unit square is the parallelogram with vertices $(0, 0)$ , $(1, 0)$ , $(3, 1)$ , $(2, 1)$ . The base on the $x$ -axis has not moved, but the top side has slid two units to the right: the square has become a slanted parallelogram.

b)

$\det(A)=(1)(1)-(2)(0)=1$ . The area of the image is $|\det(A)|\times 1=1$ : the shear PRESERVES area, exactly like a rotation. The base and the height of the parallelogram are unchanged ( $1$ and $1$ ), so its area stays $1$ , even though its shape has changed completely.

c)

No. We have $\|(0, 1)\|=1$ , while its image $(2, 1)$ has norm $\sqrt{2^{2}+1^{2}}=\sqrt{5}\approx 2.24$ . Lengths are therefore not preserved: the shear is NOT an isometry; it is neither a rotation nor a reflection. This is the central lesson of the exercise: a determinant equal to $1$ guarantees that areas are preserved, but NOT lengths. Only orthogonal matrices preserve distances, and the shear is not one of them.

d)

A point $(x, y)$ is fixed if $T(x, y)=(x, y)$ , that is $(x+2y, y)=(x, y)$ . The second coordinate is automatically satisfied, and the first gives $x+2y=x$ , so $2y=0$ , that is $y=0$ . The fixed points are exactly those of the form $(x, 0)$ : the whole $x$ -axis. Geometrically, the shear keeps its base line still, point by point, and slides the rest of the plane parallel to it, by an amount proportional to the height.

Exercise 6: Proofs: linearity, orthogonal matrices and projections

This exercise is about the structural properties of transformations. The numerical calculations serve only as checks.

a) Let $A$ be a $2\times 2$ matrix and $T(\vec{v})=A\vec{v}$ . Prove that $T$ is linear.
b) Prove that the translation $f(x, y)=(x+1, y)$ is NOT a linear transformation. Give the quickest criterion.
c) Prove that the rotation matrix $R_{\theta}$ satisfies $R_{\theta}^{T}R_{\theta}=I$ , and deduce that the rotation preserves norms.
d) A matrix $P$ is called idempotent if $P^{2}=P$ . Check that the orthogonal projection onto the $x$ -axis is idempotent, and interpret this property geometrically.

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Answers

a) Distributivity and scalars
b) $f(0, 0)=(1, 0)\neq(0, 0)$
c) $R_\theta^TR_\theta=I$ : norms preserved
d) $P^2=P$ : projecting twice changes nothing

a)

We must establish the two properties of the definition, and both follow from the rules of matrix algebra. Additivity: $T(\vec{u}+\vec{v})=A(\vec{u}+\vec{v})=A\vec{u}+A\vec{v}=T(\vec{u})+T(\vec{v})$ , by distributivity of matrix multiplication over addition. Homogeneity: $T(k\vec{u})=A(k\vec{u})=k(A\vec{u})=k\,T(\vec{u})$ , because a scalar moves freely through a matrix product. Both conditions hold, so $T$ is linear.

b)

The quickest criterion is the origin test: every linear transformation satisfies $T(\vec{0})=\vec{0}$ , since $T(\vec{0})=T(0\cdot\vec{0})=0\cdot T(\vec{0})=\vec{0}$ . Here, however, $f(0, 0)=(1, 0)\neq(0, 0)$ . The origin moves, so $f$ is not linear. One line is enough.

It can also be seen from additivity: $f(1, 0)+f(1, 0)=(2, 0)+(2, 0)=(4, 0)$ , whereas $f\big((1, 0)+(1, 0)\big)=f(2, 0)=(3, 0)$ . The two results differ. Remember the general lesson: a translation, however simple it is geometrically, is never linear. That is why the rotations and reflections of this chapter are all centred at the origin.

c)

$R_{\theta}=\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ , so $R_{\theta}^{T}=\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$ . The product $R_{\theta}^{T}R_{\theta}$ has top-left entry $\cos^{2}\theta+\sin^{2}\theta=1$ , top-right entry $-\cos\theta\sin\theta+\sin\theta\cos\theta=0$ , bottom-left entry $-\sin\theta\cos\theta+\cos\theta\sin\theta=0$ , and bottom-right entry $\sin^{2}\theta+\cos^{2}\theta=1$ . So $R_{\theta}^{T}R_{\theta}=I$ : the matrix is orthogonal, and $R_{\theta}^{-1}=R_{\theta}^{T}$ .

Norms are preserved: $\|R_{\theta}\vec{v}\|^{2}=(R_{\theta}\vec{v})\cdot(R_{\theta}\vec{v})=(R_{\theta}\vec{v})^{T}(R_{\theta}\vec{v})=\vec{v}^{T}R_{\theta}^{T}R_{\theta}\vec{v}=\vec{v}^{T}I\vec{v}=\vec{v}^{T}\vec{v}=\|\vec{v}\|^{2}$ . Since norms are non-negative, $\|R_{\theta}\vec{v}\|=\|\vec{v}\|$ . The rotation therefore preserves lengths and, by taking differences of vectors, distances between points as well: it is an isometry. The fact that $R_{\theta}^{-1}=R_{\theta}^{T}$ has a crystal-clear geometric reading: the inverse of a rotation through $\theta$ is the rotation through $-\theta$ , and replacing $\theta$ by $-\theta$ changes the sign of the sines only, which is exactly what transposition does here.

d)

$P=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ , so $P^{2}=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}=P$ . The matrix is indeed idempotent.

Geometric interpretation: projecting a point onto the $x$ -axis brings it onto that axis; projecting it a second time does not move it any more, since it is already there. Repeating the operation changes nothing, which algebra writes as $P^{2}=P$ . Notice the consequence: an idempotent matrix other than the identity cannot be invertible. Indeed, if $P$ were invertible, we could multiply $P^{2}=P$ by $P^{-1}$ and get $P=I$ . Since $P\neq I$ here, $P$ is not invertible, as $\det(P)=0$ confirms. The information lost in the projection, the $y$ -coordinate, cannot be recovered.

Exercise 7: Problem: a rotation-dilation, from the unit square to the inverse

Consider the transformation $T$ of the plane obtained by first applying the counterclockwise rotation through $45^{\circ}$ about the origin, then the dilation with factor $\sqrt{2}$ centred at the origin.

Recall that $\cos 45^{\circ}=\sin 45^{\circ}=\dfrac{\sqrt{2}}{2}$ .

a) Show that the matrix of $T$ is $A=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ .
b) Compute $\det(A)$ and deduce the factor by which areas are multiplied.
c) Find the images of the four vertices of the unit square $(0, 0)$ , $(1, 0)$ , $(1, 1)$ , $(0, 1)$ . Check that the image is a square and compute its area in two ways.
d) Compute $A^{2}$ and identify the transformation $T\circ T$ geometrically.
e) Find $A^{-1}$ and the preimage of the point $(3, 5)$ .
f) Check that $T$ acts on the point $(x, y)$ exactly as multiplying the complex number $x+iy$ by $1+i$ . What are the modulus and the argument of $1+i$ , and what do you recover?

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Answers

a) $A=\sqrt2\,R_{45}=\begin{pmatrix}1&-1\\1&1\end{pmatrix}$
b) $\det A=2$
c) Square $(0, 0)$ , $(1, 1)$ , $(0, 2)$ , $(-1, 1)$ : area 2
d) $A^2=2R_{90}$
e) $(4, 1)$
f) Modulus $\sqrt2$ , argument $45^\circ$

a)

The matrix of the rotation through $45^{\circ}$ is $R=\begin{pmatrix} \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2} \\[4pt] \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2} \end{pmatrix}$ and that of the dilation is $H=\sqrt{2}\,I$ . The rotation is applied first, so it goes on the right: $A=HR=\sqrt{2}\begin{pmatrix} \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2} \\[4pt] \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2} \end{pmatrix}=\begin{pmatrix} \dfrac{2}{2} & -\dfrac{2}{2} \\[4pt] \dfrac{2}{2} & \dfrac{2}{2} \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ .

Here the order would have changed nothing, because a dilation centred at the origin commutes with every matrix: it is a multiple of the identity. This is an exception, not the rule, and it must not become a habit.

b)

$\det(A)=(1)(1)-(-1)(1)=1+1=2$ . Areas are therefore multiplied by $|\det(A)|=2$ . Consistency with the construction: the rotation does not change areas, and the dilation with factor $\sqrt{2}$ multiplies them by $(\sqrt{2})^{2}=2$ . The two arguments agree.

c)

$T(0, 0)=(0, 0)$ , $T(1, 0)=(1, 1)$ , $T(1, 1)=(1-1, 1+1)=(0, 2)$ and $T(0, 1)=(-1, 1)$ . The image of the unit square is therefore the quadrilateral with vertices $(0, 0)$ , $(1, 1)$ , $(0, 2)$ and $(-1, 1)$ .

It is indeed a square: the four sides have length $\sqrt{1^{2}+1^{2}}=\sqrt{2}$ , and two consecutive sides from the origin, $(1, 1)$ and $(-1, 1)$ , have dot product $(1)(-1)+(1)(1)=0$ , so they are perpendicular. The unit square, rotated through $45^{\circ}$ and enlarged by a factor $\sqrt{2}$ , is still a square.

Area, first method: side $\sqrt{2}$ , so area $=(\sqrt{2})^{2}=2$ square units. Second method: $|\det(A)|\times 1=2$ square units, since the unit square has area $1$ . The two values agree.

d)

$A^{2}=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}=\begin{pmatrix} 1-1 & -1-1 \\ 1+1 & -1+1 \end{pmatrix}=\begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$ . Factor: $A^{2}=2\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ , that is, the dilation with factor $2$ composed with the rotation through $90^{\circ}$ .

This is exactly what we expected: applying a rotation through $45^{\circ}$ twice gives $90^{\circ}$ , and applying a dilation with factor $\sqrt{2}$ twice gives a factor $(\sqrt{2})^{2}=2$ . Angles add, factors multiply. Check with the determinant: $\det(A^{2})=\big(\det A\big)^{2}=4$ , and directly $(0)(0)-(-2)(2)=4$ .

e)

For a $2\times 2$ matrix: $A^{-1}=\dfrac{1}{\det(A)}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}=\dfrac{1}{2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ . Preimage of $(3, 5)$ : $A^{-1}\begin{pmatrix} 3 \\ 5 \end{pmatrix}=\dfrac{1}{2}\begin{pmatrix} 3+5 \\ -3+5 \end{pmatrix}=\dfrac{1}{2}\begin{pmatrix} 8 \\ 2 \end{pmatrix}=\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ . Check: $T(4, 1)=(4-1, 4+1)=(3, 5)$ , as required.

Geometrically, $A^{-1}$ is the rotation through $-45^{\circ}$ followed by the dilation with factor $\dfrac{1}{\sqrt{2}}$ , which exactly undoes $T$ .

f)

Compute the complex product: $(x+iy)(1+i)=x+ix+iy+i^{2}y=x+ix+iy-y=(x-y)+i(x+y)$ . This complex number has real part $x-y$ and imaginary part $x+y$ . Now $T(x, y)=A\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} x-y \\ x+y \end{pmatrix}$ . The two results are identical: multiplying by $1+i$ in the complex plane is applying $T$ in the real plane.

The modulus of $1+i$ is $\sqrt{1^{2}+1^{2}}=\sqrt{2}$ and its argument is $45^{\circ}$ , since the point $(1, 1)$ lies on the bisector of the first quadrant. We recover exactly the two parameters used to build $T$ : the modulus gives the dilation factor, the argument gives the angle of the rotation.

This is the bridge between the last two chapters of the course. Multiplying by a complex number with modulus $r$ and argument $\theta$ means composing a rotation through $\theta$ with a dilation with factor $r$ , and De Moivre's formula $\big(r\,e^{i\theta}\big)^{n}=r^{n}e^{in\theta}$ is nothing but the complex version of the computation of $A^{n}$ carried out in d). When an exam problem asks for a high power of a rotation-dilation matrix, going through complex numbers is often the shortest route.

Exercise 8: Reflections about a line and the composition of two reflections

The reflection about the line through the origin that makes an angle $\theta$ with the $x$ -axis has matrix $M(\theta)=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}$ .

a) Using this formula, recover the matrix of the reflection about the $x$ -axis ( $\theta=0$ ) and about the line $y=x$ ( $\theta=45^{\circ}$ ).
b) Show that every reflection matrix satisfies $\det(M)=-1$ and $M^{2}=I$ . What does the second equality mean geometrically?
c) Let $S_{0}$ be the reflection about the $x$ -axis and $S_{45}$ the reflection about $y=x$ . Compute the matrix of $S_{45}\circ S_{0}$ and identify the resulting transformation.
d) State and justify the general rule: the composition of two reflections whose axes make an angle $\alpha$ is a rotation through $2\alpha$ . Check that the determinants agree, then give the angle of the resulting rotation when the two axes make an angle of $30^{\circ}$ .

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Answers

a) $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $\begin{pmatrix}0&1\\1&0\end{pmatrix}$
b) $\det=-1$ , $M^2=I$ : an involution
c) Rotation through $90^\circ$
d) Rotation through $2\alpha$ ; $60^{\circ}$ for axes $30^{\circ}$ apart

a)

For $\theta=0$ : $\cos 0=1$ and $\sin 0=0$ , so $M(0)=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ , the reflection about the $x$ -axis. For $\theta=45^{\circ}$ : $2\theta=90^{\circ}$ , $\cos 90^{\circ}=0$ and $\sin 90^{\circ}=1$ , so $M(45^{\circ})=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ , the reflection about $y=x$ , which swaps the two coordinates. The two familiar matrices are indeed special cases of the general formula.

b)

$\det(M)=(\cos 2\theta)(-\cos 2\theta)-(\sin 2\theta)(\sin 2\theta)=-\cos^{2}2\theta-\sin^{2}2\theta=-1$ , using the Pythagorean identity. For the square: the top-left entry of $M^{2}$ is $\cos^{2}2\theta+\sin^{2}2\theta=1$ , the top-right entry is $\cos 2\theta\sin 2\theta-\sin 2\theta\cos 2\theta=0$ , and in the same way $M^{2}=I$ . Geometrically, $M^{2}=I$ means that applying the same reflection twice brings every point back to its place: a reflection is its own inverse, it is an involution. The determinant $-1$ confirms that each reflection reverses orientation.

c)

Applying $S_{0}$ then $S_{45}$ corresponds to the product $S_{45}S_{0}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ . This is the matrix of the counterclockwise rotation through $90^{\circ}$ . Two reflections have composed into a rotation.

d)

Rule: the composition of two reflections about axes through the origin that make an angle $\alpha$ with each other is the rotation through $2\alpha$ (in the direction that goes from the first axis to the second). Justification with matrices: $M(\beta)M(\alpha)$ is computed with the angle-sum formulas and gives exactly $R_{2(\beta-\alpha)}$ . Here the axes are the $x$ -axis ( $\theta=0$ ) and the line $y=x$ ( $\theta=45^{\circ}$ ), so the angle between them is $45^{\circ}$ , and the composition is the rotation through $2\times 45^{\circ}=90^{\circ}$ , as the computation in c) confirms. Consistency of the determinants: each reflection has determinant $-1$ , so their product has determinant $(-1)(-1)=+1$ , which is indeed that of a rotation. The orientation, reversed twice, is finally restored. For two axes $30^{\circ}$ apart, the composition is the rotation through $2\times 30^{\circ}=60^{\circ}$ .

Exercise 9: Recovering the matrix from its action on two arbitrary vectors

A linear transformation $T$ of the plane is known to satisfy $T(1, 1)=(3, 1)$ and $T(1, -1)=(1, 3)$ . The two starting vectors are NOT those of the standard basis, so the matrix has to be found some other way than by simply reading off the columns.

The starting vectors are dashed, their images are solid.

a) Setting $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , translate the two conditions into equations and find $A$ .
b) Deduce the images $T(1, 0)$ and $T(0, 1)$ of the standard basis vectors.
c) Compute $\det(A)$ , say whether $T$ is invertible and by what factor it multiplies areas.
d) Find $A^{-1}$ , then the preimage of the point $(4, 0)$ .

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Answers

a) $A=\begin{pmatrix}2&1\\2&-1\end{pmatrix}$
b) $T(1, 0)=(2, 2)$ , $T(0, 1)=(1, -1)$
c) $\det=-4$ : invertible, areas $\times 4$
d) $(1, 2)$

a)

The condition $T(1, 1)=(3, 1)$ gives $\begin{pmatrix} a+b \\ c+d \end{pmatrix}=\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ , that is $a+b=3$ and $c+d=1$ . The condition $T(1, -1)=(1, 3)$ gives $\begin{pmatrix} a-b \\ c-d \end{pmatrix}=\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ , that is $a-b=1$ and $c-d=3$ . Adding and subtracting: $a=2$ , $b=1$ , then $c=2$ , $d=-1$ . So $A=\begin{pmatrix} 2 & 1 \\ 2 & -1 \end{pmatrix}$ . Check: $A(1, 1)=(3, 1)$ and $A(1, -1)=(1, 3)$ ✓.

b)

The columns of $A$ are precisely the images of the standard basis vectors: $T(1, 0)=(2, 2)$ (first column) and $T(0, 1)=(1, -1)$ (second column). Knowing the action on two independent vectors is enough to know the action on the whole basis, hence on the whole plane: a direct consequence of linearity.

c)

$\det(A)=(2)(-1)-(1)(2)=-2-2=-4$ . It is non-zero, so $T$ is invertible. Areas are multiplied by $|\det(A)|=4$ , and the negative sign indicates that $T$ reverses the orientation of the plane.

d)

$A^{-1}=\dfrac{1}{\det(A)}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}=\dfrac{1}{-4}\begin{pmatrix} -1 & -1 \\ -2 & 2 \end{pmatrix}=\begin{pmatrix} \frac{1}{4} & \frac{1}{4} \\ \frac{1}{2} & -\frac{1}{2} \end{pmatrix}$ . Preimage of $(4, 0)$ : $A^{-1}\begin{pmatrix} 4 \\ 0 \end{pmatrix}=\begin{pmatrix} \frac{1}{4}(4) \\ \frac{1}{2}(4) \end{pmatrix}=\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ . Check: $T(1, 2)=(2+2, 2-2)=(4, 0)$ , as required.

Exercise 10: Proofs: composition, image of a line and midpoints preserved

These proofs use nothing but the definition of linearity. They explain why a linear transformation preserves the geometric structure of the plane.

a) Let $T_{1}$ and $T_{2}$ be two linear transformations with matrices $A_{1}$ and $A_{2}$ . Prove that $T_{2}\circ T_{1}$ is linear and that its matrix is the product $A_{2}A_{1}$ .
b) Prove that the image under a linear transformation $T$ of a line $\{P+t\vec{u}:t\in\mathbb{R}\}$ is a line (or a point if $T(\vec{u})=\vec{0}$ ).
c) Prove that $T$ preserves midpoints: $T\!\left(\dfrac{P+Q}{2}\right)=\dfrac{T(P)+T(Q)}{2}$ . What can you deduce about the image of a parallelogram?
d) True or false: every linear transformation preserves distances. State the correct criterion and illustrate it by comparing a rotation with the shear $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ .

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Answers

a) Matrix $A_2A_1$
b) $T(P)+t\,T(\vec u)$ : a line or a point
c) Midpoints preserved: parallelograms preserved
d) False: $A^TA=I$ is needed; $\sqrt{65}\neq 5$

a)

Linearity of the composition: for all $\vec{u},\vec{v}$ and every scalar $k$ , $(T_{2}\circ T_{1})(\vec{u}+\vec{v})=T_{2}\big(T_{1}(\vec{u}+\vec{v})\big)=T_{2}\big(T_{1}(\vec{u})+T_{1}(\vec{v})\big)=T_{2}(T_{1}\vec{u})+T_{2}(T_{1}\vec{v})$ , using the linearity of $T_{1}$ and then that of $T_{2}$ ; likewise $(T_{2}\circ T_{1})(k\vec{u})=k\,(T_{2}\circ T_{1})(\vec{u})$ . The composition is therefore linear. Its matrix: $(T_{2}\circ T_{1})(\vec{v})=T_{2}(A_{1}\vec{v})=A_{2}(A_{1}\vec{v})=(A_{2}A_{1})\vec{v}$ , by associativity of matrix multiplication. The matrix of the composition is therefore $A_{2}A_{1}$ , with the transformation applied first on the right.

b)

Take a point of the line, $P+t\vec{u}$ . Its image is $T(P+t\vec{u})=T(P)+t\,T(\vec{u})$ , by linearity. As $t$ runs through $\mathbb{R}$ , this point describes the set $\{T(P)+t\,T(\vec{u}):t\in\mathbb{R}\}$ . If $T(\vec{u})\neq\vec{0}$ , it is the line through $T(P)$ with direction vector $T(\vec{u})$ . If $T(\vec{u})=\vec{0}$ , all the points have the same image $T(P)$ : the line is collapsed onto a point. In every case, the image of a line is a line or a point, never a curve: a linear transformation does not bend lines.

c)

$T\!\left(\dfrac{P+Q}{2}\right)=\dfrac{1}{2}T(P+Q)=\dfrac{1}{2}\big(T(P)+T(Q)\big)=\dfrac{T(P)+T(Q)}{2}$ , by taking the scalar $\tfrac{1}{2}$ out and then using additivity. The midpoint of segment $PQ$ is therefore mapped to the midpoint of the segment from $T(P)$ to $T(Q)$ . Consequence: since the diagonals of a parallelogram bisect each other, this property is preserved, so the image of a parallelogram is a parallelogram (possibly flattened if the transformation is singular). More generally, $T$ preserves parallelism and ratios along a line.

d)

FALSE in general. The correct criterion: a linear transformation preserves distances if and only if its matrix is ORTHOGONAL, that is, satisfies $A^{T}A=I$ (rotations and reflections). A rotation through $\theta$ satisfies this criterion and therefore preserves lengths; for example $\|(3, 4)\|=5$ is preserved. The shear $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ , on the other hand, does NOT preserve distances even though its determinant is $1$ : the image of $(3, 4)$ is $(7, 4)$ , with norm $\sqrt{65}\neq 5$ . A determinant equal to $1$ and an orthogonal matrix are two different conditions: the first preserves areas, only the second preserves lengths.

Exercise 11: Three transformations of space

In space, consider the rotation $R$ through $90^{\circ}$ about the $z$ -axis, with matrix $\begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ , the orthogonal projection $Q$ onto the plane $z=0$ , with matrix $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ , and the reflection $S$ about the plane $x=y$ , with matrix $\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ .

a) Compute the images of the point $P(1, 2, 3)$ under $R$ , $Q$ and $S$ .
b) Compute the three determinants and interpret them in terms of volume.
c) Compute the matrix of $R\circ S$ and identify the transformation.
d) What is the smallest integer $k\geq 1$ such that $R^{k}=I$ ?
e) Which points are left fixed by $R$ ?

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Answers

a) $(-2, 1, 3)$ , $(1, 2, 0)$ , $(2, 1, 3)$
b) $1$ , $0$ , $-1$
c) Reflection about the plane $x=0$
d) $k=4$
e) The $z$ -axis

a)

$R(P)=(-2, 1, 3)$ : the quarter turn acts in the horizontal plane and leaves the $z$ -coordinate alone. $Q(P)=(1, 2, 0)$ : the $z$ -coordinate is set to zero. $S(P)=(2, 1, 3)$ : the first two coordinates are swapped.

b)

$\det R=1$ : volumes and orientation preserved. $\det Q=0$ : space is flattened onto a plane, every volume becomes zero and $Q$ is not invertible. $\det S=-1$ : volumes preserved, orientation reversed, as for a reflection in the plane.

c)

$RS=\begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ : this is the reflection about the plane $x=0$ , which changes the sign of the $x$ -coordinate. Its determinant, $-1$ , is indeed the product $1\times(-1)$ .

d)

Each application of $R$ turns through $90^{\circ}$ about the $z$ -axis: it takes four quarter turns to come back to the start, so $k=4$ . Indeed $R^{2}$ is the half turn, with matrix $\text{diag}(-1, -1, 1)$ , which is not $I$ .

e)

$R(x, y, z)=(-y, x, z)=(x, y, z)$ forces $x=-y$ and $y=x$ , so $x=y=0$ with $z$ arbitrary: the fixed points form the $z$ -axis, the axis of the rotation.

The thread: in dimension 3, the columns of the matrix are still the images of the basis vectors, and the determinant measures the effect on volumes.

Exercise 12: The image of a circle: an ellipse

Consider the transformation $T$ with matrix $A=\begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$ and the unit circle, made up of the points $(\cos t, \sin t)$ .

a) Find the image of the point $(\cos t, \sin t)$ and the equation of the image curve.
b) What are the semi-axes of this curve?
c) The area of an ellipse with semi-axes $a$ and $b$ is $\pi ab$ . Compute the area of the image, to two decimal places, and compare it with $|\det A|$ times the area of the unit disk.
d) Now apply the shear $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ to the unit disk. What is the area of the image, to two decimal places?
e) Is the image of a circle centred at the origin under a rotation a circle?

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Answers

a) $\frac{x^2}{9}+\frac{y^2}{4}=1$
b) 3 and 2
c) $6\pi\approx 18.85=|\det A|\,\pi$
d) $\pi\approx 3.14$
e) Yes

a)

$T(\cos t, \sin t)=(3\cos t, 2\sin t)$ . With $X=3\cos t$ and $Y=2\sin t$ : $\frac{X^{2}}{9}+\frac{Y^{2}}{4}=\cos^{2}t+\sin^{2}t=1$ . The image curve is the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ .

b)

The semi-axes are $3$ , along the $x$ -axis, and $2$ , along the $y$ -axis: they are the diagonal entries, which stretch each direction.

c)

$\pi\times 3\times 2=6\pi\approx 18.85$ . The unit disk has area $\pi$ , and $|\det A|=6$ : we recover $6\pi$ . The area factor given by the determinant holds for every figure, not only for polygons.

d)

The shear has determinant $1$ : the image, a tilted ellipse, has the same area as the disk, $\pi\approx 3.14$ , even though its shape is very different.

e)

Yes: a rotation preserves distances to the origin, so every point of the circle stays at the same distance. More generally, only the transformations with an orthogonal matrix map the unit circle onto itself.

The thread: a linear transformation maps a circle onto an ellipse, whose area is multiplied by the absolute value of the determinant.

Exercise 13: Undoing a composite transformation

Let $C$ be the shear with matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $R$ the counterclockwise rotation through $90^{\circ}$ , with matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ . The transformation $T$ applies $C$ first, then $R$ .

a) Compute the matrix of $T$ .
b) Compute the image of the point $(2, 3)$ under $T$ .
c) Compute $\det T$ and the matrix of $T^{-1}$ .
d) Compute $C^{-1}R^{-1}$ and $R^{-1}C^{-1}$ . Which of the two products equals $T^{-1}$ ? Explain.
e) Apply $T^{-1}$ to the point obtained in b).

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Answers

a) $T=\begin{pmatrix}0&-1\\1&1\end{pmatrix}$
b) $(-3, 5)$
c) $\det T=1$ ; $T^{-1}=\begin{pmatrix}1&1\\-1&0\end{pmatrix}$
d) $T^{-1}=C^{-1}R^{-1}$
e) $(2, 3)$

a)

The first transformation goes on the right: $T=RC=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}$ .

b)

$T\begin{pmatrix} 2 \\ 3 \end{pmatrix}=\begin{pmatrix} -3 \\ 2+3 \end{pmatrix}=\begin{pmatrix} -3 \\ 5 \end{pmatrix}$ . In two steps: $C(2, 3)=(5, 3)$ , then $R(5, 3)=(-3, 5)$ .

c)

$\det T=0\times 1-(-1)\times 1=1$ , and $T^{-1}=\begin{pmatrix} 1 & 1 \\ -1 & 0 \end{pmatrix}$ .

d)

$C^{-1}=\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ and $R^{-1}=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ . $C^{-1}R^{-1}=\begin{pmatrix} 1 & 1 \\ -1 & 0 \end{pmatrix}=T^{-1}$ , whereas $R^{-1}C^{-1}=\begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix}$ is different. To undo “ $C$ , then $R$ ”, first undo $R$ , the last one applied, then $C$ : $T^{-1}=(RC)^{-1}=C^{-1}R^{-1}$ , just as you take off your shoes before your socks.

e)

$T^{-1}\begin{pmatrix} -3 \\ 5 \end{pmatrix}=\begin{pmatrix} -3+5 \\ 3 \end{pmatrix}=\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ : we get back the starting point.

The thread: the inverse of a composition undoes the transformations in the reverse of the order in which they were done.

Exercise 14: Problem: a cast shadow, an oblique projection

The Sun's rays arrive in the direction $\vec{d}=(1, 1, -2)$ on horizontal ground, the plane $z=0$ . The shadow of a point $M(x, y, z)$ is the point where the line through $M$ with direction $\vec{d}$ meets the ground.

The orange arrow gives the direction of the Sun's rays.

a) A vertical pole joins the origin to the point $(0, 0, 4)$ . Find the shadow of its top and the length of the shadow, to two decimal places.
b) Show that the shadow of $M(x, y, z)$ is $\left(x+\frac{z}{2}, y+\frac{z}{2}, 0\right)$ and write the matrix $\Omega$ of this transformation.
c) Check that $\Omega^{2}=\Omega$ and interpret.
d) Find the kernel of $\Omega$ and its determinant.
e) Is this transformation the orthogonal projection onto the ground? Compare with the shadow of the same pole at noon under a vertical Sun.

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Answers

a) $(2, 2, 0)$ ; length $2\sqrt2\approx 2.83$
b) $\Omega=\begin{pmatrix}1&0&\frac12\\0&1&\frac12\\0&0&0\end{pmatrix}$
c) $\Omega^2=\Omega$ : a projection
d) Kernel spanned by $\vec d$ ; $\det\Omega=0$
e) No: an oblique projection

a)

The point $(t, t, 4-2t)$ reaches the ground when $t=2$ : the shadow of the top is $(2, 2, 0)$ , and the shadow of the pole is the segment from the origin to that point, of length $2\sqrt{2}\approx 2.83$ .

b)

$M+t\vec{d}=(x+t, y+t, z-2t)$ touches the ground when $t=\frac{z}{2}$ , hence the shadow $\left(x+\frac{z}{2}, y+\frac{z}{2}, 0\right)$ . Each coordinate is a linear combination of $x$ , $y$ , $z$ : $\Omega=\begin{pmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 \end{pmatrix}$ .

c)

The first row of $\Omega^{2}$ is $\left(1, 0, \frac{1}{2}+0\right)$ , the second $\left(0, 1, \frac{1}{2}\right)$ , and the third is zero: $\Omega^{2}=\Omega$ . The shadow of a shadow is itself: a point already on the ground no longer moves. This is a projection.

d)

$\Omega(x, y, z)=\vec{0}$ forces $x=-\frac{z}{2}$ and $y=-\frac{z}{2}$ : the kernel is the line spanned by $(-1, -1, 2)$ , that is, by $\vec{d}$ . Vectors parallel to the rays have a zero shadow. $\det\Omega=0$ , since the third row is zero.

e)

No: the orthogonal projection onto the ground has matrix $\text{diag}(1, 1, 0)$ and sends $(0, 0, 4)$ to the origin, a shadow of length zero, that of a Sun exactly overhead. Here the rays are oblique, and $\Omega$ is an OBLIQUE projection: same image, the ground, but a tilted kernel, the direction of the rays.

The thread: a projection is characterized by $P^{2}=P$ ; it is determined by its image and by its kernel, the direction along which it projects.

Exercise 15: Problem: recognizing a transformation matrix

Consider the matrices $M_{1}=\begin{pmatrix} 0.6 & -0.8 \\ 0.8 & 0.6 \end{pmatrix}$ , $M_{2}=\begin{pmatrix} 0.8 & 0.6 \\ 0.6 & -0.8 \end{pmatrix}$ , $M_{3}=\begin{pmatrix} 2 & -2 \\ 2 & 2 \end{pmatrix}$ and $M_{4}=\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$ . Recall that a matrix $M$ is orthogonal if $M^{T}M=I$ .

a) Among $M_{1}$ , $M_{2}$ and $M_{4}$ , which are orthogonal?
b) Show that $M_{1}$ is a rotation and give its angle, to the hundredth of a degree.
c) Compute $\det M_{2}$ . Show that $M_{2}$ is a reflection and give the angle between its axis and the $x$ -axis, to the hundredth of a degree.
d) Write $M_{3}$ as a multiple of a rotation matrix. Give the factor, to two decimal places, and the angle.
e) Compute $\det M_{4}$ . Which transformation is it?

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Answers

a) $M_1$ and $M_2$ are orthogonal, $M_4$ is not
b) Rotation through $53.13^\circ$
c) $\det=-1$ : a reflection, axis at $18.43^\circ$
d) Factor $2\sqrt2\approx 2.83$ , angle $45^\circ$
e) $\det M_4=1$ : a shear

a)

$M_{1}^{T}M_{1}$ : diagonal $0.36+0.64=1$ , off-diagonal $-0.48+0.48=0$ , so $M_{1}$ is orthogonal. Likewise $M_{2}^{T}M_{2}=I$ . For $M_{4}$ , the second column $(3, 1)$ has norm $\sqrt{10}\neq 1$ : $M_{4}$ is not orthogonal.

b)

$M_{1}$ is orthogonal with determinant $0.36+0.64=1$ : it is a rotation, of the form $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ with $\cos\theta=0.6$ and $\sin\theta=0.8$ , so $\theta\approx 53.13^{\circ}$ .

c)

$\det M_{2}=-0.64-0.36=-1$ : orthogonal with determinant $-1$ , it is a reflection, of the form $\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}$ . $\cos 2\theta=0.8$ and $\sin 2\theta=0.6$ give $2\theta\approx 36.87^{\circ}$ , and the axis makes an angle $\theta\approx 18.43^{\circ}$ with the $x$ -axis.

d)

$M_{3}=2\sqrt{2}\begin{pmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix}$ : the rotation through $45^{\circ}$ followed by the dilation with factor $2\sqrt{2}\approx 2.83$ . Check: $\det M_{3}=8=(2\sqrt{2})^{2}$ .

e)

$\det M_{4}=1$ : areas are preserved, but the matrix is not orthogonal, so lengths are not. It is the horizontal shear that slides each point by $3$ times its height.

The thread: orthogonal with determinant $1$ is a rotation; with determinant $-1$ , a reflection; a multiple of a rotation is a rotation-dilation; a determinant of $1$ without orthogonality preserves areas only.

Corrected exercises: linear transformations of the plane (201-SN4-RE)

Course recap

Calculations: images to compute

Exercise 0.1: Computing the image of a point under a transformation: ten direct calculations

Part A: the basics (/50)

Exercise 1: Matrices of the standard geometric transformations

Exercise 2: Rotation through 60 degrees and powers of a matrix

Exercise 3: Composition of transformations and non-commutativity

Exercise 4: Recovering the matrix, image of a figure and area factor

Exercise 5: The shear: a transformation that preserves area without being a rotation

Part B: exam level (/50)

Exercise 6: Proofs: linearity, orthogonal matrices and projections

Exercise 7: Problem: a rotation-dilation, from the unit square to the inverse

Exercise 8: Reflections about a line and the composition of two reflections

Exercise 9: Recovering the matrix from its action on two arbitrary vectors

Exercise 10: Proofs: composition, image of a line and midpoints preserved

Part C: the classics (/50)

Exercise 11: Three transformations of space

Exercise 12: The image of a circle: an ellipse

Exercise 13: Undoing a composite transformation

Exercise 14: Problem: a cast shadow, an oblique projection

Exercise 15: Problem: recognizing a transformation matrix

See also

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