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Describing geometric transformations algebraically

We have now seen several geometric plane transformations. We have also seen how matrices can describe some transformations (linear ones) in a simple algebraic way.

We now investigate these geometric operations from an algebraic point of view. We ask whether, and how, they can be represented by matrices. This gives us a very interesting way to relate everyday geometric operations to algebra.

Rotations

As discussed previously, given a point $P$ in the plane, and an angle $\phi$, we can consider the rotation around $P$ by angle $\phi$. By convention, we measure angles anticlockwise and in radians. We denote³ this rotation by

\[ \text{Rot}_{P, \; \phi} \; : \; \mathbb{R}^2 \rightarrow \mathbb{R}^2. \]

Note the following preliminary facts about rotations:

1. A rotation by angle $0$ is just the identity transformation, as is rotation by $2\pi$.
2. Rotation by $\phi + 2\pi$ is the same transformation as rotation by $\phi$.
3. Rotations of angles $\theta$ and $\phi$ about the same point $P$ give a rotation of $\theta + \phi$ about $P$.
4. Any rotation is bijective. Indeed, a rotation by $\phi$ about $P$ can be undone by rotating by $-\phi$, so the transformation $\text{Rot}_{P, \; \phi}$ has inverse $\text{Rot}_{P, \; -\phi}$.

That is,

\[ \text{Rot}_{P, \; 0} = \text{Rot}_{P, \; 2\pi} = I, \quad \text{Rot}_{P, \; \phi + 2\pi} = \text{Rot}_{P, \; \phi}, \] \[ \text{Rot}_{P, \; \phi} \circ \text{Rot}_{P, \; \theta} = \text{Rot}_{P, \; \theta + \phi}, \quad \left( \text{Rot}_{P, \; \phi} \right)^{-1} = \text{Rot}_{P, \; -\phi}. \]

Now, when we rotate around the origin ${\bf 0}$, it turns out that $\text{Rot}_{{\bf 0}, \; \phi}$ is a linear transformation. We will show why this is true, and find the corresponding matrix.

To show that $\text{Rot}_{{\bf 0}, \; \phi}$ is linear, we will show that it satisfies the "distributive" laws

\[ \text{Rot}_{{\bf 0}, \phi} (c {\bf v}) = c \text{Rot}_{{\bf 0}, \phi} ({\bf v}) \quad \text{and} \quad \text{Rot}_{{\bf 0}, \phi} \left( {\bf v} + {\bf w} \right) = \text{Rot}_{{\bf 0}, \phi} \left( {\bf v} \right) + \text{Rot}_{{\bf 0}, \phi} \left( {\bf w} \right), \]

for any vectors ${\bf v}, {\bf w}$ and real number $c$.

For the first law, note that the two vectors $c{\bf v}$ and ${\bf v}$ point in the same direction; $c{\bf v}$ is $c$ times longer than ${\bf v}$. After rotation by $\phi$ about ${\bf 0}$, the two resulting vectors still point in the same direction and one is still $c$ times longer than the other, so

\[ \text{Rot}_{{\bf 0}, \phi} \left( c {\bf v} \right) = c \text{Rot}_{{\bf 0},\phi} ( {\bf v} ) \quad \text{as desired}. \]

For the second law, place ${\bf v}$ at the origin and place ${\bf w}$ head-to-tail with it, as shown, to obtain the vector ${\bf v} + {\bf w}$. After rotating these vectors all by $\phi$, we have head-to-tail vectors $\text{Rot}_{{\bf 0}, \; \phi} ({\bf v})$ and $\text{Rot}_{{\bf 0}, \; \phi} ({\bf w})$, summing to $\text{Rot}_{{\bf 0}, \; \phi} ({\bf v} + {\bf w})$. Thus

\[ \text{Rot}_{{\bf 0}, \; \phi} ({\bf v}) + \text{Rot}_{{\bf 0}, \; \phi} ({\bf w}) = \text{Rot}_{{\bf 0}, \; \phi} ( {\bf v} + {\bf w} ) \quad \text{as desired}. \]

Now that we know $\text{Rot}_{{\bf 0}, \; \phi}$ is a linear transformation, we know that it corresponds to a matrix: the next question is to find the matrix.

To find the matrix, consider where the standard basis vectors $(1,0)$ and $(0,1)$ go under the rotation: this will give us the two columns of the matrix, as discussed earlier.

Rotating $(1,0)$ by an angle of $\phi$ about the origin, we obtain the point $(\cos \phi, \sin \phi)$, by elementary trigonometry in the right-angle triangle shown.

Thus we have found the columns of the matrix of $\text{Rot}_{{\bf 0}, \; \phi}$. The matrix is

\[ \begin{bmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{bmatrix}. \]

Such a matrix is often called a rotation matrix with angle $\phi$.

The formula for a rotation about the origin can also be derived directly. We can do this using polar coordinates. Instead of describing a point by its Cartesian coordinates $(x,y)$, we can use its distance from the origin $r$, and its direction $\theta$ from the origin. The angle $\theta$ is measured anticlockwise from the positive $x$-axis.

Basic trigonometry in the triangle shown gives $x = r \cos \theta$ and $y = r \sin \theta$. When we apply a rotation about the origin of angle $\phi$, the point $(x,y)$ moves an angle of $\phi$ around the origin, but its distance from the origin $r$ is unchanged. Hence the point $(x,y) = (r \cos \theta, r \sin \theta)$ is mapped to $(x',y') = (r \cos (\theta + \phi), r \sin (\theta + \phi) )$. We can now use the sine and cosine addition formulae to rewrite this expression:

\begin{align*} \text{Rot}_{{\bf 0}, \; \phi} \begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix} r \cos (\theta + \phi) \\ r \sin (\theta + \phi) \end{bmatrix} = \begin{bmatrix} r \cos \theta \cos \phi - r \sin \theta \sin \phi \\ r \sin \theta \cos \phi + r \cos \theta \sin \phi \end{bmatrix} \\ &= \begin{bmatrix} (\cos \phi) x - (\sin \phi) y \\ (\sin \phi) x + (\cos \phi ) y \end{bmatrix} = \begin{bmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \end{align*}

This gives us a formula for $\text{Rot}_{{\bf 0}, \; \phi}$ and the rotation matrix directly.

While a rotation about the origin gives a linear transformation, a rotation about any other point is not linear, and hence is not given by a matrix. It is a plane transformation, but not a linear one. You can prove this in the next exercise.

In the section on Translations, you can find a formula for a rotation about any point.

Translations

Picking up an object and moving it is a very familiar concept. As discussed previously, given a vector ${\bf v}$, we can consider the translation of the plane by the vector ${\bf v}$. We denote this transformation by⁵ \[ \text{Trans}_{{\bf v}} : \mathbb{R}^2 \rightarrow \mathbb{R}^2. \]

Let ${\bf v} = (a,b)$. When we translate a point, with position vector ${\bf x} = (x,y)$, by ${\bf v}$, we simply add the two vectors, as shown below. Thus

\[ \text{Trans}_{{\bf v}} ({\bf x}) = {\bf x} + {\bf v}, \quad \text{or equivalently,} \quad \text{Trans}_{{\bf v}} (x,y) = (x+a, y+b). \]

Translation by ${\bf 0}$ does not move points very far! It is the identity: $\text{Trans}_{{\bf 0}} = I$.

When ${\bf v} \neq {\bf 0}$, translation by ${\bf v}$ is not a linear transformation. The constant terms $a,b$ in the formula $(x+a,y+b)$ are not allowed under our definition of linear transformation. Another way to see that $\text{Trans}_{{\bf v}}$ is not linear is to note that $\text{Trans}_{{\bf v}} ({\bf 0}) = {\bf v}$, but a linear transformation must take the origin ${\bf 0}$ to ${\bf 0}$ (exercise 2). (Transformations which allow both linear and constant terms are called affine transformations.)

Projections

As discussed earlier, given a line $L$ on the plane, we can consider the plane transformation which projects points onto $L$. We can denote⁶ this transformation by

\[ \text{Proj}_L \; : \; \mathbb{R}^2 \rightarrow \mathbb{R}^2. \]

Such projections are very familiar from everyday life. If the plane is a map, $L$ is a road, and $P$ is your position, then $\text{Proj}_L (P)$ is the point on the road closest to you. If the line $L$ is horizontal, representing the ground, $P$ is above ground, and the sun is directly above, then $\text{Proj}_L (P)$ is the shadow of $P$ on the ground.

We have some preliminary facts about projections:

1. A projection $\text{Proj}_L$ is not surjective. Its image consists only of points on $L$.
2. A projection $\text{Proj}_L$ is not injective. Many points project to any given point of $L$.
3. If $P$ is a point on $L$, then $\text{Proj}_L (P) = P$.
4. After projecting a point onto $L$, projecting onto $L$ again has no further effect. That is, $\text{Proj}_L \circ \text{Proj}_L = \text{Proj}_L$.

A function $f$ such that $f \circ f = f$ is called an idempotent. The last property above says a projection is an idempotent.

As it turns out, when the line $L$ passes through the origin, $\text{Proj}_L$ is a linear transformation. We will show why, and find the corresponding matrix.

First of all, suppose $L$ is the $x$-axis. Projection onto $L$ in this case just sets the $y$-coordinate of a point to $0$, and leaves the $x$-coordinate unchanged: $\text{Proj}_{\text{$x$-axis}} (x,y) = (x,0)$. Similarly $\text{Proj}_{\text{$y$-axis}} (x,y) = (0,y)$. These are linear transformations: \[ \text{Proj}_{\text{$x$-axis}} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}, \quad \text{Proj}_{\text{$y$-axis}} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ y \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \]

Now suppose that $L$ is any line through the origin, making an angle of $\phi$ with the positive $x$-axis (as always, measured anticlockwise in radians), and let ${\bf v} = (x,y)$.

The key idea to calculate $\text{Proj}_L ({\bf v})$ is to perform the projection in several steps. We first rotate the plane by $-\phi$ about the origin. This rotates $L$ to the $x$-axis, and rotates ${\bf v}$ too. Then, we project onto the $x$-axis. Finally, we rotate back by $\phi$ around the origin. The result is that ${\bf v}$ has been projected onto $L$. In symbols,

\[ \text{Proj}_L = \text{Rot}_{{\bf 0}, \phi} \circ \text{Proj}_{\text{$x$-axis}} \circ \text{Rot}_{{\bf 0}, -\phi}. \]

Why did we do this? Because we know all the matrices for the transformations on the right hand side! The matrix for $\text{Proj}_{\text{$x$-axis}}$ is given above, and

\[ \text{Rot}_{{\bf 0}, \phi} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}, \quad \text{Rot}_{{\bf 0}, -\phi} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi \\ - \sin \phi & \cos \phi \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \]

Hence $\text{Proj}_L$ is linear, and its matrix is given by multiplying these matrices together:

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

So projection onto $L$ is a linear transformation corresponding to this matrix:

\[ \text{Proj}_L \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \sin \phi \cos \phi & \sin^2 \phi \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} %= %\begin{bmatrix} x \cos^2 \phi + y \sin \phi \cos \phi \\ x \sin \phi \cos \phi + y \sin^2 \phi \end{bmatrix}. \]

When $\phi = 0$, this matrix agrees with our calculation for projection onto the $x$-axis.

An alternative derivation of the projection matrix is given in the following exercise.

It's now possible to see some facts about projections from the algebraic properties of a projection matrix.

While projection onto a line through the origin is linear, projection onto a line not through the origin is not linear, and hence not given by a matrix, as you can prove now.

Reflections

As discussed earlier, given a line $L$ in the plane, we can consider the plane transformation which reflects points in $L$.

We denote⁷ this transformation by

\[ \text{Ref}_L : \mathbb{R}^2 \rightarrow \mathbb{R}^2. \]

Here are some elementary facts about reflections:

1. Reflecting a point $P$ in $L$, and reflecting it in $L$ again, returns you to $P$. Hence,
\[ \text{Ref}_L \circ \text{Ref}_L = I. \]
2. A reflection is bijective; in fact, the inverse of a reflection is itself:
\[ \left( \text{Ref}_L \right)^{-1} = \text{Ref}_L. \]
3. In $P$ is a point on $L$, then its reflection in $L$ is just $P$ again; $\text{Ref}_L (P) = P$.

A function $f$ such that $f \circ f$ is the identity —- equivalently, a function which is its own inverse —- is called an involution. So reflections are involutions.

When $L$ is the $x$-axis, reflection in $L$ fixes the $x$-coordinate and "flips" the $y$-coordinate. Similarly, reflection in the $y$-axis "flips" the $x$-coordinate:

\[ \text{Ref}_{\text{$x$-axis}} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ -y \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}, \quad \text{Ref}_{\text{$y$-axis}} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -x \\ y \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \]

So reflection in either axis is a linear transformation.

In fact, for any line $L$ passing through the origin, $\text{Ref}_L$ is a linear transformation. Letting $\phi$ be the angle between $L$ and the positive $x$-axis, the corresponding matrix is given by

\[ \text{Ref}_L \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos (2\phi) & \sin (2\phi) \\ \sin (2\phi) & -\cos (2\phi) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \]

In the following two exercises, we outline two methods to derive this result, similar to those seen previously for projections.

Reflection in one particular line $y=x$ has a useful effect: it swaps coordinates.

The following exercise gives an interesting relationship between reflections and projections, and leads to a third way to derive the reflection matrix.

(The above exercise holds for any line $L$ —- it need not pass through the origin!)

In the next exercise, we observe properties of reflections in the algebra of reflection matrices.

We have seen that reflection in a line passing through the origin is linear. But, as with projections, if $L$ does not pass through the origin, then $\text{Ref}_L$ is not linear.

Dilations

The plane transformation of a dilation has a similar effect to Alice's "DRINK ME" potion or "EAT ME" cake: it stretches or shrinks objects in the plane.

As discussed previously, given a line $L$ and a real number $k$, we can consider the dilation from $L$ by factor $k$. We write⁸ this transformation as

\[ \text{Dil}_{L,k} : \mathbb{R}^2 \rightarrow \mathbb{R}^2. \]

The effect of such a dilation is to stretch lengths by a factor of $k$ in the direction perpendicular to $L$. If $k$ is negative, then points are flipped to the other side of $L$.

Here are some basic properties of dilations:

1. Dilating from $L$ with factor $m$, and then $k$, gives a dilation from $L$ with factor $km$.
\[ \text{Dil}_{L,k} \circ \text{Dil}_{L,m} = \text{Dil}_{L,km} \]
2. If $k \neq 0$, then dilation from $L$ with factor $k$ is bijective. Its inverse is given by dilation from $L$ with factor $1/k$.
\[ \left( \text{Dil}_{L,k} \right)^{-1} = \text{Dil}_{L, \frac{1}{k}} \]
3. If $P$ is a point on $L$, then for any $k$, $\text{Dil}_{L,k} (P) = P$.

These facts are true even for dilation by negative factors.

We will analyse dilations similarly to projections and reflections. First, dilation from the $x$-axis with factor $k$ leaves the $x$-coordinate unchanged, and multiplies the $y$-coordinate by $k$: $\text{Dil}_{\text{$x$-axis},k} (x,y) = (x,ky)$. Similarly for the $y$-axis, $\text{Dil}_{\text{$y$-axis},k} (x,y) = (kx,y)$. Hence

\[ \text{Dil}_{\text{$x$-axis},k} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ ky \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & k \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}, \quad \text{Dil}_{\text{$y$-axis},k} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} kx \\ y \end{bmatrix} = \begin{bmatrix} k & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \]

Thus a dilation from either axis is a linear transformation.

In fact, for any line $L$ passing through the origin and any real $k$, $\text{Dil}_{L,k}$ is a linear transformation. Letting $\phi$ denote the angle $L$ makes with the positive $x$-axis, we have

\[ \text{Dil}_{L,k} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 + (k-1) \sin^2 \phi & (1-k) \sin \phi \cos \phi \\ (1-k) \sin \phi \cos \phi & 1 + (k-1) \cos^2 \phi \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \]

You can check that $\phi = 0$ or $\pi/2$ gives the matrices above for dilations in the axes.

Dilation matrices are more complicated than rotation, reflection or projection matrices. Nonetheless, geometric properties of dilations are reflected in the algebra of dilation matrices.

Now for some particular values of $k$, dilation from a line $L$ by factor $k$ reduces to some transformations we have seen previously.

1. Dilation from $L$ with factor $k=-1$ flips every point to the corresponding point on the opposite side of $L$: it is reflection in $L$.
2. Dilation from $L$ with factor $k=0$ is just projection onto $L$.
"Stretching from $L$ by factor zero" simply projects points onto $L$.
3. Dilation from $L$ with factor $k=1$ leaves every point where it is.

In this sense, dilations form a generalisation of reflections and projections. We have

\[ \text{Dil}_{L,0} = \text{Proj}_L, \quad \text{Dil}_{L,-1} = \text{Ref}_L, \quad \text{Dil}_{L,1} = I. \]

Spiral symmetries

A spiral symmetry is another interesting type of plane transformation. The spiral symmetry about a point $P$ with factor $r$ and angle $\theta$ is given by by dilating the whole plane by a factor $r$ from $P$, followed by a rotation by $\theta$ around $P$.

A spiral symmetry about the origin thus dilates a point $(x,y)$ to $(rx,ry)$ and then rotates it by $\theta$, to give the result

\[ \begin{bmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} rx \\ ry \end{bmatrix} = \begin{bmatrix} rx \cos \theta - ry \sin \theta \\ rx \sin \theta + ry \cos \theta \end{bmatrix} = \begin{bmatrix} r\cos \theta & - r \sin \theta \\ r \sin \theta & r \cos \theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \]

One reason spiral symmetries are interesting is because they arise naturally with complex numbers. Consider a complex number $w $ in polar form $w = r \; \text{cis} \theta = r (\cos \theta + i \sin \theta)$, where $r = |w|$ and $\theta = \arg w$. As the complex numbers form the Argand plane $\mathbb{C} \cong \mathbb{R}^2$, we may regard multiplication by $w$ as a plane transformation $\text{Mult}_w : \mathbb{R}^2 \rightarrow \mathbb{R}^2$, given by $\text{Mult}_w (z) = wz$. That is, $\text{Mult}_w$ multiplies complex numbers by $w$. If $z = x+yi$ then

\[ wz = r (\cos \theta + i \sin \theta) (x+yi) = \left( rx \cos \theta - ry \sin \theta \right) + i \left( rx \sin \theta + ry \cos \theta \right). \]

If we write out the real and imaginary parts as components of a vector, this means

\[ \text{Mult}_w \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} rx \cos \theta - ry \sin \theta \\ rx \sin \theta + ry \cos \theta \end{bmatrix} = \begin{bmatrix} r \cos \theta & - r \sin \theta \\ r \sin \theta & r \cos \theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \]

This is exactly a spiral symmetry! Hence, the effect of multiplication by a complex number $w$ on the complex plane is a spiral symmetry about the origin $0$ with factor $|w|$ and angle $\arg \theta$.

³Please note: This is not a standard notation. We find it useful, but it may not be used by others!
⁴Not to be confused with $\mathbb{R} \rightarrow \mathbb{R}$!
⁵Again, this is not a standard notation
⁶Again, beware that this notation is not standard.
⁷Again, notation not standard.
⁸Again, not a standard notation.

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