10.06 Matrices as transformations

Matrices as transformations

Multiplying coordinates on the plane of the form $\left(x,y\right)$(x,y) as a matrix 

by special $2\times2$2×2 matrices, can result in specific transformations on the plane.

Before we go too much further, have a play with this applet and see if you can identify special transformational matrices that perform any of these actions

• reflection
• rotation

Or specific reflections like

• reflect a point/figure across the $x$x-axis
• reflect a point/figure across the $y$y-axis
• reflect a point/figure across the origin
• reflect a point/figure on the line $y=x$y=x
• reflect a point/figure on the line $y=-x$y=−x

Or specific rotations

• rotate a point/figure clockwise $90^\circ$90°  or counter-clockwise $270^\circ$270°
• rotate a point/figure clockwise $270^\circ$270° or counter-clockwise $90^\circ$90°

 

 

The construction of these transformational matrices are quite specific, they contain $0$0's, $1$1's or $-1$−1's and you can quickly become familiar with how the placement of them affects the transformation that occurs. 

By exploring the way the matrix multiplication works, we can see that for transformation matrix 

and the point $A\left(x,y\right)$A(x,y) we can see that 

There are a number of reflections that may interest us.  

Reflections across an axis

Were you able to identify the matrices that effect a transformation equivalent to the reflection across the $x$x, or $y$y axis?  

This applet will show you what they are.

 

Reflections across the origin

Were you able to identify the matrices that effect a transformation equivalent to the reflection through the origin?   

So we know we want to change the point $A\left(x,y\right)$A(x,y) to its reflection $A'\left(-x,-y\right)$A′(−x,−y)

This applet demonstrates this reflection.

 

 

Reflection on lines y = x or y = -x

Were you able to identify the matrices that effect a transformation equivalent to the reflection across the lines y = x or $y=-x$y=−x?

This means we want to change the point $A\left(x,y\right)$A(x,y) to its reflection $A'\left(y,x\right)$A′(y,x)

This applet demonstrates this reflection.

 

Rotation $90^\circ$90° clockwise or counter-clockwise

A $90^\circ$90° clockwise rotation will have the same result as a $270^\circ$270° counterclockwise rotation. Similarly a $90^\circ$90° counterclockwise rotation will have the same results as a $270^\circ$270° clockwise.   You just need to be able to think in both directions and realize they are the same.  Sometimes you will be asked questions that use specific language, it is a common mistake of students to always assume a clockwise rotation. 

Did you find a rotation matrix in your investigation at the beginning? 

This applet will help us see the rotations we are most interested in. $90^\circ$90° , $180^\circ$180°, $270^\circ$270° counterclockwise (but remember they could also be described using clockwise rotations). 

 

 

Practice questions

Question 1

Question 2

Matrices as translations

To translate a point or set of points, we add a fixed value to the $x$x-coordinates and another fixed value to the $y$y-coordinates of each point. This slides the points to a new position without changing the scale or rotating or reflecting them.

If $(x,y)$(x,y) is a point in the plane, a particular translation moves the point to $(x+a,y+b)$(x+a,y+b) for some values of $a$a and $b$b. To perform this translation, we have added $(a,b)$(a,b) to the point $(x,y)$(x,y) in the plane. We can express this using column vectors.

$x$x

$+$+

$a$a

$=$=

$x+a$x+a

$y$y

$b$b

$y+b$y+b

 

Graphically speaking, the point $\left(x,y\right)$(x,y) moves in the direction of the translation vector. More specifically, the point $\left(x,y\right)$(x,y) translates horizontally by $a$a units and vertically by $b$b units.

 

Worked example

Question 3

The four points $(0,0)$(0,0), $(2,1)$(2,1), $(1,2)$(1,2) and $(-1,1)$(−1,1) are the vertices of a parallelogram in the plane.

Where will the vertices be after being translated by the vector		$2$2		?
		$3$3

Think: We want to add the translation vector to each point, represented as a column vector.

Do:

$0$0     $+$+     $2$2     $=$=     $2$2         $0$0       $3$3       $3$3

$2$2     $+$+     $2$2     $=$=     $4$4         $1$1       $3$3       $4$4

$1$1     $+$+     $2$2     $=$=     $3$3         $2$2       $3$3       $5$5

$-1$−1     $+$+     $2$2     $=$=     $1$1         $1$1       $3$3       $4$4

As coordinates, the new vertices are $\left(2,3\right)$(2,3), $\left(4,4\right)$(4,4), $\left(3,5\right)$(3,5) and $\left(1,4\right)$(1,4).

Reflect: Since the components of the translation vector are both positive, each point is translated to the right and then up. How would this look if the components of the translation vector were negative?

 

Practice questions

question 4

question 5

question 6

Constructing a transformation matrix

Any vector in the plane can be expressed as a linear combination of the vectors $\mathbf{i}=(1,0)$i=(1,0) and $\mathbf{j}=(0,1)$j=(0,1). These are called standard basis vectors. For example, the vector $(2,7)$(2,7) can be expressed as $2\mathbf{i}+7\mathbf{j}$2i+7j. If this vector were to be multiplied by the matrix $\mathbf{M}$M, we would have, in accord with the linearity properties, 

$\mathbf{M}\left(2\mathbf{i}+7\mathbf{j}\right)=2\mathbf{Mi}+7\mathbf{Mj}$M(2i+7j)=2Mi+7Mj.

You should check that $\mathbf{Mi}$Mi is just the first column of matrix $\mathbf{M}$M and $\mathbf{Mj}$Mj is the second column. Conversely, in designing a transformation matrix we need only consider the desired effects on vectors $\mathbf{i}$i and $\mathbf{j}$j and these will form the columns of the transformation matrix.

 

Rotation

We wish to rotate all the points in the plane counterclockwise through $60^\circ$60° by a matrix multiplication.

The point $\mathbf{i}=(1,0)$i=(1,0) moves to $(\cos60^\circ,\sin60^\circ)$(cos60°,sin60°) and the point $\mathbf{j}=(0,1)$j=(0,1) moves to $(-\sin60^\circ,\cos60^\circ)$(−sin60°,cos60°). These form the columns of the transformation matrix. So, 

 

Dilation

Consider the points $(x,y)$(x,y) in the plane such that $y=x^2$y=x2. The set of points forms a parabola.

Suppose we dilate the parabola by a factor of $3$3 in the $y$y-direction and $2$2 in the $x$x-direction. The dilation moves the basis vector $\mathbf{i}$i to $(2,0)$(2,0) and the basis vector $\mathbf{j}$j is moved to $(0,3)$(0,3).

So, the transformation matrix is  .

We calculate .

If we now put $x=\frac{u}{2}$x=u2​ and therefore, $y=\frac{u^2}{4}$y=u24​, we have pairs of points $(u,\frac{3}{4}u^2)$(u,34​u2) and these again form a parabola shape. This suggests that under a dilation by matrix multiplication a parabola remains a parabola.

 

Reflection

We wish to reflect the points on the plane through the origin so that each point is twice as far from the origin as it was originally but in the opposite direction. What is the new location of the point $(2,-1)$(2,−1)?

The point $(1,0)$(1,0) is transformed to $(-2,0)$(−2,0) and the point $(0,1)$(0,1) goes to $(0,-2)$(0,−2). These give us the columns of the transformation matrix.

 

Observe that multiplication of the zero vector $(0,0)$(0,0) by any transformation matrix leaves the origin point unchanged.

 

Composition of linear transformations

Here we will explore whether a composition of linear transformations is again a linear transformation and, if so, how it might be represented by a matrix multiplication.

The composition of two linear transformations means we perform one of them followed by the other. If $\mathbf{T}_1$T1​ and $\mathbf{T}_2$T2​ are linear transformations, then $\mathbf{T}_2\circ\mathbf{T}_1\left(\mathbf{v}\right)$T2​∘T1​(v) will mean perform $\mathbf{T}_1$T1​ on vector $\mathbf{v}$v followed by $\mathbf{T}_2$T2​ on the result.

To check whether the composition $\mathbf{T}_2\circ\mathbf{T}_1$T2​∘T1​ is a linear transformation, we need to check whether 

$\mathbf{T}_2\circ\mathbf{T}_1\left(\mathbf{u+v}\right)=\mathbf{T}_2\circ\mathbf{T}_1\left(\mathbf{u}\right)+\mathbf{T}_2\circ\mathbf{T}_1\left(\mathbf{v}\right)$T2​∘T1​(u+v)=T2​∘T1​(u)+T2​∘T1​(v), and whether

$k\left(\mathbf{T}_2\circ\mathbf{T}_1\right)\left(\mathbf{v}\right)=\mathbf{T}_2\circ\mathbf{T}_1\left(k\mathbf{v}\right)$k(T2​∘T1​)(v)=T2​∘T1​(kv), where $k$k is a scalar.

1. For the first criterion, we know that 

$\mathbf{T}_1\left(\mathbf{u+v}\right)=\mathbf{T}_1\left(\mathbf{u}\right)+\mathbf{T}_1\left(\mathbf{v}\right)$T1​(u+v)=T1​(u)+T1​(v) because $\mathbf{T}_1$T1​ is a linear transformation.

Then, 

$\mathbf{T}_2\circ\mathbf{T}_1\left(\mathbf{u+v}\right)=\mathbf{T}_2\left(\mathbf{T}_1\left(\mathbf{u}\right)+\mathbf{T}_1\left(\mathbf{v}\right)\right)=\mathbf{T}_2\circ\mathbf{T}_1\left(\mathbf{u}\right)+\mathbf{T}_2\circ\mathbf{T}_1\left(\mathbf{v}\right)$T2​∘T1​(u+v)=T2​(T1​(u)+T1​(v))=T2​∘T1​(u)+T2​∘T1​(v)

because $\mathbf{T}_2$T2​ is a linear transformation, and this is what we had to show.

2. For the second criterion, we have

$\mathbf{T}_1\left(k\mathbf{v}\right)=k\mathbf{T}_1\left(\mathbf{v}\right)$T1​(kv)=kT1​(v)

because $\mathbf{T}_1$T1​ is a linear transformation. Therefore,

$\mathbf{T}_2\circ\mathbf{T}_1\left(k\mathbf{v}\right)=\mathbf{T}_2\left(k\mathbf{T}_1\left(\mathbf{v}\right)\right)=k\mathbf{T}_2\left(\mathbf{T}_1\left(\mathbf{v}\right)\right)$T2​∘T1​(kv)=T2​(kT1​(v))=kT2​(T1​(v))because $\mathbf{T}_2$T2​ is a linear transformation, and this is the same as $k\left(\mathbf{T}_2\circ\mathbf{T}_1\right)\left(\mathbf{v}\right)$k(T2​∘T1​)(v), as required.

We have shown that the composition of two linear transformations is also a linear transformation. It follows that there is a matrix representation of the composed transformations.

We apply multiplication by the matrix $\mathbf{M}_1$M1​ to carry out the transformation $\mathbf{T}_1$T1​ followed by multiplication of the result by $\mathbf{M}_2$M2​ to carry out $\mathbf{T}_2$T2​. 

Matrix multiplication is associative so that instead of $\mathbf{M}_2\left(\mathbf{M}_1\mathbf{v}\right)$M2​(M1​v) we can multiply the matrices first as in $\mathbf{M}_2\mathbf{M}_1\left(\mathbf{v}\right)$M2​M1​(v).

We have been careful about keeping the correct order in which the matrices are multiplied because there are cases where carrying out the transformations in the reversed order will give a different result. 

Worked examples

Question 7

An counterclockwise rotation through $90^\circ$90° is achieved by multiplication by the matrix $\mathbf{R}$R and a reflection in the horizontal axis by matrix $\mathbf{A}$A given below.

Compare the effects on a point $\mathbf{X}$X of multiplication by $\mathbf{RA}$RA and by $\mathbf{AR}$AR.

Question 8

Explore what transformation matrices can be obtained by repeated application of matrices $\mathbf{R}$R and $\mathbf{A}$A from Example 1.

Two applications of $\mathbf{R}$R can be written $\mathbf{R}^2$R2 and we might label this composition $\mathbf{R}_2$R2​.

Applying $\mathbf{R}$R a third time gives

A fourth application of $\mathbf{R}$R results in the identity matrix and further applications will repeat this cycle. This is as expected since four rotations of $90^\circ$90° brings every point back to its starting position.

Similarly, two reflections in the horizontal axis is equivalent to doing nothing. So, $\mathbf{A}^2=\mathbf{I}$A2=I.

Next, we can try $\mathbf{A}\mathbf{RA}$ARA. This gives

And, $\mathbf{R}\mathbf{AR}$RAR.

Finally, we can form 

If we label $\mathbf{AR}=\mathbf{B}_1$AR=B1​ and $\mathbf{RA}=\mathbf{B}_2$RA=B2​, we can form a table of compositions as follows.

$\circ$∘	$\mathbf{I}$I	$\mathbf{A}$A	$\mathbf{R}$R	$\mathbf{R}_2$R2​	$\mathbf{R}_3$R3​	$\mathbf{B}_1$B1​	$\mathbf{B}_2$B2​	$\mathbf{C}$C
$\mathbf{I}$I	$I$I	$A$A	$R$R	$R_2$R2​	$R_3$R3​	$B_1$B1​	$B_2$B2​	$C$C
$\mathbf{A}$A	$A$A	$I$I	$B_1$B1​	$C$C	$B_2$B2​	$R$R	$R_3$R3​	$R_2$R2​
$\mathbf{R}$R	$R$R	$B_2$B2​	$R_2$R2​	$R_3$R3​	$I$I	$A$A	$C$C	$B_1$B1​
$\mathbf{R}_2$R2​	$R_2$R2​	$C$C	$R_3$R3​	$I$I	$R$R	$B_2$B2​	$B_1$B1​	$A$A
$\mathbf{R}_3$R3​	$R_3$R3​	$B_1$B1​	$I$I	$R$R	$R_2$R2​	$C$C	$A$A	$B_2$B2​
$\mathbf{B}_1$B1​	$B_1$B1​	$R_3$R3​	$C$C	$B_2$B2​	$A$A	$I$I	$R_2$R2​	$R$R
$\mathbf{B}_2$B2​	$B_2$B2​	$R$R	$A$A	$B_1$B1​	$C$C	$R_2$R2​	$I$I	$R_3$R3​
$\mathbf{C}$C	$C$C	$R_2$R2​	$B_2$B2​	$A$A	$B_1$B1​	$R_3$R3​	$R$R	$I$I

You should check that no further transformations can be generated from $\mathbf{A}$A and $\mathbf{R}$R. Other possible compositions lead to the matrices that are already in the table. These eight transformations with the composition operation form a mathematically important structure known as a group.

If other linear transformations are chosen to begin this process of generating further transformations by composition, other group structures result that have different numbers of elements.

Inverse of linear transformations

The inverse of a linear transformation, if it exists, is the transformation that reverses or undoes the effect of the linear transformation. So,

$\mathbf{T}^{-1}\left(\mathbf{T}(\mathbf{x})\right)=\mathbf{x}$T−1(T(x))=x

The vector $\mathbf{x}$x is moved by the transformation $\mathbf{T}$T and then returned to its original value by the inverse transformation $\mathbf{T}^{-1}$T−1.

Linear transformations are carried out by matrix multiplications. For each linear transformation there is a corresponding matrix, and there is a linear transformation for each matrix that can multiply a vector $\mathbf{v}$v.

The inverse of a linear transformation is carried out by multiplication by the inverse of the transformation matrix, if it exists.

A matrix fails to have an inverse when a row can be expressed as a linear combination of the other rows or when the matrix is not square. A linear transformation fails to have an inverse when the transformation maps many points of the space to the same transformed point. An example of such a transformation is one that maps all the points in to a single line. 

In this case, it is impossible to determine how to reverse the transformation.

 

However, linear transformations that are dilations, reflections or rotations map every point of the space to a unique target point, and every point in the space is the image of some point in the space from which it was transformed. In the terminology of functions, we say that such a transformation is one-to-one and onto. These transformations are invertible. 

Worked examples

Question 9

A transformation and its inverse are given by matrices 

 

It is easy to check that these matrices are inverses of one another by performing the multiplication $\mathbf{M}\cdot\mathbf{M}^{-1}$M·M−1, which gives the identity matrix.

Calculate $\mathbf{M}^{-1}\left(\mathbf{M}\mathbf{X}\right)$M−1(MX) where $\mathbf{X}$X is an arbitrary point $(x,y)$(x,y) in the plane. Thus, confirm that the transformation and its inverse behave as expected.

Multiplication of the transformed point by the inverse matrix restores the point to its original position.

To find the inverse of a $2\times2$2×2 matrix, we swap the elements on the main diagonal, reverse the signs of the elements on the other diagonal, and multiply the resulting matrix by the reciprocal of its determinant.

Question 10

A linear transformation is given by the matrix  .

The transformation moves a point $(x,y)$(x,y) to a new point $(\mathbf{X,Y})$(X,Y). We can show that whatever the values of $x$x and $y$y, the new point $(\mathbf{X,Y})$(X,Y) under this transformation always lies on a particular line through the origin.

We see that $\mathbf{Y}=-2\mathbf{X}$Y=−2X for all possible values of $x$x and $y$y. We say that the matrix $\mathbf{A}$A has mapped all of  onto the line $\mathbf{Y}=-2\mathbf{X}$Y=−2X, which is regarded as a $1$1-dimensional subspace of the $2$2-dimensional space.

We cannot know which particular $(x,y)$(x,y) a point $(\mathbf{X,Y})$(X,Y) on the line came from. Hence, the transformation can have no inverse.

For example, it is easy to confirm that every point $(x,y)$(x,y) that lies on the line $y=\frac{2}{3}x$y=23​x is mapped by the linear transformation to the point $(0,0)$(0,0).  

Accordingly, the matrix $\mathbf{A}$A is not invertible. You should check that its second row is a multiple of the first.

Determinants and transformations

Every square matrix has a number associated with it, called its determinant. For a matrix $\mathbf{A}$A, we write $\left|\mathbf{A}\right|$|A| or $\text{det}\left(\mathbf{A}\right)$det(A) for the determinant function.

In the case of a $2\times2$2×2 matrix,  .

This generalizes to more than $2$2 dimensions. The determinant of a matrix of dimension $n\times n$n×n is a sum of $n!$n! terms each of which is a signed product of $n$n elements from distinct rows and columns of the matrix. The sign, $+$+ or $-$−, of a product in the sum comes from a consideration of the ordering of the row numbers in the term.

For example, the terms for a $3\times3$3×3 matrix contain elements $r_1,r_2$r1​,r2​ and $r_3$r3​ from rows $1,2$1,2 and $3$3. A particular term in the sum, say $r_2r_3r_1$r2​r3​r1​ from columns $1,2,3$1,2,3 respectively, has $r_1$r1​ disordered with respect to $r_2$r2​ and $r_3$r3​ so its sign is that of is $(-1)^2$(−1)2. That is, '$+$+'.

If there is an odd number of reversals in the ordering of the subscripts, the term is negative. Otherwise, the term is positive. So, with the column numbers kept in the order $1,2,3$1,2,3, we would have

$r_1r_2r_3-r_1r_3r_2-r_2t_1r_3+r_2r_3r_1+r_3r_1r_2-r_3r_2r_1$r1​r2​r3​−r1​r3​r2​−r2​t1​r3​+r2​r3​r1​+r3​r1​r2​−r3​r2​r1​

 

This may seem somewhat arbitrary and not particularly intuitive but, historically, determinants have appeared as part of the theory of linear algebra. They can be used, for example, in solving systems of linear equations by Cramer's Method and we have seen that in the case of a $2\times2$2×2 matrix, the determinant is used in finding the inverse of the matrix.

For larger matrices, row operation methods are the preferred way of finding a matrix inverse although knowledge of the determinant shows whether or not the inverse exists. A matrix with determinant zero has no inverse.

 

Area

Consider two vectors $\mathbf{x}$x and $\mathbf{y}$y in the plane. Depicted as arrows with their tails at the origin, they form a parallelogram.

The area of the parallelogram is $A=|\mathbf{x}||\mathbf{y}|\sin\theta$A=|x||y|sinθ.

We can express the area in terms of the coordinates without referring to the angle $\theta$θ by first writing the dot product of $\mathbf{x}$x and $\mathbf{y}$y. We have, $\mathbf{x\cdot y}=x_1y_1+x_2y_2$x·y=x1​y1​+x2​y2​ but this is also equal to $|\mathbf{x}||\mathbf{y}|\cos\theta$|x||y|cosθ. Hence, 

$\cos\theta=\frac{x_1y_1+x_2y_2}{|\mathbf{x}||\mathbf{y}|}$cosθ=x1​y1​+x2​y2​|x||y|​ and so,

$\sin\theta=\sqrt{1-\frac{\left(x_1y_1+x_2y_2\right)^2}{|\mathbf{x}|^2|\mathbf{y}|^2}}$sinθ=√1−(x1​y1​+x2​y2​)2|x|2|y|2​

When this expression for $\sin\theta$sinθ is substituted into the area formula, a simplification can occur and it eventually turns out that,

$A=x_1y_2-x_2y_1$A=x1​y2​−x2​y1​

which is just the determinant of the matrix $\mathbf{P}$P formed by letting the vectors $\mathbf{x}$x and $\mathbf{y}$y be the columns. Thus, .

Now, suppose a linear transformation, represented by a matrix $\mathbf{M}$M, is applied to the two vectors $\mathbf{x}$x and $\mathbf{y}$y. We might ask how the area of the parallelogram will change.

In the worked example below, we show that $\text{det}\left(\mathbf{MP}\right)=\text{det}\left(\mathbf{M}\right)\text{det}\left(\mathbf{P}\right)$det(MP)=det(M)det(P). 

Therefore, when the columns of $\mathbf{P}$P are transformed by the matrix $\mathbf{M}$M, a new parallelogram is formed whose area is the determinant of $\mathbf{MP}$MP which is just the product of the separate determinants.

This multiplicative property leads us to believe that when any collection of points in the plane forming the vertices of a region is subjected to a linear transformation, the transformed points outline a region whose area is changed from the original area by multiplication by the determinant of the transformation matrix. We can think of any region as being approximated by a large number of small parallelograms each of which will be rescaled by the determinant.

Worked examples

Question 11

Show that:

$\text{det}\left(\mathbf{MP}\right)=\text{det}\left(\mathbf{M}\right)\text{det}\left(\mathbf{P}\right)$det(MP)=det(M)det(P)

Let 

Then, 

Now,

$\text{det}(\mathbf{MP})$det(MP)	$=$=	$(ax_1+bx_2)(cy_1+dy_2)-(ay_1+by_2)(cx_1+dx_2)$(ax1​+bx2​)(cy1​+dy2​)−(ay1​+by2​)(cx1​+dx2​)
	$=$=	$acx_1y_1+adx_1y_2+bcx_2y_1+bdx_2y_2-acx_1y_1-adx_2y_1-bcx_1y_2-bdx_2y_2$acx1​y1​+adx1​y2​+bcx2​y1​+bdx2​y2​−acx1​y1​−adx2​y1​−bcx1​y2​−bdx2​y2​
	$=$=	$adx_1y_2+bcx_2y_1-adx_2y_1-bcx_1y_2$adx1​y2​+bcx2​y1​−adx2​y1​−bcx1​y2​
	$=$=	$ad(x_1y_2-x_2y_1)-bc(x_1y_2-x_2y_1)$ad(x1​y2​−x2​y1​)−bc(x1​y2​−x2​y1​)
	$=$=	$(x_1y_2-x_2y_1)(ad-bc)$(x1​y2​−x2​y1​)(ad−bc)
	$=$=	$\text{det}(\mathbf{P})\text{det}(\mathbf{M})$det(P)det(M)

 

question 12

A circle in the plane with radius $1$1 is subject to a linear transformation in the form of a dilation in the horizontal direction by a factor of $k$k and in the vertical direction by $m$m to make a shape that we claim is an ellipse. What should the area of the ellipse be?

The circle has area $\pi\times1^2=\pi$π×12=π.

The transformation matrix is .

So, the area of the ellipse is $\pi km$πkm, where $k$k and $m$m are the semi-axes of the ellipse.

Establish geometric results using matrix multiplications

We have seen that the composition of two linear transformations can be expressed as the product of the matrices that produce the transformations.

The composition of two linear transformations is always another linear transformation.  It can happen that repeated compositions of a small number of transformations generate a limited number of new transformations. Further compositions of pairs of transformations within the set are then also in the set.

The diagram below illustrates two transformations performed on an initial configuration. First, the plane is reflected horizontally, and then it is reflected in the vertical axis. The final position of the colored regions is equivalent to a single rotation counterclockwise through $180^\circ$180°.

The matrices that bring about these transformations are: $\mathbf{H}$H reflects points across the horizontal axis, $\mathbf{V}$V reflects in the vertical axis, $\mathbf{R}$R is the $180^\circ$180° rotation, and we have shown the identity matrix $\mathbf{I}$I.

In the following table, the matrices in the left column are understood to be composed with the matrices in the top row. The correctness of the entries in the table can be checked by thinking about the colored squares diagram above or by carrying out the matrix multiplications.

$\circ$∘	$\mathbf{I}$I	$\mathbf{H}$H	$\mathbf{V}$V	$\mathbf{R}$R
$\mathbf{I}$I	$I$I	$H$H	$V$V	$R$R
$\mathbf{H}$H	$H$H	$I$I	$R$R	$V$V
$\mathbf{V}$V	$V$V	$R$R	$I$I	$H$H
$\mathbf{R}$R	$R$R	$V$V	$H$H	$I$I

This table of four linear transformations under composition represents a mathematical structure called a group.

The elements of the group are said to be a closed set under the group operation in the sense that no other elements can be formed from them.

1. The group includes the 'do nothing' element, called the identity element $\mathbf{I}$I.
2. Each element has an inverse element that undoes its effect and under composition gives the identity element.
3. If three elements are in a composition, $\left(\mathbf{A}\circ\mathbf{B}\right)\circ\mathbf{C}$(A∘B)∘C is the same as $\mathbf{A}\circ\left(\mathbf{B}\circ\mathbf{C}\right)$A∘(B∘C) and the parentheses can be left out.

It turns out that there are two different group structures that have four elements. The one generated by the reflections and rotations shown above is called the $4$4-group.

Worked examples

Question 13

Consider the $90^\circ$90° counterclockwise rotation of the plane. The corresponding matrix is 

By considering the effect of this linear transformation on points in the plane or by carrying out matrix multiplications, we see that there are again just four elements that can be generated from $\mathbf{A}$A.  

They correspond to rotations through $90^\circ$90°, $180^\circ$180°, $270^\circ$270° and $0^\circ$0°. The matrices are $\mathbf{A}$A, $\mathbf{A}^2$A2, $\mathbf{A}^3$A3 and $\mathbf{I}$I.

The group table for this set of elements under composition is

$\circ$∘	$\mathbf{I}$I	$\mathbf{A}$A	$\mathbf{A}^2$A2	$\mathbf{A}^3$A3
$\mathbf{I}$I	$I$I	$A$A	$A^2$A2	$A^3$A3
$\mathbf{A}$A	$A$A	$A^2$A2	$A^3$A3	$I$I
$\mathbf{A}^2$A2	$A^2$A2	$A^3$A3	$I$I	$A$A
$\mathbf{A}^3$A3	$A^3$A3	$I$I	$A$A	$A^2$A2

You should compare this table with the one for the $4$4-group and check that they are indeed different in structure.

Setting out the matrices explicitly we have

The matrices $\mathbf{H}$H and $\mathbf{V}$V are not in this list. Therefore, we have confirmed that it is not possible to achieve the horizontal and vertical reflection transformations if we are restricted to only performing the $90^\circ$90° rotations. Similarly, $\mathbf{A}$A and $\mathbf{A}^3$A3 are unavailable if we can only do the horizontal and vertical reflections.