Knowing all the different types of transformation, we shall now get to do the algebra of transformations. Let’s begin with a simple example:
Singular transformation in 2 dimensions maps all shapes are transformed into either a point or a line, and a line is transformed into a single point. In other words, the area of the object is destroyed. Consider the matrices below:
The first one maps all shapes to the line y = x. The second matrix maps all points to the x-axis, while the last one maps everything to the origin. You will know that a matrix M is a singular matrix when | M | = 0. There is a way to tell whether a matrix maps to a line or to a point. Consider a singular matrix
If the column vector (a, b) = (c, d), then the matrix maps all shapes to a point. If the column vector (a, b) ≠ (c, d) but (a, b) // (c, d), then the matrix maps all shapes to a line.
AREA SCALE-FACTOR AND THE DETERMINANT
Throughout our discussion on transformations, we haven’t discussed on how the transformation affects the area of an object. We want to know whether a certain transformation makes a certain object enlarged or diminished. It turns out that the determinant of the matrix of transformation tells us information on how the area would be in the end. With the matrix of transformation M, We see that
Area of object × det (M) = Area of image
In the case when | M | = 0, the transformation maps lines or shapes to a point, and the area is destroyed, in which agrees with the part earlier on.
If you might have noticed, this reminds you on the chapter about eigenvalues and eigenvectors, in which this situation, the eigenvalue is one. To find the invariant points for the transformation M, for example
You substitute it into the equation above, then you get
x = x
y = –y
So this tells us that the invariant points of this transformation are any points (x, 0), or simply just the points on the x-axis. Verify yourself to see whether this is true.
An invariant line, maps a line to the same line, but not necessarily mapping all the points to the same points. In our study, all invariant lines must pass through the origin, and even if there were invariant lines that do not pass through the origin, it must be parallel (has the same gradient) to another invariant line which passes through the origin. To find the invariant lines under a certain transformation, we make use of the parametric form of the line, x = t, y = at. We substitute the variable t into x and y and we have
Note that the variable x maps to another variable X, but not to itself. I’ll show you an example:
Dividing both the equations, we get a quadratic equation
m2 + 4m – 5 = 0, m = 1, –5
We have the lines y = x, y = –5x.
You might want to test whether the lines y = x + c or y = –5x + c are invariant too. Substitute it back into the equation,
For m = 1,
x + c = X
5x – 4x – 4c = X + c
We get c = 0, ∴ the lines y = x + c are not invariant.
For m = –5,
-5x + c = X
25x = –5X + 5c
Since both are just 1 equation, c is dependent of x and X, and thus y = –5x + c are invariant lines.
∴ The invariant lines are y = x, y = –5x + c, where c is an arbitrary constant.
Knowing how to transform points, we shall now learn how to transform lines. As in the part on invariant lines, we substitute the parametric equation of x and y, then we solve the equation in terms of X & Y, as the equation below
2x + 2 – 2x = X = 2
4x + 4 – 4x = Y = 4
∴ The line transform into the point (x, y) = (2, 4).
Notice that in this case, the line is transformed into a point. In other cases if it transforms into another line, remember to find an equation that relates X with Y. You should be aware that this is the very same method you will do if you were to find the transformation of circles, parabolas, hyperbolas, ellipses or other curves. Make use of their parametric equations and substitute them into the equation. Recall the parametric forms of these curves.
I think I don’t need to elaborate too much on this. An inverse transformation helps us to find the object if the image is given. You find the inverse of the matrix of transformation, and the equation will become
From here you should recall that a singular transformation has no inverse. In other words, you can’t find a matrix that transform a single point to 4 other points, or transform a line into a pentagon.
ADDITION, SUBTRACTION, SCALAR MULTIPLICATION, COMPOSITION
The addition and the subtraction of transformations M and N,
M(x) + N(x) = (M + N) (x)
M(x) – N(x) = (M – N) (x)
Although is defined so, has no geometrical meaning. For example, I add a matrix of rotation of 45 degrees with a matrix of reflection along the line y = x, gives you some awkward transformation, which doesn’t really have a relation to both. But the scalar multiplication of a matrix does mean something,
(cM) (x) = c(M) (x)
as it has the effect of scaling. Both these operations, I assume you already know how to do so, as this is covered in the chapter Matrices in Maths T. We are more interested in the composition of transformations. Given two transformation M and N, If the an object undergoes transformation M, then transformation N, it can be written as
Or we could also write it as (N ∘ M) (x) = x.
I think you probably remembered in form 4 that the transformation NM means “transform M first, then transform N”. This is quite straightforward, I think. In exams, you will be asked to find the matrix of the combined transformation of 2 or more transformations. If not, you will be given the points of the object and image, with half of the transformation, then ask you to find the other missing transformation, as well as describing it. Just make use of what you learnt about Matrices.
I think this chapter shouldn’t be a problem for you. It is those weird definition questions that you might probably don’t know how to solve. A scoring chapter, yet again.